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

**Example 2:**

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

**Example 3:**

Input:nums = [3,3], target = 6Output:[0,1]

**Constraints:**

`2 <= nums.length <= 10`

^{4}`-10`

^{9}<= nums[i] <= 10^{9}`-10`

^{9}<= target <= 10^{9}**Only one valid answer exists.**

**Follow-up: **Can you come up with an algorithm that is less than `O(n`

time complexity?^{2})

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;
numbers.add(num);
}
return false;
}
}
```

Time complexity O(N)

## Top comments (0)