title: "Algorithm basics and sample problems" description: "" added: "" top: true order: 4
const binarySearch = (nums, target) => {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
// parseInt(i + (j - i) / 2)
let mid = Math.floor((left + right) / 2);
if (nums[mid] === target) {
return mid;
} else if (nums[mid] > target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
// If the target is not found,
// return the index where it would be if it were inserted in order.
return left;
}
function bubbleSort(nums) {
for (let i = nums.length - 1; i > 0; i--) {
for (let j = 0; j < i; j++) {
if (nums[j] > nums[j + 1]) {
let tmp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = tmp;
}
}
}
}
function selectionSort(arr) {
for (let i = 0; i < arr.length; i++) {
let min = i;
for (let j = i + 1; j < arr.length; j++) {
if (arr[min] > arr[j]) {
min = j;
}
}
if (min !== i) {
let tmp = arr[i];
arr[i] = arr[min];
arr[min] = tmp;
}
}
return arr;
}
function insertionSort(arr) {
for (let i = 1; i < arr.length; i++) {
let key = arr[i];
let j = i - 1;
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j = j - 1;
}
arr[j + 1] = key;
}
return arr;
}
function quickSort(arr, lo, hi) {
if (lo >= hi) {
return;
}
const pivot = partition(arr, lo, hi);
quickSort(arr, lo, pivot - 1);
quickSort(arr, pivot + 1, hi);
}
function partition(nums, left, right) {
let pivot = nums[right];
let i = left;
for (let j = left; j < right; j++) {
if (nums[j] <= pivot) {
swap(nums, i, j);
i++;
}
}
swap(nums, i, right);
return i;
}
Quick sort is an in-place algorithm, but the stack due to recursive calls adds additional storage space proportional to the recursive depth.
It's not recommended to choose the first or last element to be the pivot, your pivot value is always the largest value on already sorted or nearly sorted arrays. So rather than splitting the array into two roughly equal subarrays, you split it into a single sub array that has only one fewer element than you started with. One way to choose the pivot to avoid this is to pick the pivot randomly (or choose the median of the first, middle and last element). This makes it unlikely to hit the worst case, and so on average will work well.
function merge(arr1, arr2) {
const merged = [];
let i = 0;
let j = 0;
while (i < arr1.length && j < arr2.length) {
if (arr1[i] <= arr2[j]) {
merged.push(arr1[i]);
i++;
} else {
merged.push(arr2[j]);
j++;
}
}
merged.push(...arr1.slice(i), ...arr2.slice(j));
return merged;
}
function mergeSort(arr) {
if (arr.length < 2) {
return arr.slice(0);
}
const split = Math.floor(arr.length / 2);
const left = arr.slice(0, split);
const right = arr.slice(split, arr.length);
return merge(mergeSort(left), mergeSort(right));
}
Count the number of reverse pairs in an array using the merge sort algorithm. A reverse pair is a pair of numbers (i, j) where i < j and nums[i] > nums[j].
function merge(arr1, arr2, count = 0) {
const merged = [];
let i = 0;
let j = 0;
while (i < arr1.length && j < arr2.length) {
if (arr1[i] <= arr2[j]) {
merged.push(arr1[i]);
i++;
} else {
merged.push(arr2[j]);
j++;
// Count reverse pairs
count += arr1.length - i;
}
}
merged.push(...arr1.slice(i), ...arr2.slice(j));
return merged;
}
class ListNode {
constructor(value) {
this.value = value;
this.next = null;
}
}
// Basic operations for a linked list:
// 1. Find the middle node of a linked list
// 2. Check if a linked list has a cycle
// 3. Reverse a linked list
function getMiddleNode(head) {
if (head === null) {
return null;
}
let slow = head;
let fast = head;
while (fast !== null && fast.next !== null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
function hasCycle(head) {
if (head === null || head.next === null) {
return false;
}
let slow = head;
let fast = head;
while (fast !== null && fast.next !== null) {
slow = slow.next;
fast = fast.next.next;
if (slow === fast) {
return true;
}
}
return false;
}
function reverseList(head) {
if (head === null || head.next === null) {
return head;
}
let current = head;
let prev = null;
while (current !== null) {
const next = current.next;
current.next = prev;
prev = current;
current = next;
}
return prev;
}
// Implement Queue using Stacks
class MyQueue {
constructor() {
this.s1 = [];
this.s2 = [];
}
push(x) {
while (this.s1.length > 0) {
this.s2.push(this.s1.pop());
}
this.s1.push(x);
while (this.s2.length > 0) {
this.s1.push(this.s2.pop());
}
}
pop() {
return this.s1.pop();
}
}
class UndoRedoManager {
constructor() {
this.undoStack = [];
this.redoStack = [];
}
doAction(action) {
action();
this.undoStack.push(action);
this.redoStack = [];
}
undo() {
if (this.undoStack.length > 0) {
const lastAction = this.undoStack.pop();
lastAction();
this.redoStack.push(lastAction);
}
}
redo() {
if (this.redoStack.length > 0) {
const lastUndoneAction = this.redoStack.pop();
lastUndoneAction();
this.undoStack.push(lastUndoneAction);
}
}
}
The Art of Computer Programming, Vol. 2, section 3.4.2 “Random sampling and shuffling” describes two solutions:
5! = 120 rows./*
Fisher-Yates Algorithm
To shuffle an array a of n elements (indices 0..n-1):
for i from n−1 downto 1 do
j ← random integer such that 0 ≤ j ≤ i
exchange a[j] and a[i]
*/
function shuffleArray(arr) {
for (let i = arr.length - 1; i > 0; i--) {
const j = Math.floor(Math.random() * (i + 1));
[arr[i], arr[j]] = [arr[j], arr[i]];
}
}
// Note that sorting is an `O(N log N)` operation where the Fisher-Yates algorithm is `O(N)`.
var preorderTraversal = function(root) {
if (root == null) return [];
let stack = [];
let result = [];
stack.push(root);
while (stack.length > 0) {
let node = stack.pop();
result.push(node.val);
if (node.right) {
stack.push(node.right);
}
if (node.left) {
stack.push(node.left);
}
}
return result;
};
var inorderTraversal = function(root) {
if (root == null) return [];
let result = [];
let stack = [];
let current = root;
while (stack.length > 0 || current !== null) {
if (current !== null) {
stack.push(current);
current = current.left;
} else {
current = stack.pop();
result.push(current.val);
current = current.right;
}
}
return result;
};
var postorderTraversal = function (root) {
if (root == null) return [];
let result = [];
let stack = [root];
let cur = null;
while (stack.length) {
cur = stack.pop();
result.push(cur.val);
cur.left && stack.push(cur.left);
cur.right && stack.push(cur.right);
}
return result.reverse();
};
var levelOrderTraversal = function(root) {
if (!root) return [];
let nodeQueue = [root];
let result = [];
while (nodeQueue.length) {
let size = nodeQueue.length;
let level = [];
for (let i = 0; i < size; i++) {
let curNode = nodeQueue.shift();
level.push(curNode.val);
if (curNode.left) {
nodeQueue.push(curNode.left);
}
if (curNode.right) {
nodeQueue.push(curNode.right);
}
}
// Reverse the order for zigzag pattern
// if (!isLeftToRight) {
// level.reverse();
// }
// isLeftToRight = !isLeftToRight;
result.push(level);
}
return result;
};
// Max and Min depth of Binary Tree
var maxDepth = function(root) {
if (!root) return 0;
const left = maxDepth(root.left);
const right = maxDepth(root.right);
return Math.max(left, right) + 1;
}
var minDepth = function(root) {
if (!root) return 0;
const nodeQueue = [root];
let depth = 0;
while (nodeQueue.length) {
const size = nodeQueue.length;
depth++;
for (let i = 0; i < size; i++) {
const curNode = nodeQueue.shift();
// If it's a leaf node, return the depth
if (!curNode.left && !curNode.right) {
return depth;
}
if (curNode.left) {
nodeQueue.push(curNode.left);
}
if (curNode.right) {
nodeQueue.push(curNode.right);
}
}
}
return depth;
}
function dfs(i, j, visited, params) {
// 1. Base Case: Check boundaries, visited state, or other termination conditions
if (i < 0 || i >= rows || j < 0 || j >= cols || visited[i][j] || !isValid(i, j, params)) {
return;
}
// 2. Mark the current node as visited
visited[i][j] = true;
// 3. Perform any necessary operations for the current node
processNode(i, j, params);
// 4. Recur for all possible directions
dfs(i - 1, j, visited, params);
dfs(i + 1, j, visited, params);
dfs(i, j - 1, visited, params);
dfs(i, j + 1, visited, params);
// 5. Backtrack: Unmark the current node after exploring all paths from that node
visited[i][j] = false;
}
The visited array is global or shared across all recursive calls. If you are searching for one valid path (e.g., finding a path in a maze), there's no need to restore the visited array. Once a node is marked as visited, it remains visited throughout the traversal. If you are searching for all possible paths (e.g., finding all routes in a graph), you must restore the visited array. After exploring a path, you backtrack by marking the node as unvisited to allow its reuse in other paths.
class DisjointSet {
constructor(size) {
this.parent = Array.from({ length: size }, (_, index) => index);
this.rank = Array(size).fill(1);
}
find(x) {
if (this.parent[x] !== x) {
this.parent[x] = this.find(this.parent[x]);
}
return this.parent[x];
}
union(x, y) {
const rootX = this.find(x);
const rootY = this.find(y);
if (rootX !== rootY) {
if (this.rank[rootX] > this.rank[rootY]) {
this.parent[rootY] = rootX;
} else if (this.rank[rootX] < this.rank[rootY]) {
this.parent[rootX] = rootY;
} else {
this.parent[rootY] = rootX;
this.rank[rootX] += 1;
}
}
}
connected(x, y) {
return this.find(x) === this.find(y);
}
}
function Heap () {
this.items = [];
}
Heap.prototype.swap = function (index1, index2) {
var temp = this.items[index1];
this.items[index1] = this.items[index2];
this.items[index2] = temp;
}
Heap.prototype.parentIndex = function (index) {
return Math.floor((index - 1) / 2);
}
Heap.prototype.leftChildIndex = function (index) {
return index * 2 + 1;
}
Heap.prototype.rightChildIndex = function (index) {
return index * 2 + 2;
}
Heap.prototype.parent = function (index) {
return this.items[this.parentIndex(index)];
}
Heap.prototype.leftChild = function (index) {
return this.items[this.leftChildIndex(index)];
}
Heap.prototype.rightChild = function (index) {
return this.items[this.rightChildIndex(index)];
}
Heap.prototype.peek = function (index) {
return this.items[0];
}
Heap.prototype.size = function () {
return this.items.length;
}
function MinHeap () {
this.items = [];
}
MinHeap.prototype = Object.create(Heap.prototype);
// add at the last position and then re-order the heap
MinHeap.prototype.add = function (item) {
this.items[this.items.length] = item;
this.bubbleUp();
}
// remove the minimum element and need to keep the heap order
MinHeap.prototype.poll = function () {
let item = this.items[0];
this.items[0] = this.items[this.items.length - 1];
this.items.pop();
this.bubbleDown();
return item;
}
MinHeap.prototype.bubbleUp = function () {
let index = this.items.length - 1;
while (this.parent(index) && this.parent(index) > this.items[index]) {
this.swap(this.parentIndex(index), index);
index = this.parentIndex(index);
}
}
MinHeap.prototype.bubbleDown = function () {
let index = 0;
while (this.leftChild(index)) {
let smallerIndex = this.leftChildIndex(index);
if (this.rightChild(index) && this.rightChild(index) < this.leftChild(index)) {
smallerIndex = this.rightChildIndex(index);
}
if (this.items[index] <= this.items[smallerIndex]) break;
this.swap(index, smallerIndex);
index = smallerIndex;
}
}
/*
You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps.
How many distinct ways can you climb to the top?
*/
let climbStairs = function(n) {
const dp = Array(n+1).fill(0);
dp[1] = 1;
dp[2] = 2;
for (let i = 3; i < dp.length; i++) {
dp[i] = dp[i-1] + dp[i-2];
}
return dp[n];
};
/*
Given a list of non-negative integers representing the amount of money of each house, and the adjacent houses have security system,
determine the maximum amount of money you can rob without alerting the police.
*/
let rob = function(nums) {
const dp = Array(nums.length).fill(0);
dp[0] = nums[0];
dp[1] = Math.max(nums[0], nums[1]);
for (let i = 2; i < nums.length; i++) {
dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
}
return dp[nums.length - 1];
};
// All houses at this place are arranged in a circle.
let rob2 = function(nums) {
const len = nums.length;
if (len === 1) return nums[0];
// the first house and the last house cannot be robbed together
return Math.max(robRange(nums.slice(0, -1)), robRange(nums.slice(1)));
function robRange(range) {
const n = range.length;
const dp = Array(n).fill(0);
dp[0] = range[0];
dp[1] = Math.max(range[0], range[1]);
for (let i = 2; i < n; i++) {
dp[i] = Math.max(dp[i - 2] + range[i], dp[i - 1]);
}
return dp[n - 1];
};
};
// Compute the fewest number of coins that you need to make up that amount. You have an infinite number of each kind of coin.
let coinChange = function(coins, amount) {
if (amount === 0) {
return 0;
}
const dp = Array(amount + 1).fill(Number.MAX_VALUE)
dp[0] = 0;
for (let i = 1; i < dp.length; i++) {
for (let j = 0; j < coins.length; j++) {
// e.g. dp[11] = min(dp[10], dp[9], dp[6]) + 1
if (i - coins[j] >= 0) {
dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
}
}
}
return dp[dp.length - 1] === Number.MAX_VALUE ? -1 : dp[dp.length - 1];
};
/*
LRU cache is implemented using a doubly linked list and hash table.
- Map is used to search the node in O(1);
- List maintains the order in O(1) with head and tail.
*/
function DLLNode(key, data) {
this.key = key;
this.data = data;
this.next = null;
this.prev = null;
}
function LRUCache(capacity) {
this.map = new Map();
this.capacity = capacity;
this.head = new DLLNode("", null);
this.tail = new DLLNode("", null);
this.head.next = this.tail;
this.tail.prev = this.head;
}
LRUCache.prototype.addNode = function(node) {
let realTail = this.tail.prev;
realTail.next = node;
this.tail.prev = node;
node.prev = realTail;
node.next = this.tail;
}
LRUCache.prototype.removeNode = function(node) {
let prev = node.prev;
let next = node.next;
prev.next = next;
next.prev = prev;
}
LRUCache.prototype.get = function(key) {
let node = this.map.get(key);
if (node === undefined) {
return -1;
} else {
this.removeNode(node);
this.addNode(node);
return node.data;
}
}
LRUCache.prototype.put = function(key, value) {
let node = this.map.get(key);
if (node) {
this.removeNode(node);
}
let newNode = new DLLNode(key, value);
this.addNode(newNode);
this.map.set(key, newNode);
if (this.map.size > this.capacity) {
var realHead = this.head.next;
this.removeNode(realHead);
this.map.delete(realHead.key)
}
}
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.
var threeSum = function(nums) {
let res = [];
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length; i++) {
if (i > 0 && nums[i] === nums[i-1]) {
continue;
}
let j = i + 1;
let k = nums.length - 1;
while (j < k) {
let total = nums[i] + nums[j] + nums[k];
if (total === 0) {
res.push([nums[i], nums[j], nums[k]]);
// remove the duplicate result
while (nums[j] === nums[j+1] && j < k) {
j++;
}
while (nums[k] === nums[k-1] && j < k) {
k--;
}
j++;
k--;
} else if (total > 0) {
k--;
} else {
j++;
}
}
}
return res;
};
Given a non-empty array of integers, return the k most frequent elements.
var topKFrequent = function(nums, k) {
const res = [];
const map = new Map();
const bucket = Array(nums.length + 1);
// key is nums[i], value is its count
for (let n of nums) {
if (map.has(n)) {
map.set(n, map.get(n) + 1);
} else {
map.set(n, 1);
}
}
// the same frequency in the same bucket
for (let [key, val] of map.entries()) {
if (!bucket[val]) {
bucket[val] = [];
}
bucket[val].push(key);
}
// collect buckets from the end
for (let pos = bucket.length - 1; pos >= 0 && res.length < k; pos--) {
if (bucket[pos]) {
res.push(...bucket[pos]);
}
}
return res;
};
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth largest distinct element. You may assume k is always valid.
var findKthLargest = function(nums, k) {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
let pos = partition(nums, left, right);
// if pos is 3, it is the 4th largest element
if (pos === k - 1) return nums[pos];
else if (pos > k - 1) right = pos - 1;
else left = pos + 1;
}
// elements before i >= pivot; every elements after i < pivot
function partition(nums, low, high) {
let pivot = nums[high];
let i = low;
for (let j = low; j < high; j++) {
if (nums[j] >= pivot) {
swap(nums, i, j);
i++;
}
}
swap(nums, i, high);
return i;
}
function swap(nums, i, j) {
let tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
};
Given an array of strings, group anagrams together. i.e. Input: ["eat", "tea", "tan", "ate", "nat", "bat"], Output: [ ["ate","eat","tea"], ["nat","tan"], ["bat"] ]
var groupAnagrams = function(strs) {
const hashTable = new Map();
function sort(str) {
return str.split("").sort().join("");
}
for (let i = 0; i < strs.length; i++) {
const str = strs[i];
const key = sort(str);
/**
* Another way to construct the key:
*
* const counts = Array(26).fill(0);
* for (let j = 0; j < str.length; j++) {
* counts[str[j].charCodeAt(0) - 'a'.charCodeAt(0)]++;
* }
* const key = counts.join("");
**/
if (!hashTable.has(key)) {
hashTable.set(key, [str]);
} else {
hashTable.get(key).push(str);
}
}
return [...hashTable.values()];
};
Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false otherwise.
var canConstruct = function(ransomNote, magazine) {
let cnt = Array(26).fill(0);
for (let c of magazine) {
cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
}
for (let c of ransomNote) {
cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)]--;
if (cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)] < 0) {
return false;
}
}
return true;
};
Given a string, find the length of the longest substring without repeating characters.
// slide window
var lengthOfLongestSubstring = function(s) {
let max = 0;
let windowSet = new Set();
let i = 0;
for (let j = 0; j < s.length; j++) {
while (windowSet.has(s[j])) {
windowSet.delete(s[i]);
i++;
}
windowSet.add(s[j]);
max = Math.max(max, j - i + 1);
}
return max;
};
Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose sum is greater than or equal to target.
var minSubArrayLen = function(target, nums) {
let minLength = Infinity;
let sum = 0;
let left = 0;
for (let right = 0; right < nums.length; right++) {
sum += nums[right];
while (sum >= target) {
minLength = Math.min(minLength, right - left + 1);
sum -= nums[left];
left++;
}
}
return minLength === Infinity ? 0 : minLength;
};
Find All Anagrams in a String. Given two strings s and p, return an array of all the start indices of p's anagrams in s. For example, Input: s = "cbaebabacd", p = "abc"; Output: [0,6].
var findAnagrams = function(s, p) {
let res = [];
let dict = Array(26).fill(0);
for (let i = 0; i < p.length; i++) {
dict[p.charCodeAt(i) - 'a'.charCodeAt(0)]++;
}
let windowSet = Array(26).fill(0);
let i = 0;
for (let j = 0; j < s.length; j++) {
windowSet[s.charCodeAt(j) - 'a'.charCodeAt(0)]++;
if (isSame(windowSet, dict)) {
res.push(i);
}
if (j >= p.length - 1) {
windowSet[s.charCodeAt(i) - 'a'.charCodeAt(0)]--;
i++;
}
}
return res;
}
function isSame(dict1, dict2) {
for (let i = 0; i < 26; i++) {
if (dict1[i] !== dict2[i]) {
return false;
}
}
return true;
}
Given two strings s1 and s2, return true if s2 contains a permutation of s1 (one of s1's permutations is the substring of s2).
var checkInclusion = function(s1, s2) {
let len1 = s1.length;
let len2 = s2.length;
if (len1 > len2) return false;
let count = Array(26).fill(0);
for (let i = 0; i < len1; i++) {
// 97 -> 'a'.charCodeAt(0)
count[s1.charCodeAt(i) - 97]++;
count[s2.charCodeAt(i) - 97]--;
}
if (count.every(i => i === 0)) {
return true;
}
for (let i = len1; i < len2; i++) {
count[s2.charCodeAt(i) - 97]--;
// the character leaving the window, always maintaining a window of size `len1`
count[s2.charCodeAt(i - len1) - 97]++;
if (count.every(e => e === 0)) {
return true;
}
}
return false;
};
You are given a string s and an integer k. You can choose any character of the string and change it to any other character. You can perform this operation at most k times. Return the length of the longest substring containing the same letter you can get after performing the above operations.
var characterReplacement = function(s, k) {
let left = 0;
let maxCharCount = 0;
let visited = {};
for (let right = 0; right < s.length; right++) {
let char = s[right];
visited[char] = (visited[char] || 0) + 1;
maxCharCount = Math.max(maxCharCount, visited[char]);
// right - left + 1: current window size
// # of char in the window that are different from the most frequent character
if (right - left + 1 - maxCharCount > k) {
visited[s[left]]--;
left++;
// moving left does not make our current window shorter
}
}
// the size of the final valid window
return s.length - left;
};
You are given a string s. We want to partition the string into as many parts as possible so that each letter appears in at most one part. Return a list of the size of these parts.
var partitionLabels = function(s) {
let last = Array(26).fill(-1);
let partitions = [];
let left = 0;
let end = 0;
// lastIndexOf
for (let i = 0; i < s.length; i++) {
last[s.charCodeAt(i) - 'a'.charCodeAt(0)] = i;
}
for (let i = 0; i < s.length; i++) {
end = Math.max(end, last[s.charCodeAt(i) - 'a'.charCodeAt(0)]);
if (i === end) {
partitions.push(i - left + 1);
left = i + 1;
}
}
return partitions;
};
Given a string s, return the longest palindromic substring in s.
var longestPalindrome = function(s) {
let left = 0, right = 0, maxLength = 0;
const extend = (s, i, j, n) => {
while (i >= 0 && j < n && s[i] === s[j]) {
if (j - i + 1 > maxLength) {
left = i;
right = j;
maxLength = j - i + 1;
}
i--;
j++;
}
}
for (let i = 0; i < s.length; i++) {
extend(s, i, i, s.length); // i is the center
extend(s, i, i + 1, s.length); // i and i + 1 are the center
}
return s.slice(left, right + 1);
};
Given an array nums of size n, return the majority element. The majority element is the element that appears more than ⌊n / 2⌋ times. (assume it always exists)
var majorityElement = function(nums) {
let count = 0;
let candidate = 0;
for (let num of nums) {
if (count === 0) {
candidate = num;
}
if (num === candidate) {
count += 1;
} else {
count -= 1;
}
}
return candidate;
};
Given a string containing just the characters '(' and ')', return the length of the longest valid parentheses substring.
var longestValidParentheses = function(s) {
let stack = [];
let maxLength = 0;
let start = 0;
for (let i = 0; i < s.length; i++) {
if (s[i] === '(') {
stack.push(i);
} else {
if (stack.length === 0) {
start = i + 1;
} else {
stack.pop();
if (stack.length === 0) {
maxLength = Math.max(maxLength, i - start + 1);
} else {
maxLength = Math.max(maxLength, i - stack[stack.length - 1]);
}
}
}
}
};
Given a string s containing only three types of characters: '(', ')' and '*', return true if s is valid. '*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string "".
var checkValidString = function(s) {
// the min and max # of open parentheses that must be matched
let leftMin = 0, leftMax = 0;
for (let c of s) {
if (c === '(') {
leftMin++;
leftMax++;
} else if (c === ')') {
leftMin--;
leftMax--;
} else {
leftMin--; // treating as ')'
leftMax++; // treating as '('
}
if (leftMax < 0) return false;
// we must have seen '*' earlier in the string
// use '*' to balance the excess of closing parentheses
if (leftMin < 0) leftMin = 0;
}
return leftMin === 0;
};
Write a function to find the longest common prefix string amongst an array of strings. If there is no common prefix, return an empty string "".
var longestCommonPrefix = function(strs) {
strs.sort();
for (let i = 0; i < strs[0].length; i++) {
if (strs[0][i] !== strs[strs.length - 1][i]){
return strs[0].slice(0, i);
}
}
return strs[0];
};
Given an unsorted integer array nums. Return the smallest positive integer that is not present in nums.
var firstMissingPositive = function(nums) {
const numSet = new Set(nums.filter(num => num > 0));
let i = 1;
while (numSet.has(i)) {
i++;
}
return i;
};
Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. Take [10, 4, 20, 1, 3, 2] as an example, the longest consecutive elements sequence is [1, 2, 3, 4].
var longestConsecutive = function(nums) {
let numSet = new Set(nums);
let longest = 0;
for (let n of nums) {
// check if n starts a new sequence
if (!numSet.has(n - 1)) {
let length = 1;
while (numSet.has(n + length)) {
length++;
}
longest = Math.max(longest, length);
}
}
return longest;
};
Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].
var productExceptSelf = function(nums) {
let n = nums.length;
let output = [];
let leftProducts = [];
let rightProducts = [];
leftProducts[0] = 1;
rightProducts[n - 1] = 1;
for (let i = 1; i < n; i++) {
leftProducts[i] = leftProducts[i - 1] * nums[i - 1];
}
for (let i = n - 2; i >= 0; i--) {
rightProducts[i] = rightProducts[i + 1] * nums[i + 1];
}
for (let i = 0; i < n; i++) {
output[i] = leftProducts[i] * rightProducts[i];
}
return output;
};
Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead.
var dailyTemperatures = function(temperatures) {
let stack = [];
let res = new Array(temperatures.length).fill(0);
for (let i = 0; i < temperatures.length; i++) {
while (stack.length > 0 && temperatures[i] > temperatures[stack[stack.length - 1]]) {
let preIndex = stack.pop();
res[preIndex] = i - preIndex;
}
// push the index, not the value
stack.push(i);
}
return res;
};
The next greater element of x in an array is the first greater element that is to the right of x in the same array. You are given two distinct arrays, where nums1 is a subset of nums2. For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.
var nextGreaterElement = function(nums1, nums2) {
// key is a number from nums2
// value is its next greater number
let map = new Map();
let stack = [];
for (let num of nums2) {
while (stack.length && num > stack[stack.length - 1]) {
map.set(stack.pop(), num);
}
stack.push(num);
}
for (let i = 0; i < nums1.length; i++) {
nums1[i] = map.has(nums1[i]) ? map.get(nums1[i]) : -1;
}
return nums1;
};
Given string num representing a non-negative integer num, and an integer k, return the smallest possible integer after removing k digits from num.
var removeKdigits = function(num, k) {
let stack = [];
for (let c of num) {
while (stack.length && k > 0 && c < stack[stack.length-1]) {
stack.pop();
k--;
}
stack.push(c);
}
while (stack.length && k > 0) {
stack.pop();
k--;
}
// remove all the leading zeros
while (stack.length && stack[0] === '0') {
stack.shift();
}
return stack.join('') || "0";
};
Given an array of integers nums, calculate the pivot index of this array. The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index's right.
var pivotIndex = function(nums) {
let total = nums.reduce((a, b) => a + b, 0);
let leftTotal = 0;
for (let i = 0; i < nums.length; i++) {
let rightTotal = total - leftTotal - nums[i];
if (rightTotal === leftTotal) {
return i;
}
leftTotal += nums[i];
}
return -1;
};
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. For example, Input: nums = [1,1,1], k = 2; Output: 2
// the sum of subarray can be get by `sum[0,j] - sum[0,i]`
var subarraySum = function(nums, k) {
let sum = 0, res = 0;
let map = new Map(); // key is the sum, value is the # of way to get that sum
map.set(0, 1);
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
if (map.has(sum - k)) {
res += map.get(sum - k);
}
if (map.get(sum)) {
map.set(sum, map.get(sum) + 1);
} else {
map.set(sum, 1);
}
}
return res;
};
Given an array of integers nums and an integer k. A continuous subarray is called nice if there are k odd numbers on it. Return the number of nice sub-arrays.
var numberOfSubarrays = function(nums, k) {
let oddCount = 0;
let result = 0;
let count = new Map();
// key is the count of odd numbers encountered so far
// value is the # of times we've encountered that particular count of odd numbers.
count.set(0, 1);
for (const num of nums) {
if (num % 2 === 1) {
oddCount += 1;
}
if (oddCount >= k && count.has(oddCount - k)) {
result += count.get(oddCount - k);
}
count.set(oddCount, (count.get(oddCount) || 0) + 1);
}
return result;
};
Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr.
var sumOddLengthSubarrays = function(arr) {
let ans = 0;
// preSum[i] stores the sum of elements from index 0 to i-1.
// if arr = [1, 2, 3], then preSum will be [0, 1, 3, 6].
let preSum = Array(arr.length + 1).fill(0);
for (let i = 0; i < arr.length; i++) {
preSum[i + 1] = preSum[i] + arr[i];
}
for (let i = 0; i < arr.length; i++) {
for (let j = 1; i + j <= arr.length; j += 2) {
ans += preSum[i + j] - preSum[i];
}
}
return ans;
};
Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.
var containsNearbyDuplicate = function(nums, k) {
const map = new Map(); // key is nums[i], value is its index
for (let i = 0; i < nums.length; i++) {
const num = nums[i];
if (map.has(num) && i - map.get(num) <= k) {
return true;
}
map.set(num, i);
}
return false;
};
Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. You must do this in-place without making a copy of the array.
var moveZeroes = function(nums) {
let index = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] !== 0) {
nums[index] = nums[i];
index++;
}
}
for(let i = index; i < nums.length; i++) {
nums[i] = 0;
}
};
Given an integer array nums, find the subarray with the largest sum, and return its sum.
var maxSubArray = function(nums) {
let sum = 0;
let max = -Infinity;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
max = Math.max(max, sum);
// It's better to start a fresh subarray from the next element
// rather than carrying forward a negative sum.
if (sum < 0) {
sum = 0;
}
}
return max;
};
Given a non-empty array of integers nums, the degree of this array is defined as the maximum frequency of any one of its elements. Your task is to find the smallest possible length of a subarray of nums, that has the same degree as nums.
var findShortestSubArray = function(nums) {
let firstOccurrence = new Map();
let lastOccurrence = new Map();
let frequencyMap = new Map();
nums.forEach((num, index) => {
if (firstOccurrence.get(num) === undefined) {
firstOccurrence.set(num, index);
}
lastOccurrence.set(num, index);
frequencyMap.set(num, (frequencyMap.get(num) || 0) + 1);
});
let ans = nums.length;
let degree = Math.max(...frequencyMap.values());
for (let [num, frequency] of frequencyMap) {
if (frequency === degree) {
ans = Math.min(ans, lastOccurrence.get(num) - firstOccurrence.get(num) + 1);
}
}
return ans;
};
Given an integer array containing 0's and 1's, where 0 means empty and 1 means not empty, and an integer n, return true if n new flowers can be planted in the flowerbed. Flowers cannot be planted in adjacent plots.
var canPlaceFlowers = function(flowerbed, n) {
for (let i = 0; i < flowerbed.length && n > 0;) {
if (flowerbed[i] === 1) {
i += 2;
}
// current position is empty
else if (i === flowerbed.length - 1 || flowerbed[i + 1] === 0) {
n--;
i += 2;
}
// current position is empty but the next position has a flower
else {
i += 3;
}
}
return n <= 0;
};
You are given an integer array nums. Each element in the array represents your maximum jump length at that position. Return true if you can reach the last index, or false otherwise.
var canJump = function(nums) {
// Greedy algorithm
let reachable = nums[0];
for (let i = 1; i < nums.length; i++) {
if (i > reachable) {
return false;
}
if (i + nums[i] > reachable) {
reachable = i + nums[i];
}
}
return true;
};
There are n gas stations along a circular route. Your car costs cost[i] of gas to travel from the ith station to its next station. You begin the journey with an empty tank at one of the gas stations. Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1.
var canCompleteCircuit = function(gas, cost) {
// gas: [1, 2, 3, 4, 5]
// cost: [3, 4, 5, 1, 2]
// First, can we know if a solution exists?
// If `sum of gas - sum of cost < 0`, there is no way to complete a round trip.
let totalTank = 0;
let currentTank = 0;
let start = 0;
for (let i = 0; i < gas.length; i++) {
const netCost = gas[i] - cost[i];
totalTank += netCost;
currentTank += netCost;
if (currentTank < 0) {
start = i + 1;
currentTank = 0;
}
}
return totalTank < 0 ? -1 : start;
// The key insight here is that if you can't reach a station j from station i,
// you also can't reach it from any station between i and j.
};
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]). You are given a target value to search. If found in the array return its index, otherwise return -1. You may assume no duplicate exists in the array.
var search = function(nums, target) {
let start = 0;
let end = nums.length - 1;
while (start <= end) {
const mid = Math.floor((start + end) / 2);
if (nums[mid] === target) return mid;
// [mid, end] is ascending order
if (nums[mid] < nums[end]) {
if (target > nums[mid] && target <= nums[end]) {
start = mid + 1;
} else {
end = mid - 1;
}
}
// [start, mid] is ascending order
else {
if (target >= nums[start] && target < nums[mid]) {
end = mid - 1;
} else {
start = mid + 1;
}
}
}
return -1;
};
You are given an m x n integer matrix. Each row is sorted in non-decreasing order. The first integer of each row is greater than the last integer of the previous row. Given an integer target, return true if target is in matrix or false otherwise.
var searchMatrix = function(matrix, target) {
let left = 0;
let right = matrix.length - 1;
let mid;
while (left <= right) {
mid = Math.floor((left + right ) / 2);
if (target >= matrix[mid][0] && target <= matrix[mid][matrix[mid].length - 1]) {
break;
}
if (target > matrix[mid][0]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return binarySearch(matrix[mid], target);
};
Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value. If target is not found in the array, return [-1, -1].
var searchRange = function(nums, target) {
const binarySearch = (nums, target, isSearchingLeft) => {
let left = 0;
let right = nums.length - 1;
// index of either the leftmost or rightmost occurrence of the target
let idx = -1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] > target) {
right = mid - 1;
} else if (nums[mid] < target) {
left = mid + 1;
} else {
idx = mid;
if (isSearchingLeft) {
right = mid - 1;
} else {
left = mid + 1;
}
}
}
return idx;
};
let left = binarySearch(nums, target, true);
let right = binarySearch(nums, target, false);
return [left, right];
};
a) Find the integer square root; b) Check if a number is the perfect square number.
function getSquareRoot(num) {
let i = 0;
let j = num;
while (i <= j) {
let mid = Math.floor((i + j) / 2);
if (mid * mid === num) {
return mid;
} else if (mid * mid < num) {
i = mid + 1;
} else {
j = mid - 1;
}
}
return -1;
}
function checkPerfectSquare(num) {
let i = 0;
let j = num;
while (i <= j) {
let mid = Math.floor((i + j) / 2);
if (mid * mid === num) {
return true;
}
if (mid * mid < num) {
i = mid + 1;
} else {
j = mid - 1;
}
}
return false;
}
Given a collection of intervals, merge all overlapping intervals. i.e. Input: [[1,3],[2,6],[8,10],[15,18]]; Output: [[1,6],[8,10],[15,18]].
var merge = function(intervals) {
intervals.sort((a, b) => a[0] - b[0]);
for (let i = 0; i < intervals.length - 1; i++) {
const cur = intervals[i];
const next = intervals[i + 1];
if (cur[1] >= next[0]) {
intervals[i] = undefined;
intervals[i + 1] = [Math.min(cur[0], next[0]), Math.max(cur[1], next[1])];
}
}
return intervals.filter(q => q); // fiter to pass undefined value
};
Given an array of meeting time intervals consisting of start and end times, find the minimum number of conference rooms required. For example, Given [[0, 30],[5, 10],[15, 20]], return 2.
var minMeetingRooms = function(intervals) {
if (intervals.length === 0) return 0;
intervals.sort((a, b) => a[0] - b[0]);
let rooms = [intervals[0][1]];
for (let i = 1; i < intervals.length; i++) {
// check if it can reuse the existing room
if (intervals[i][0] >= rooms[0]) {
rooms[0] = intervals[i][1];
} else {
rooms.push(intervals[i][1]);
}
// mock PriorityQueue to find the earliest ending time
rooms.sort((a, b) => a - b);
}
return rooms.length;
}
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.
var trap = function(height) {
const stack = [];
let res = 0;
for (let i = 0; i < height.length; i++) {
const curH = height[i];
while (stack.length && curH > height[stack[stack.length - 1]]) {
const preIndex = stack.pop();
if (stack.length) {
// i -> right boundary
// stack[stack.length - 1] -> left boundary
const heightVal = Math.min(height[stack[stack.length - 1]], curH) - height[preIndex];
const length = i - stack[stack.length - 1] - 1;
res += heightVal * length;
}
}
stack.push(i);
}
return res;
};
You are given an integer array height of length n. There are n vertical lines. Find two lines that together with the x-axis form a container, such that the container contains the most water.
var maxArea = function(height) {
let i = 0, j = height.length - 1;
let res = 0;
while (i < j) {
const cap = Math.min(height[i], height[j]) * (j - i);
res = Math.max(res, cap);
if (height[i] < height[j]) {
i += 1;
} else {
j -= 1;
}
}
return res;
};
The next permutation of an array of integers is the next lexicographically greater permutation of its integer. If such arrangement is not possible, the array must be rearranged as the lowest possible order.
var nextPermutation = function(nums) {
// Find the first pair from the right where nums[i] < nums[i+1]
let i;
for (i = nums.length - 2; i >= 0; i--) {
if (nums[i] < nums[i + 1]) {
break;
}
}
if (i >= 0) {
// Find the first number on the right side of nums[i] that is greater than nums[i]
let j;
for (j = nums.length - 1; j > i; j--) {
if (nums[j] > nums[i]) {
break;
}
}
[nums[i], nums[j]] = [nums[j], nums[i]];
}
// The subarray to the right of 'i' is guaranteed to be in descending order
reverse(nums, i + 1, nums.length - 1);
}
function reverse(nums, start, end) {
for (let left = start, right = end; left < right; left++, right--) {
[nums[left], nums[right]] = [nums[right], nums[left]];
}
}
Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
var permute = function(nums) {
const result = [];
function backtrack(path) {
if (path.length === nums.length) {
result.push([...path]);
return;
}
for (let i = 0; i < nums.length; i++) {
const num = nums[i];
if (path.includes(num)) continue;
path.push(num);
backtrack(path); // console.log(`backtrack(${path});`)
path.pop();
}
}
backtrack([]);
return result;
};
// Time complexity: O(n!)
// Space complexity: O(n)
Given an integer array nums of unique elements, return all possible subsets.
var subsets = function(nums) {
const result = [];
const subset = [];
function backtrack(start) {
result.push([...subset]);
for (let i = start; i < nums.length; i++) {
subset.push(nums[i]);
backtrack(i + 1);
subset.pop();
}
}
backtrack(0);
return result;
};
Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. A mapping of digits to letters (just like on the telephone buttons) is 2->abc ... 9->wxyz.
var letterCombinations = function(digits) {
if (digits.length === 0) {
return [];
}
const phone_map = ["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"];
const output = [];
backtrack("", digits, phone_map, output);
return output;
function backtrack(combination, next_digits, phone_map, output) {
if (next_digits.length === 0) {
output.push(combination);
return;
}
const letters = phone_map[next_digits[0] - '2'];
for (const letter of letters) {
backtrack(combination + letter, next_digits.slice(1), phone_map, output);
}
}
};
Given a binary search tree (BST), find the lowest common ancestor node of two given nodes in the BST.
var lowestCommonAncestor = function(root, p, q) {
// If root is the LCA node, p and q should be at different side
// (root.val - p.val) * (root.val - q.val) > 0, means p, q at same side
// (root.val - p.val) * (root.val - q.val) < 0, means p, q at different side
while ((root.val - p.val) * (root.val - q.val) > 0) {
if (p.val < root.val) {
root = root.left;
} else {
root = root.right;
}
}
return root;
};
// LCA of a binary tree (not BST)
var lowestCommonAncestor = function(root, p, q) {
if (!root || root === p || root === q) return root;
let left = lowestCommonAncestor(root.left, p, q);
let right = lowestCommonAncestor(root.right, p, q);
// p and q are in the right subtree
if (!left) return right;
// p and q are in the left subtree
if (!right) return left;
// p and q at different side
return root;
};
Given the root of a binary tree, invert the tree, and return its root.
var invertTree = function(root) {
if (root === null) return root;
[root.left, root.right] = [root.right, root.left];
invertTree(root.left);
invertTree(root.right);
return root;
};
Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.
var hasPathSum = function(root, targetSum) {
if (root === null) {
return false;
}
if (root.val === targetSum && (root.left === null && root.right === null)) {
return true;
}
return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);
};
Given the root of a binary tree, return the maximum path sum of any non-empty path. A path can start and end at any nodes in the tree.
var maxPathSum = function(root) {
let max = -Infinity;
function findSums(node) {
if (!node) return 0;
let left = findSums(node.left),
right = findSums(node.right),
allSum = left + right + node.val,
leftNodeSum = left + node.val,
rightNodeSum = right + node.val;
// Tracks all possible paths, include both left and right children
max = Math.max(max, node.val, allSum, leftNodeSum, rightNodeSum);
// Return value can only use ONE child. This is what we "offer up" to parent nodes
return Math.max(leftNodeSum, rightNodeSum, node.val);
};
findSums(root);
return max;
};
Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.
var isSubtree = function(root, subRoot) {
if (subRoot === null) return true;
if (root === null) return false;
if (isSameTree(root, subRoot)) {
return true;
}
// if not same, continue to check subtrees
return (isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot));
};
function isSameTree(root, subRoot) {
// can't access `root.val` when its null
if (root === null && subRoot === null) {
return true;
}
if ((root && subRoot === null) || (root === null && subRoot)) {
return false;
}
if (root.val !== subRoot.val) {
return false;
}
return (isSameTree(root.left, subRoot.left) && isSameTree(root.right, subRoot.right));
}
Given the root of a binary tree, determine if it is a valid binary search tree.
var isValidBST = function(root) {
// Check range of each node
const helper = (root, min, max) => {
if (root === null) return true;
if (root.val < min || root.val > max) return false;
return helper(root.left, min, root.val) && helper(root.right, root.val, max);
}
return helper(root, -Infinity, Infinity);
};
Construct binary tree from pre-order and in-order traversal.
var buildTree = function(preorder, inorder) {
if (preorder.length === 0) {
return null;
}
const root = new TreeNode(preorder[0]);
const mid = inorder.indexOf(preorder[0]);
// the # of elements in the left subtree is the same in both traversals
root.left = buildTree(preorder.slice(1, mid + 1), inorder.slice(0, mid));
root.right = buildTree(preorder.slice(mid + 1), inorder.slice(mid + 1));
return root;
};
Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.
var kthSmallest = function(root, k) {
// BST -> in-order traversal
let stack = [];
let current = root;
let count = 0;
while (stack.length > 0 || current !== null) {
if (current !== null) {
stack.push(current);
current = current.left;
} else {
current = stack.pop();
count++;
if (count === k) {
return current.val;
}
current = current.right;
}
}
return null; // If k is greater than the number of nodes in the BST
};
Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
var rightSideView = function(root) {
if (!root) return [];
let result = [];
let queue = [root];
while (queue.length) {
let levelSize = queue.length;
for (let i = 0; i < levelSize; i++) {
let node = queue.shift();
if (i === levelSize - 1) {
result.push(node.val);
}
if (node.left) {
queue.push(node.left);
}
if (node.right) {
queue.push(node.right);
}
}
}
return result;
};
Given the head of a linked list, remove the nth node from the end of the list and return its head.
// fast pointer moves forward `n` steps first,
// then fast and slow pointers move forward together,
// until the fast pointer reaches the end of the linked list.
var removeNthFromEnd = function(head, n) {
let dummy = new ListNode(0, head);
let fast = dummy;
let slow = dummy;
for (let i = 0; i < n; i++) {
fast = fast.next;
}
while (fast.next) {
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next;
return dummy.next;
};
Given a linked list, swap every two adjacent nodes and return its head.
var swapPairs = function(head) {
if (head == null || head.next == null) {
return head;
}
const nextHead = head.next;
const skipHead = head.next.next;
nextHead.next = head;
head.next = swapPairs(skipHead);
return nextHead;
};
You are given the heads of two sorted linked lists. Merge the two lists into one sorted list.
var mergeTwoLists = function(list1, list2) {
let dummy = new ListNode();
let cur = dummy;
while (list1 && list2) {
if (list1.val > list2.val) {
cur.next = list2;
list2 = list2.next;
} else {
cur.next = list1;
list1 = list1.next;
}
cur = cur.next;
}
cur.next = list1 || list2;
return dummy.next;
};
Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges are all surrounded by water.
var numIslands = function(grid) {
let count = 0;
let rows = grid.length;
if (rows === 0) return 0;
let cols = grid[0].length;
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
if (grid[i][j] === '1') {
dfs(grid, i, j, rows, cols);
count++;
}
}
}
return count;
};
function dfs(grid, i, j, rows, cols) {
if (i < 0 || j < 0 || i > rows - 1 || j > cols - 1 || grid[i][j] === '0') {
return;
}
grid[i][j] = '0';
dfs(grid, i + 1, j, rows, cols);
dfs(grid, i, j + 1, rows, cols);
dfs(grid, i - 1, j, rows, cols);
dfs(grid, i, j - 1, rows, cols);
}
You are given an m x n binary matrix grid. An island is a group of 1's connected 4-directionally. The area of an island is the number of cells with a value 1. Return the maximum area of an island in grid.
var maxAreaOfIsland = function(grid) {
let n = grid.length;
let m = grid[0].length;
let ans = 0;
function dfs(i, j) {
if (i < 0 || j < 0 || i > n-1 || j > m-1 || !grid[i][j]) {
return 0;
}
grid[i][j] = 0;
return 1 + dfs(i-1, j) + dfs(i, j-1) + dfs(i+1, j) + dfs(i, j+1);
}
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
if (grid[i][j]) {
ans = Math.max(ans, dfs(i, j));
}
}
}
return ans;
};
Given an m x n grid of characters board and a string word, return true if word exists in the grid. The word can be constructed from letters of sequentially adjacent cells (horizontally or vertically).
var exist = function(board, word) {
let m = board.length;
let n = board[0].length;
let visited = Array.from(Array(m), () => Array(n).fill(false));
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
if (dfs(board, word, i, j, 0, visited))
return true;
}
}
return false;
};
// k is the current position in the word
function dfs(board, word, i, j, k, visited) {
let m = board.length;
let n = board[0].length;
if (k === word.length) {
return true;
}
if (i < 0 || i > m-1 || j < 0 || j > n-1 || visited[i][j] || board[i][j] !== word.charAt(k)) {
return false;
}
visited[i][j] = true;
if (dfs(board, word, i + 1, j, k + 1, visited) ||
dfs(board, word, i - 1, j, k + 1, visited) ||
dfs(board, word, i, j + 1, k + 1, visited) ||
dfs(board, word, i, j - 1, k + 1, visited))
return true;
visited[i][j] = false;
return false;
}
Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists. Every adjacent pair of words differs by a single letter.
var ladderLength = function(beginWord, endWord, wordList) {
let wordSet = new Set(wordList)
let queue = [beginWord];
let steps = 1;
// bfs
while (queue.length) {
// words in the same "level"
let n = queue.length;
for (let i = 0; i < n; i++) {
let word = queue.shift();
if (word === endWord) {
return steps;
}
for (let j = 0; j < word.length; j++) {
// replace the char with letters from [a - z]
for (let k = 0; k < 26; k++) {
let newWord = word.slice(0, j) + String.fromCharCode(k + 97) + word.slice(j + 1);
if (wordSet.has(newWord)) {
queue.push(newWord);
wordSet.delete(newWord);
}
}
}
}
steps++;
}
return 0;
};
There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai. Return true if you can finish all courses.
// Topological Sort
var canFinish = function(numCourses, prerequisites) {
if (numCourses <= 0) {
return true;
}
let inDegree = new Array(numCourses).fill(0);
let graph = Array.from({ length: numCourses }, () => []);
prerequisites.forEach(edge => {
let parent = edge[1];
let child = edge[0];
graph[parent].push(child);
inDegree[child]++;
});
// Initialize the queue with courses having no prerequisites (inDegree = 0)
let queue = [];
for (let i = 0; i < numCourses; i++) {
if (inDegree[i] === 0) {
queue.push(i);
}
}
// Process nodes with no prerequisites
let counter = 0;
while (queue.length > 0) {
let course = queue.shift();
counter++;
graph[course].forEach(child => {
inDegree[child]--;
if (inDegree[child] === 0) {
queue.push(child);
}
});
}
return counter === numCourses;
};
The message is decoded via the following mapping: "1" -> 'A' ... "26" -> 'Z'. There are many different ways you can decode the message because some codes are contained in other codes ("2" and "5" vs "25"). Given a string s containing only digits, return the number of ways to decode it.
var numDecodings = function (s) {
let dp = Array(s.length + 1).fill(0);
// there is only one way to decode an empty string
dp[0] = 1;
if (s[0] !== '0') {
dp[1] = 1;
} else {
return 0;
}
for (let i = 2; i <= s.length; i++) {
if (s[i - 1] !== '0') {
dp[i] += dp[i - 1];
}
if (s[i - 2] === '1' || (s[i - 2] === '2' && s[i - 1] <= '6')) {
dp[i] += dp[i - 2];
}
}
return dp[s.length];
};
Given an integer array nums, return the length of the longest strictly increasing subsequence. For example, [10,9,2,5,3,7,101,18]'s longest increasing subsequence is [2,3,7,101].
var lengthOfLIS = function(nums) {
if (nums.length === 0) {
return 0;
}
// dp[i] represents the length of the LIS ending at index i
let dp = Array(nums.length).fill(1);
let maxLength = 1;
for (let i = 1; i < nums.length; i++) {
// iterate index before i
for (let j = 0; j < i; j++) {
if (nums[j] < nums[i]) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
maxLength = Math.max(maxLength, dp[i]);
}
return maxLength;
};
Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0. For example, text1 = "abcde", text2 = "ace", then the longest common subsequence is "ace" and its length is 3.
var longestCommonSubsequence = function(text1, text2) {
let m = text1.length;
let n = text2.length;
// dp[i][j] means the LCS of first i characters in text1 and first j characters in text2
// 1. X[m-1] == Y[n-1] -> LCS(Xm-1，Yn-1)
// 2. X[m-1] != Y[n-1] -> max(LCS(Xm-1, Yn), LCS(Xm, Yn-1))
let dp = Array.from(Array(m + 1), () => Array(n + 1).fill(0));
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (text1[i - 1] === text2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
};
Best Time to Buy and Sell Stock. You are given an array prices where prices[i] is the price of a given stock on the ith day.
// only one transaction
var maxProfit = function(prices) {
let min = prices[0];
let maxProfit = 0;
for (let i = 0; i < prices.length; i++) {
if (prices[i] < min) {
min = prices[i];
} else if (prices[i] - min > maxProfit) {
maxProfit = prices[i] - min;
}
}
return maxProfit;
};
// multiple transactions (greedy)
var maxProfit = function(prices) {
let maxProfit = 0;
for (let i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1]) {
maxProfit += prices[i] - prices[i - 1];
}
}
return maxProfit;
};
// multiple transactions but with transaction fee for each one
var maxProfit = function(prices, fee) {
// dp[i] represents two states
// 1. dp[i][0]: max profit on day i if you hold a stock.
// 2. dp[i][1]: max profit on day i if you do not hold a stock.
let dp = Array.from(Array(prices.length), () => Array(2).fill(0));
dp[0][0] = 0 - prices[0];
dp[0][1] = 0;
for (let i = 1; i < prices.length; i++) {
dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
dp[i][1] = Math.max(dp[i - 1][0] + prices[i] - fee, dp[i - 1][1]);
}
return dp[prices.length - 1][1];
};
// After you sell your stock, you cannot buy stock on the next day
var maxProfit = function(prices) {
// dp[i][0]: Holding a stock after day i.
// dp[i][1]: Not holding a stock after day i without entering a cooldown.
// dp[i][2]: Just sold a stock on day i.
// dp[i][3]: In a cooldown after selling stock.
let dp = Array.from(Array(prices.length), () => Array(4).fill(0));
dp[0][0] = -prices[0];
for (let i = 1; i < prices.length; ++i) {
dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i], dp[i - 1][3] - prices[i]);
dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][3]);
dp[i][2] = dp[i - 1][0] + prices[i];
dp[i][3] = dp[i - 1][2];
}
return Math.max(dp[prices.length - 1][1], dp[prices.length - 1][2], dp[prices.length - 1][3]);
};