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## Problem

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[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:

• $2 <= nums.length <= 10^4$
• $-10^9 <= nums[i] <= 10^9$
• $-10^9 <= target <= 10^9$
• Only one valid answer exists.

## Code

class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> map = new HashMap<>();

for(int i = 0; i < nums.length; i++){
int num = nums[i];
if(map.containsKey(target - num)){
return new int[]{map.get(target - num), i};
}

map.put(num, i);
}

return new int[]{-1, -1};
}
}

class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
nun_map = defaultdict(int)

for index, num in enumerate(nums):
other = target - num
if other in nun_map:
return [nun_map[other], index]

nun_map[num] = index

return [-1, -1]

func twoSum(nums []int, target int) []int {
var res []int
numsMap := make(map[int]int)

for i, num := range nums {
remain := target - num
if j, ok := numsMap[remain]; ok {
res = append(res, i, j)
return res
}
numsMap[num] = i
}

return res
}