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LeetCode 23. Merge k Sorted Lists (javascript)

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

Example 1:

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2:

Input: lists = []
Output: []

Example 3:

Input: lists = [[]]
Output: []

Constraints:

  • k == lists.length
  • 0 <= k <= 10^4
  • 0 <= lists[i].length <= 500
  • -10^4 <= lists[i][j] <= 10^4
  • lists[i] is sorted in ascending order.
  • The sum of lists[i].length won’t exceed 10^4.

Idea:

Divide and Conquer

Solution:

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode[]} lists
 * @return {ListNode}
 */
var mergeKLists = function(lists) {
    return merge(lists, 0, lists.length - 1);
};

function merge(lists, l , r) {
    if (lists.length === 0) return null;
    if (l === r) return lists[l];
    let mid = Math.floor((l + r) / 2);
    let l1 = merge(lists, l, mid);
    let l2 = merge(lists, mid + 1, r);
    return mergeTwoLists(l1, l2);
}

var mergeTwoLists = function(l1, l2) {
    // If one of the list is empty, return the other one.
      if (l1 === null) {
        return l2;
      }
      if (l2 === null) {
        return l1;
      }
    
    // The smaller one becomes the head.
    if (l1.val < l2.val) {
        l1.next = mergeTwoLists(l1.next, l2);
        return l1;
    } else {
        l2.next = mergeTwoLists(l1, l2.next);
        return l2;
    }
};

LeetCode 1325. Delete Leaves With a Given Value (javascript)

Given a binary tree root and an integer target, delete all the leaf nodes with value target.

Note that once you delete a leaf node with value targetif it’s parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you can’t).

Example 1:

Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left). 
After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).

Example 2:

Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]

Example 3:

Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.

Example 4:

Input: root = [1,1,1], target = 1
Output: []

Example 5:

Input: root = [1,2,3], target = 1
Output: [1,2,3]

Constraints:

  • 1 <= target <= 1000
  • The given binary tree will have between 1 and 3000 nodes.
  • Each node’s value is between [1, 1000].

Solution:

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} target
 * @return {TreeNode}
 */
var removeLeafNodes = function(root, target) {
    if (root === null) return null;
    root.left = removeLeafNodes(root.left, target);
    root.right = removeLeafNodes(root.right, target);
    // return null to remove that leaf
    if (root.val === target && root.left === null && root.right === null) {
        return null;
    }
    return root;
};

LeetCode 236. Lowest Common Ancestor of a Binary Tree (javascript)

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:

  • The number of nodes in the tree is in the range [2, 105].
  • -109 <= Node.val <= 109
  • All Node.val are unique.
  • p != q
  • p and q will exist in the tree.

Solution:

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
    if (root === null || root === p || root === q) return root;
    let left = lowestCommonAncestor(root.left, p, q);
    let right = lowestCommonAncestor(root.right, p, q);
    return left === null ? right : (right === null ? left : root);
};

LeetCode 226. Invert Binary Tree (javascript)

Given the root of a binary tree, invert the tree, and return its root.

Example 1:

Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]

Example 2:

Input: root = [2,1,3]
Output: [2,3,1]

Example 3:

Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100

Solution:

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
    if (root === null) return root;
    
    let temp = new TreeNode();
    temp = root.right;
    root.right = root.left;
    root.left = tmp;
    
    root.left = invertTree(root.left);
    root.right = invertTree(root.right);
    return root;
};

LeetCode 104. Maximum Depth of Binary Tree (javascript)

Given the root of a binary tree, return its maximum depth.

A binary tree’s maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 3

Example 2:

Input: root = [1,null,2]
Output: 2

Example 3:

Input: root = []
Output: 0

Example 4:

Input: root = [0]
Output: 1

Constraints:

  • The number of nodes in the tree is in the range [0, 104].
  • -100 <= Node.val <= 100

Solution:

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxDepth = function(root) {
    if (root === null) return 0;
    return 1 + Math.max(maxDepth(root.left) , maxDepth(root.right));
};

LeetCode 101. Symmetric Tree (javascript)

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

Constraints:

  • The number of nodes in the tree is in the range [1, 1000].
  • -100 <= Node.val <= 100

Solution:

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    return isSameTree(root, root); 
};

function isSameTree(t1, t2) {
    if (t1 === null && t2 === null) return true;
    if (t1 === null || t2 === null) return false;
    return t1.val === t2.val && isSameTree(t1.left, t2.right) && isSameTree(t1.right, t2.left); 
}

LeetCode 21. Merge Two Sorted Lists (javascript)

Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists.

Example 1:

Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: l1 = [], l2 = []
Output: []

Example 3:

Input: l1 = [], l2 = [0]
Output: [0]

Constraints:

  • The number of nodes in both lists is in the range [0, 50].
  • -100 <= Node.val <= 100
  • Both l1 and l2 are sorted in non-decreasing order.

Solution: (Recursion)

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function(l1, l2) {
    // If one of the list is emptry, return the other one.
    if (!l1 || !l2) return l1 ? l1 : l2;
    
    // The smaller one becomes the head.
    if (l1.val < l2.val) {
        l1.next = mergeTwoLists(l1.next, l2);
        return l1;
    } else {
        l2.next = mergeTwoLists(l1, l2.next);
        return l2;
    }
};

LeetCode 347. Top K Frequent Elements (javascript)

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

Constraints:

  • 1 <= nums.legth <= 105
  • k is in the range [1, the number of unique elements in the array].
  • It is guaranteed that the answer is unique.

Solution 1: Bucket Sort O(n); O(n)

  • Use a HashMap to store each number and its frequency.
  • Use bucket array to save numbers into different bucket whose index is the frequency
/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
var topKFrequent = function(nums, k) {
    let counts = new Map();
    let buckets = [];
    for (let i = 0; i <= nums.length; i++)
        buckets.push([]);
    
    // count frequent of the elements
    for (let num of nums) {
        if (counts.has(num)) {
            counts.set(num, counts.get(num) + 1);
        } else {
            counts.set(num, 1);
        }
    } 
    // put them into buckets by frequent
    for (let [key, value] of counts) {
        buckets[value].push(key);
    }
    // fetch the larget frequest bucket first, until reach k
    let ans = [];
    for (let i = buckets.length - 1; i > 0 && ans.length < k; i--) {
        if (buckets[i] !== null) ans.push(...buckets[i]);
    }
    return ans;
};

Solution 2: maxHeap O(n log n); O(n)

Javascript does not have a standard heap / priority queue data structure that you can use out of the box.

See Java solution:

public List<Integer> topKFrequent(int[] nums, int k) {
        List<Integer> res = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return res;
        }

        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }

        PriorityQueue<Map.Entry<Integer, Integer>> maxHeap = new 
            PriorityQueue<>((a, b) -> b.getValue() - a.getValue());
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            maxHeap.offer(entry);
        }

        while (res.size() < k) { //important
            res.add(maxHeap.poll().getKey());
        }
        return res;
    }

Solutoin 3: TreeMap (n log n); O(n)

Use treeMap. Use freqncy as the key so we can get all freqencies in order

See Java solution:

public List<Integer> topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int n: nums){
            map.put(n, map.getOrDefault(n,0)+1);
        }

        TreeMap<Integer, List<Integer>> freqMap = new TreeMap<>();
        for(int num : map.keySet()){
           int freq = map.get(num);
           if(!freqMap.containsKey(freq)){
               freqMap.put(freq, new LinkedList<>());
           }
           freqMap.get(freq).add(num);
        }

        List<Integer> res = new ArrayList<>();
        while(res.size()<k){
            Map.Entry<Integer, List<Integer>> entry = freqMap.pollLastEntry();
            res.addAll(entry.getValue());
        }
        return res;
    }

LeetCode 316. Remove Duplicate Letters (javascript)

Given a string s, remove duplicate letters so that every letter appears once and only once. You must make sure your result is the smallest in lexicographical order among all possible results.

Note: This question is the same as 1081: https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/

Example 1:

Input: s = "bcabc"
Output: "abc"

Example 2:

Input: s = "cbacdcbc"
Output: "acdb"

Constraints:

  • 1 <= s.length <= 104
  • s consists of lowercase English letters.

Idea:

Hash + Stack

Solution:

/**
 * @param {string} s
 * @return {string}
 */
var removeDuplicateLetters = function(s) {
    let chars = s.split('');
    // record the index of last occurrence for each character
    let lastIndex = new Map();
    for (let i = 0; i < chars.length; i++) {
        lastIndex.set(chars[i], i);
    }
    let stack = [];
    let used = new Map();
    for (let i = 0; i < chars.length; i++) {
        // if the current character has been used, skip it
        if (used.has(chars[i]) && used.get(chars[i])) continue;
        
        // if the current character is smaller than stack[stack.length - 1]
        // and the last index of stack[stack.length - 1] is larger than i, it means we can use it later,
        // so we pop it out and mark used as false
        while (stack.length !== 0 && chars[i] < stack[stack.length - 1] && lastIndex.get(stack[stack.length - 1]) > i)
            used.set(stack.pop(), false);
        
        stack.push(chars[i]);
        used.set(chars[i], true);
    }
    let ans = [];
    while (stack.length !== 0) ans.push(stack.pop());
    return ans.reverse().join('');
};

LeetCode 224. Basic Calculator (javascript)

Given a string s representing an expression, implement a basic calculator to evaluate it.

Example 1:

Input: s = "1 + 1"
Output: 2

Example 2:

Input: s = " 2-1 + 2 "
Output: 3

Example 3:

Input: s = "(1+(4+5+2)-3)+(6+8)"
Output: 23

Constraints:

  • 1 <= s.length <= 3 * 105
  • s consists of digits, '+''-''('')', and ' '.
  • s represents a valid expression.

Solution:

/**
 * @param {string} s
 * @return {number}
 */
var calculate = function(s) {
    let len = s.length;
    if (len === 0) return 0;
    let stack = [];
    let res = 0;
    let sign = 1;
    
    // remove space
    s.replace(" ", "");
    for (let i = 0; i < len; i++) {
        let n = s.charAt(i);
        if (!isNaN(parseInt(n))) {
            let cur = parseInt(n);
            while (i + 1 < len && !isNaN(parseInt(s.charAt(i + 1)))) {
                cur = cur * 10 + parseInt(s.charAt(i + 1));
                i++;
            }
            res += cur * sign;
        } else if (n === '-') {
            sign = -1;
        } else if (n === '+') {
            sign = 1;
        } else if (n === '(') {
            stack.push(res);
            res = 0;
            stack.push(sign);
            sign = 1;
        } else if (n === ')') {
            res = res * stack.pop() + stack.pop();
        }
    }
    
    return res;
};