# Two Sum LeetCode problem Solution Two Sum LeetCode problem Solution

In this post, we are going to Two Sum LeetCode problem which is related to arrays and Hash Table. Let’s have a look at the problem statement first and then try to solve the problem.

## Two Sum LeetCode 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]
Explanation: Because nums + nums == 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.

## Two Sum LeetCode Problem Solutions

I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.

### Problem Solution in C++

``````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
}
};``````

### Problem Solution in Java

``````class Solution {
public int[] twoSum(int[] nums, int target) {
int n = nums.length;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (nums[i] + nums[j] == target) {
return new int[]{i, j};
}
}
}
return new int[]{}; // No solution found
}
}``````

### Problem Solution in Python

``````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``````