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‎.idea/leet code.iml

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‎Easy/1. Two Sum.js

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/*
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Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
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You may assume that each input would have exactly one solution, and you may not use the same element twice.
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You can return the answer in any order.
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*/
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/*
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https://leetcode.com/problems/two-sum/discuss/2388388/100-fastest-solution-java-script-solution-0n
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*/
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/* TIME COMPLEXITY IS O(logn) */
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/*
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Example 1:
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Input: nums = [2,7,11,15], target = 9
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Output: [0,1]
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Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
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Example 2:
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Input: nums = [3,2,4], target = 6
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Output: [1,2]
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Example 3:
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Input: nums = [3,3], target = 6
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Output: [0,1]
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*/
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/**
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* @param {number[]} nums
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* @param {number} target
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* @return {number[]}
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*/
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var twoSum = function (nums, target) {
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var s = 0; //initilize a pointer to 0
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for (let i = 1; i < nums.length; i++) { //itterate 1 to length
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if ((nums[s] + nums[i]) == target) { //check the pointer is equal to the target number or not
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return [s, i]; //return if both sum are equal to the target
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}
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if (i == nums.length - 1) { //if i equal to the length-1 then reinitilize it's to s+1
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i = s + 1;
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s++;
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}
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}
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}
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var nums = [3, 2, 4], target = 6;
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console.log(twoSum(nums, target));

‎Easy/125. Valid Palindrome.js

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/*
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125. Valid Palindrome
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https://leetcode.com/problems/valid-palindrome/
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*/
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/* TIME COMPLEXITY O(N) */
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/*
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Example 1:
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Input: s = "A man, a plan, a canal: Panama"
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Output: true
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Explanation: "amanaplanacanalpanama" is a palindrome.
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Example 2:
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Input: s = "race a car"
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Output: false
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Explanation: "raceacar" is not a palindrome.
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*/
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/**
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* @param {string} s
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* @return {boolean}
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*/
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//Create a function to check that the whole string is palindrom or not
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//This function is the main function to check our string is valid palindrom or not
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function check_palindrom(s) {
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let pointer = s.length - 1; // Make a pointer which is start form length
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for (let i = 0; i < s.length / 2; i++, pointer--) {
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if (s[i] != s[pointer]) { // Check is the Ith possition value and pointer possition value is equal or not
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return false; // return false if any of character is not equal we don't require to itterate whole string
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}
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}
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return true; // Return true if all condition is satisfy
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}
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// Method 1
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// This Below method is basically used for non-regex method.
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var isPalindrome = function (s) {
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// Make a array which is used to store all non-alphanumeric characters.
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s = s.toLowerCase(); // Convert whole string to lower case.
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var ansArray = [];
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for (let i = 0; i < s.length; i++) {
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// The bellow condition is used to check that the number should be a-z or 0-9
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if ((s.charCodeAt(i) >= 97 && s.charCodeAt(i) <= 122) || (s.charCodeAt(i) >= 48 && (s.charCodeAt(i) <= 57))) {
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ansArray.push(s[i]) // If the condition is true then add this value to the array
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}
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}
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// Here The join function is used to convert array to string.
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return check_palindrom(ansArray.join(""));
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};
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// Method 2
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// Below method is basically used for use regex
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var isPalindrome = function (s) {
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// the below replace method is basically used to remove all non-alphanumeric character
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// And the lower function is used for convert whole string to lower case
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s = s.replace(/[^A-Za-z0-9]/g, '').toLowerCase();
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// Convert whole string to lower case.
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// Here The join function is used to convert array to string.
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return check_palindrom(s);
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};
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console.log(isPalindrome("0P"));
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console.log(isPalindrome("A man, a plan, a canal: Panama"));
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// char Code
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// a = 97
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// z = 122
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// A = 65
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// Z = 90
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// 0 = 48
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// 9 = 57

‎Easy/13. Roman to Integer.js

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/* 13. Roman to Integer
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https://leetcode.com/problems/roman-to-integer/
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*/
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/* TIME COMPLEXITY IS O(N) */
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/*
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Example 1:
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Input: s = "III"
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Output: 3
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Explanation: III = 3.
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Example 2:
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Input: s = "LVIII"
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Output: 58
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Explanation: L = 50, V= 5, III = 3.
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Example 3:
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Input: s = "MCMXCIV"
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Output: 1994
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Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
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*/
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var romanToInt = function (s) {
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// Create A Map
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const map = {
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'I': 1,
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'V': 5,
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'X': 10,
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'L': 50,
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'C': 100,
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'D': 500,
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'M': 1000
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}
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let ans = 0;
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for (let i = 0; i < s.length; i++) {
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const cur = map[s[i]], next = map[s[i + 1]];
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//check that the current indext value is smaller than the next index value or not if the current value smaller than next value return current value as minus(-) value.
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ans += cur < next ? -cur : cur;
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}
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return ans;
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};
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romanToInt("MCMXCIV");
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//Input :MCMXCIV
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//Output:1994

‎Easy/14. Longest Common Prefix.js

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/*
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14. Longest Common Prefix
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https://leetcode.com/problems/longest-common-prefix/
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https://leetcode.com/problems/longest-common-prefix/discuss/2396181/easiest-java-script-solution-on2
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*/
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/* TIME COMPLEXITY IS O(N2) */
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/*
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Example 1:
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Input: strs = ["flower","flow","flight"]
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Output: "fl"
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Example 2:
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Input: strs = ["dog","racecar","car"]
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Output: ""
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Explanation: There is no common prefix among the input strings.
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*/
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/**
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* @param {string[]} strs
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* @return {string}
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*/
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var longestCommonPrefix = function (strs) {
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var firstIndexStr = strs[0], ans = "";
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if (strs[0] == "") return ""; //if string contain blank
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if (strs.length == 1) return strs[0] //return if string contain only one word
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for (let i = 0; i < firstIndexStr.length; i++) { //this loop is used for check the firstWord all character with it's respective to other word character.
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var count = 1; //use count veriable to check that all character satisfy with all word character or not.
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for (let j = 1; j < strs.length; j++) {
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if (strs[j][i] == firstIndexStr[i]) { //here we check string First Word with other character words
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count++; //if the character match with other character than increment to +1
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}
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}
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if (count == strs.length) { //check that all word character satisfy with all character or not.
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ans += firstIndexStr[i];
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} else {
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return ans.length > 0 ? ans : ""; //chect that the ans length should be greater than 1 than return ans otherwise return "";
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}
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}
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return ans;
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};
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console.log(longestCommonPrefix(["flower", "flower", "flower", "flower"]));

‎Easy/15. 3Sum.js

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/**
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* @param {number[]} nums
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* @return {number[][]}
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*/
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var threeSum = function (nums) {
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nums.sort((a, b) => a - b);
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let res = [];
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for (let i = 0; i < nums.length - 2; i++) {
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// skipping the duplicate elements
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if (i > 0 && nums[i] == nums[i - 1]) continue;
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let j = i + 1;
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let k = nums.length - 1;
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// Two sum approach
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while (j < k) {
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let sum = nums[j] + nums[k];
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if (sum == -nums[i]) {
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res.push([nums[i], nums[j], nums[k]]);
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// skipping the duplicate elements
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while (nums[j] == nums[j + 1]) j++;
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while (nums[k] == nums[k - 1]) k--;
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k--;
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j++;
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}
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else if (sum > -nums[i]) {
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k--;
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}
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else {
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j++;
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}
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}
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}
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return res;
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};
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console.log(threeSum([-1, 0, 1, 2, -1, -4]));

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