Problem name: Two Sum

Problem statement:

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]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Constraints:

• 2 <= nums.length <= 103
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

Solution:

#Solution1 : Simple solution with n^2 time complixity.

class Solution {
public:
     vector<int> twoSum(vector<int>& nums, int target) {
         vector<int> final;
         for(int i=0; i<nums.size();i++){
             for(int j=i+1; j<nums.size();j++){
                 if(nums[i] + nums[j] == target){
                     final.push_back(i);
                     final.push_back(j);
                 }
             }
         }
         return final;
     }
};

#Solution2 : Another solution using unordered map with better time complexity.

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> m;
        vector<int> final;
        for(int i=0; i<nums.size();i++){
            int complement = target - nums[i];
            if(m.find(complement) != m.end()){
                final.push_back(m[complement]);
                final.push_back(i);
                return final;
            }
            m[nums[i]] = i;
        }
        return final;
    }
};