problem solving

three_sum.js

/**
 * Three Sum
 *
 * Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique
 * triplets in the array which gives the sum of zero.
 *
 * Note:
 *
 * The solution set must not contain duplicate triplets.
 *
 * Example:
 *
 * Given array nums = [-1, 0, 1, 2, -1, -4],
 *
 * A solution set is:
 * [
 *   [-1, 0, 1],
 *   [-1, -1, 2]
 * ]
 */

 /**
  * triplets in the array which gives the sum of zero
  * @param {number[]} nums 
  * @returns {[[number]]} array of arrays
  */
function threeSum(nums) {

    //Corner case
    if(!nums || nums.length<3) return ret;


    var returned = []

    var n = nums.length
    nums = nums.sort(function (a, b) {
        return a - b
    })

    for (let i = 0; i < n - 2; i++) {

        //Check if value is sorted
        if(i !== 0 && nums[i] === nums[i-1]){
            continue
        }

        let j = i + 1 //next pos
        let k = n - 1 //last pos

        while (j < k) {


            let sum = nums[i] + nums[j] + nums[k]
            if (sum === 0) {
                returned.push([nums[i], nums[j], nums[k]])
                j++
                k--

                //skip duplicates from j iterator                   
                while (j < k && nums[j] == nums[j - 1])
                    j++;

                //skip duplicates from k iterator
                while (j < k && nums[k] == nums[k + 1])
                    k--;

            } else if (sum < 0) { //remember array is sorted if sum is less then increment j else decrement j
                j++
            } else {
                k--
            }
        }

    }

    return returned
}

two_sum.js

/**
 * Problem : Two Sum
 *
 * Given an array of integers, return indices of the two numbers such that they add up to a specific target.
 *
 * You may assume that each input would have exactly one solution, and you may not use the same element twice.
 *
 * Example:
 *
 * Given nums = [2, 7, 11, 15], target = 9,
 *
 * Because nums[0] + nums[1] = 2 + 7 = 9,
 * return [0, 1].
 */

/**
 * 
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]} indices
*/
function twoSum(nums , target){
    
    //To avoid inner forloop we can use map to store the values 
    const map = {}

    for (let i = 0; i < nums.length; i++) {
  
        var diff = target-nums[i] 
        //Check difference is in the 
        if(diff in map){
            return [map[diff] , i ]
        }

        map[nums[i]] = i
    }
}


var nums = [2, 7, 11, 15, 10]
var target = 17

console.log(twoSum(nums , target))