Problem

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution

// Solution:
// The optimal solution is to use a hash table to store the numbers in the array as keys and their indices as values.
// Then, we can iterate through the array and check if the target minus the current number is in the hash table.
// If it is, we return the indices of the current number and the target minus the current number.
// This solution is optimal because it has a time complexity of O(n) and a space complexity of O(n).

const twoSum = (nums, target) => {
  const hashTable = {};
  for (let i = 0; i < nums.length; i++) {
    if (hashTable[target - nums[i]] !== undefined) {
      return [hashTable[target - nums[i]], i];
    }
    hashTable[nums[i]] = i;
  }
};

console.log(twoSum([2, 7, 11, 15], 9)); // [0, 1]