Leetcode ANSWER for 1. Two Sum

 1. Two Sum

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]
Explanation: 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 <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
Only one valid answer exists.
Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

SOLUTION:

                                                                                                                                                                    function twoSum(nums, target) {
  let map = {};
  for (let i = 0; i < nums.length; i++) {
    let complement = target - nums[i];
    if (complement in map) {
      return [map[complement], i];
    }
    map[nums[i]] = i;
  }
}

In this code, we create a dictionary (map) to store the numbers we've seen so far, along with their indices. Then, we loop through the array of numbers (nums) and check if the complement (i.e. the difference between the target and the current number) is already in the map. If it is, we've found the two numbers we're looking for, so we return their indices. If it's not, we add the current number to the map so we can check for its complement later.
You can call this function with an array of numbers and a target value like this:
let nums = [2, 7, 11, 15];
let target = 9;
console.log(twoSum(nums, target)); // output: [0, 1]

In this example, the two numbers that add up to the target value of 9 are located at indices 0 and 1 in the nums array.