Two-sum solution in JavaScript [closed]

Question

I was given this question during an interview. It is similar to two sum problems and I would like to ask for feedback for my solution.

Merging 2 Packages

Given a package with a weight limit limit and an array arr of item weights, implement a function getIndicesOfItemWeights that finds two items whose sum of weights equals the weight limit limit. Your function should return a pair [i, j] of the indices of the item weights, ordered such that i > j. If such a pair doesn’t exist, return an empty array.

Example:

input: arr = [4, 6, 10, 15, 16], lim = 21

output: [3, 1] # since these are the indices of the # weights 6 and 15 whose sum equals to 21 Constraints:

[time limit] 5000ms

[input] array.integer arr

0 ≤ arr.length ≤ 100 [input] integer limit

[output] array.integer

function getIndicesOfItemWeights(arr, limit) { let weights = {}; let result = [] for ( let i = 0; i < arr.length; i++){ let compliment = limit - arr[i]; if (typeof weights[compliment]!= "undefinded"){ if (i > (weights[compliment])){ result.push(weights[compliment], i); //return [i, weights[compliment]]; } else { return [weights[compliment] , i]; } } else { weights[arr[i]] = i; // put new weight into weight obj } } return []; } let arr = [4, 6, 10, 15, 16]; let limit = 21; console.log(getIndicesOfItemWeights(arr, limit)) // output: [3, 1] # since these are the indices of the // weights 6 and 15 whose sum equals to 21

Answer 1

there are many ways to aproach this, somehow i find your solution alittle bit difficult to read but i like the idea of using compliment = limit - arr[i];

a simpler way would be either using nested for loops :

const getIndicesOfItemWeights = (arr, limit) => { let result = []; for (let i = 0; i < arr.length; i++) { if (result[0] && result[1]) break; // break out of the loop if the result is full. for (let j = i; j < arr.length; j++) { if (arr[i] + arr[j] === limit) { // if the sum of two elements is eqaul to the limit. result.push(i, j); // push the indices to the result array. break; // break out of the second loop } } } return result.sort((a, b) => b - a); } const arr = [4, 6, 10, 15, 16]; const limit = 21; const x = getIndicesOfItemWeights(arr, limit); console.log(x)

or take the shortcut of using the compliment = limit - arr[i];

const arr = [4, 6, 10, 15, 16]; const limit = 21; const getIndicesOfItemWeights = (arr, limit) => { let result = []; arr.forEach(e => { const a = arr.find(x => x === limit - e); if (a && result.length === 0) { // if the element is found and the array is empty, push the indices to the result array result.push(arr.indexOf(e), arr.indexOf(a)); return; } }) return result.sort((a, b) => b - a); } const x = getIndicesOfItemWeights(arr, limit); console.log(x)

Answer 2

this is my updated solution after the code review:

let getIndicesOfItemWeights = function(arr, limit) { let map = new Map() let indices = [] for(let i = 0; i < arr.length; i++){ let difference = limit-arr[i] if(map.has(difference)){ indices.push(i, map.get(difference)) return indices } else { map.set(arr[i], i) } } return indices }