C++ Program to find the indices of two numbers in an array that add up to the target (Logic Explained)

 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

Code:

class Solution {

public:

    vector<int> twoSum(vector<int>& nums, int target) {

        unordered_map<int, int> map;

        vector<int> result;

        for (int i = 0; i < nums.size(); i++) {

            int complement = target - nums[i];

            if (map.count(complement)) {

                result.push_back(map[complement]);

                result.push_back(i);

                return result;

            }

            map[nums[i]] = i;

        }

        return result;

    }

};

Explanation:

• The solution uses an unordered_map to store the elements and their indices in the array.
• In each iteration of the for loop, we calculate the complement of the current element and search for it in the map. If the complement is found, we return the indices of the two elements in the result vector.
• If the complement is not found, we insert the current element and its index into the map.
• After the loop, the result vector will contain the indices of the two numbers that add up to the target.