Given an array of integers nums&nbsp;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.

&nbsp;
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]


&nbsp;
Constraints:


	2 &lt;= nums.length &lt;= 104
	-109 &lt;= nums[i] &lt;= 109
	-109 &lt;= target &lt;= 109
	Only one valid answer exists.


&nbsp;
Follow-up:&nbsp;Can you come up with an algorithm that is less than&nbsp;O(n2)&nbsp;time complexity?
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        n = len(nums)
        for i in range(n - 1):
            for j in range(i + 1, n):
                if nums[i] + nums[j] == target:
                    return [i, j]
        return []  # No solution found
class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] answer = new int[2];
		// Two for loops for selecting two numbers and check sum equal to target or not
		
        for(int i = 0; i < nums.length; i++){
            for(int j = i+1; j < nums.length; j++) {
			 // j = i + 1; no need to check back elements it covers in i;
                if(nums[i] + nums[j] == target) {
				// Check sum == target or not
                    answer[0] = i;
                    answer[1] = j;
                    return answer;
                }
            } 
        }
          return null;
    }
}
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int n = nums.size();
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums[i] + nums[j] == target) {
                    return {i, j};
                }
            }
        }
        return {}; // No solution found
    }
};
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    var sol = [];
    var found = 0;
    for(let i = 0; i < nums.length; i ++) {
        for(let j = i + 1; j < nums.length; j ++) {
            if(nums[i] + nums[j] === target) {
                sol.push(i, j);
                found = 1;
                break;
            }
        }
        if(found == 1) return sol;
    }
    return sol;
};
Two Sum