Each day I solve several coding challenges and puzzles from Codr's ranked mode. The goal is to reach genius rank, along the way I explain how I solve them. You do not need any programming background to get started, and you will learn a ton of new and interesting things as you go.

```
function threeSum(nums) {
if (nums.length < 3) return [];
const list = [];
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length; i++) {
if (nums[i] > 0) break;
if (i > 0 && nums[i] === nums[i - 1]) continue;
let left = i;
let right = nums.length - 1;
while (left < right) {
if (left === i) {
left++;
} else if (right === i) {
right--;
} else if (nums[left] + nums[right] + nums[i] === 0) {
list.push([nums[left], nums[right], nums[i]]);
while(nums[left] === nums[left + 1]) {
left++;
}
left++;
while(nums[right] === nums[right - 1]) {
right--;
}
right--;
continue;
} else if (nums[left] + nums[right] + nums[i] > 0) {
right--;
} else {
left++;
}
}
}
return list;
};
let A = threeSum([-0,1,-1,1,-0,0]);
A = A.length;
// A = ? (number)
```

In today's challenge we are dealing with a function `threeSum`

, I have no clue what it does but is has something to do with three and sums.

The challenge wants us to solve for `A`

which is the `length`

of the output of `threeSum`

. This function returns `list`

which is an array.

We have no clue what this function does, but we know its output. Let's figure out how this `list`

array is being filled. The only place where we find an operation adding items to this array is here:

```
} else if (nums[left] + nums[right] + nums[i] === 0) {
list.push([nums[left], nums[right], nums[i]]);
...
}
```

As we can see, it pushes an item (array) into `list`

when the sum of three numbers from `nums`

(the input) is equal to zero. In a nutshell this algorithm is designed to find triplets whose sum is zero.

When we analyze the code fully, we see that the input array `nums`

is sorted in ascending order; the outer most for-loop iterates over each number in `nums`

indexed by `i`

; followed by `left`

and `right`

pointers which are to the right of `i`

. In this fashion the algorithm is searching for only unique triplets whose sum is zero. Here's some pseudo-code to illustrate the process:

```
nums = [-0, 1, -1, 1, -0, 0]
-> sort
nums = [-1, -0, -0, 0, 1, 1]
----------
i = 0
left = 0
right = 5
...
(-0) + 1 + (-1) = 0 --> push
i = 0
left = 1
right = 5
----------
i = 1
left = 1
right = 5
...
(-0) + 0 + (-0) = 0 --> push
i = 1
left = 2
right = 3
----------
for all other attempts:
i + left + right !== 0
list.length == 2
```

By solving these challenges you train yourself to be a better programmer. You'll learn newer and better ways of analyzing, debugging and improving code. As a result you'll be more productive and valuable in business. Get started and become a certified Codr today at https://nevolin.be/codr/

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