# Tow Sum

## 1. Two Sum

### Easy

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.

Follow-up: Can you come up with an algorithm that is less than `O(n2) `time complexity?

Python solution using HashMap

``````class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
new_hash_map = {}

for i, n in enumerate(nums):
difference = target - n
if difference in new_hash_map:
return(i, new_hash_map[difference])
else:
new_hash_map[n] = i

``````

Time complexity O(N)

Java Solution

``````class Solution {
public boolean containsDuplicate(int[] nums) {
Set<Integer> numbers = new HashSet<Integer>();
for (int num : nums) {
if (numbers.contains(num)) return true;