# Two Sum – LeetCode

1. Two Sum

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]
```

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.

## Solution

### Approach 1: Brute Force

```class Solution {
public int[] twoSum(int[] nums, int target) {
for(int i = 0;i < nums.length; i++){
for(int j = i+1;j < nums.length; j++){
if(nums[i]+nums[j] == target)
return new int[]{i,j};
}
}
throw new IllegalArgumentException("No two sum solution");
}
}```

Time complexity – O(n2) Space complexity – O(1)

### Approach 2: Using HashMap to store difference with index

```class Solution {
public int[] twoSum1(int[] nums, int target) {
HashMap<Integer,Integer> preComputed = new HashMap<>();
for(int i = 0;i< nums.length;i++){
if(preComputed.containsKey(target-nums[i])){
return new int[]{preComputed.get(target-nums[i]), i};
}
preComputed.put(nums[i],i);
}
throw new IllegalArgumentException("No two sum solution");
}
}```

Time complexity – O(n) Space complexity – O(n)