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Expanding how this Two Sum algorithm works

I'm currently practicing JS and came across this problem on Leetcode:

Two Sum

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

I solved it with this:

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

But then I checked the discussion for other solutions and stumbled upon this one:

var twoSum = function(nums, target) {
    for(let [k,v] of nums.entries()){
        if(nums.slice(k+1).lastIndexOf(target-v)>-1) return [k,nums.lastIndexOf(target-v)]}};

https://leetcode.com/problems/two-sum/discuss/1543367/3-Lines-JavaScript-Two-Sum

I'm confused about how it's supposed to work especially on line 3. Tried console logging each step but ended up confusing myself even more.

Would appreciate any help. Thank you!

about 4 years ago · Juan Pablo Isaza
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