001. 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.

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:

My Solution

I've done this one a few times. This is the oldest solution I found, and it's a doozy.

const twoSum = (nums, target) => {
let list = [nums[0]]
for (let i = 1, j = nums.length; i < j; i++) {
for (let k = 0, l = list.length; k < l; k++) {
if (nums[i] + list[k] === target) {
return [k, i];
}
}
list.push(nums[i]);
}
}

This solution is bad. I've taken a problem with a single input array and somehow made it take O(n2)O(n{^2}) time and O(n)O(n) space. (If I'd just looped through nums itself twice looking for complements rather than store previously seen numbers in the list array, I could've at least gotten space complexity of O(1)O(1).) I can only assume I didn't know about hashmaps when I wrote this.

The next attempt I found was much better.

const twoSum = function(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;
}
};

Here I use a JavaScript object to keep track of numbers I've seen and their indices. This gets runtime back down to O(n)O(n) with space complexity of O(n)O(n).

My most recent solution is basically the same, except that it uses the newer built-in Map in JavaScript.

const twoSum = function(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);
}
};