The very first technical question I got asked was the classic two sum algorithm problem. I was fresh to algorithms and could solve it but I couldn't optimize it to a requested time complexity. Turns out this is a super common problem and you will likely encounter it either in an interview or down the road practicing algorithms.

It's a pattern useful to recognize and it shows up in different types of problems so it's ideal to know how to tackle it if it ever rears its head.

### The Problem

So the general gist of a two sum is that you have a list or an array of numbers and a target sum to hit. You're looking to return the indexes of the two numbers that when added together hit the target sum. There should only be one solution to the problem from the list of numbers and a number can not be used twice.

### My first solution

Let's assume the input is:

- array = [1, 3, 10, 11, 14]
- goal = 13

```
const twoSum = (array, goal) => {
let indexes = [];
for(let i = 0; i < array.length; i++){
for(let j = i + 1; j < array.length; j++){
if (array[i] + array[j] === goal) {
indexes.push(i);
indexes.push(j);
}
}
}
return indexes;
}
```

This will return an array of [1, 2].

It works but if you check it out you'll notice it's running a loop inside of a loop to figure out which two numbers add up to the goal. That puts us at a time complexity of O(n^2). Pretty slow. Not a big deal for a small array like ours but it's far from optimized and I could say without a shadow of a doubt that if you're doing this type of problem they will be looking for you to improve the time complexity.

### A more optimized solution

Lets assume the same input:

- array = [1, 3, 10, 11, 14]
- goal = 13

```
const twoSum = (array, goal) => {
let mapOfNumbers = {};
let twoIndexes = [];
for (let i = 0; i < array.length; i++) {
mapOfNumbers[array[i]] = i;
}
for (let i = 0; i < array.length; i++) {
let target = goal - arr[i];
if(mapOfNumbers[target] !== null && mapOfNumbers[target] !== i) {
twoIndexes.push(i);
twoIndexes.push(mapOfNumbers[target]);
}
}
return twoIndexes;
}
```

Okay so let's talk about what's going on in this.

First thing is I mapped out the numbers and their indexes. I used the first for loop to accomplish this. Notice I assigned the number as the key and the index as its value.

Second thing is a ran a second for loop through the array and first determined what the value would have to be to equal to goal at that index.

Then inside that loop I do an if statement to check two things. One being if that target value exists in the map we created. The second making sure it's not at the same index as the one we are currently working with.

If they both pass then I push those two indexes into my output array and simply return it.

So while this one has two loops they're not nested so the time complexity lands at O(n) power. Much better.

## Wrap Up

Okay so that's all I'm covering for today but if you have any questions at all feel free to reach out to me. I hope this is helpful to somebody in solving this problem you will no doubt encounter at some point. Happy Coding!

## Top comments (18)

Great explanation. Under a time constraint pressure I would have undoubtedly gone with the nested arrays solution. The optimized solution you have there is better (considering the problems rules we must abide by). But after thinking about it, the first loop you have there is unnecessary because you are just remapping it to quickly find the index of the difference between the array number and the goal.

`Array.prototype.indexOf`

will do that for you. Here is an even more optimized solution coupled with an early return.const input = [1, 3, 10, 11, 13];

function twoSum(array, target) {

for (let i = 0; i < array.length; i++) {

const diffIndex = array.indexOf(target - array[i]);

if (diffIndex >= 0 && diffIndex !== i) {

return [i, diffIndex];

}

}

return []; // no solution found

}

console.log(twoSum(input, 13)); // outputs [1,2]

codesandbox.io/s/keen-browser-hmsp...

Nice solution! I like the early return, that way the solution doesn't double up on the numbers.

However, I think that

`indexOf`

actually runs in`O(n)`

time, because it has to search through the array for the specified number. That means your solution is still`O(n^2)`

, because it has two nested loops.I think the most optimized would be a mix of both, so:

Thanks for your reply! Since it was more of a conceptual answer, I didn't run it before commenting. I have now; I figured out the error and fixed it.

For those interested, the issue was that a

`return`

statement in a`forEach`

loop actually just returns inside the loop, and doesn't return in the function. To fix it, I converted it to a plain`for`

loop.As for the 5 people who liked it, I'm sure they appreciated the concept behind the comment and could see past any logic errors. I don't think that quite makes them fools.

Your solution will fail with test input

[1, 5, 5, 4, 9]

10

since method indexOf stops immediately at the first match

No it does not. It correctly outputs [0, 4]. But I see where you are going with it, change that last value from 9 to 10 so that only the two 5 values will work. It still outputs [2, 1] which are the indices of both 5 values. Why? Because even though it won't catch it on the first five it comes across it will catch it on the second. But you bring up a good point. Can we optimize this so that you don't have to keep going through the array if the solution is a duplicate value that is AFTER the index you are currently on? Yes. And here is my solution for that.

indexOf uses forloop internally so the time complexity of your code becomes O(n^2).

Instead use map for this solution.

Awesome post! Using an

`Object`

is a great way to ensure it runs in`O(n)`

time.Also, a tip I learned for making code blocks a lot easier to read is to take advantage of syntax highlighting; i.e. instead of using just the three backticks, put the language after it. It's hard to show in markdown, but something like this:

```

```

``javascript`

.That should give something like this:

Thanks so much for sharing! Your code is super readable and your explanations are very understandable.

Oh that's amazing I will definitely be using that thank you!

Bravo, so simple and powerful,

The result of the previous example will be [1, 2, 2, 1]

So we need an solution for it,

We can break the array once the goal is achieved, Or to union the returned array using following expression

return [... new Set(...twoIndexes)]

another thing along side the above improvment which can improve performance of your solution is like you know JS return false for falsy values and 0 considered falsy value even it's a real value like 0

So we need to replace the first part of the IF condition to be

mapOfNumbers[target] != null Instead of mapOfNumbers[target]

Because what if index of the target is 0 ? it will return false

so we continue looping through the array for no reason

Great catch and a super good point! I'm going to edit that into the post much appreciation.

What about something like this?

best solution with just one for loop

thumb up for your readable code and explanation !

Thank you!

test

Love it, thank you!

I think you can add a break after executing the if condition to avoid duplication of indices