# 1. Two Sum

## The Problem

Link to original problem on Leetcode.

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Examples

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].


Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]


Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints
• 2 <= nums.length <= 103
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

## My Solution

### Naïve Solution

The simplest approach is nested for loops, solving in $O(n{^2})$ time and $O(1)$ space.

const twoSum = (nums, target) => {    for (let i = 0; i < nums.length; i++) {        for (let j = 0; j < nums.length; j++) {            if (i !== j && nums[i] + nums[j] === target) {                return [i, j];            }        }    }}

### Best Solution

We can reduce the time complexity to $O(n)$ by increasing the space complexity to $O(n)$ in the form of a hashmap. Here I use a JavaScript object to keep track of numbers I've seen and their indices.

const twoSum = (nums, target) => {  let reference = {};    for (let i = 0, j = nums.length; i<j; i++) {        if (target - nums[i] in reference) {            return [reference[target - nums[i]], i];        }        reference[nums[i]] = i;    }};

You could also use the newer built-in Map in JavaScript.

const twoSum = (nums, target) => {    let complements = new Map();    for (let i = 0; i < nums.length; i++) {        if (complements.has(target - nums[i])) {            return [complements.get(target - nums[i]), i];        }        complements.set(nums[i], i);    }};