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# Two Sum LeetCode Solution

## Problem – Two Sum LeetCode Solution

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 Solution in Java

``````public int[] twoSum(int[] numbers, int target) {
int[] result = new int;
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < numbers.length; i++) {
if (map.containsKey(target - numbers[i])) {
result = i;
result = map.get(target - numbers[i]);
return result;
}
map.put(numbers[i], i);
}
return result;
}
``````

### Two Sum LeetCode Solution in Python

``````class Solution(object):
def twoSum(self, nums, target):
buffer_dictionary = {}
for i in rangenums.__len()):
if nums[i] in buffer_dictionary:
return [buffer_dictionary[nums[i]], i] #if a number shows up in the dictionary already that means the
#necesarry pair has been iterated on previously
else: # else is entirely optional
buffer_dictionary[target - nums[i]] = i
# we insert the required number to pair with should it exist later in the list of numbers
``````

### Two Sum LeetCode Solution in C++

``````vector<int> twoSum(vector<int> &numbers, int target)
{
//Key is the number and value is its index in the vector.
unordered_map<int, int> hash;
vector<int> result;
for (int i = 0; i < numbers.size(); i++) {
int numberToFind = target - numbers[i];

//if numberToFind is found in map, return them
if (hash.find(numberToFind) != hash.end()) {
//+1 because indices are NOT zero based
result.push_back(hash[numberToFind] + 1);
result.push_back(i + 1);
return result;
}

//number was not found. Put it in the map.
hash[numbers[i]] = i;
}
return result;
}
``````

### Two Sum LeetCode Solution in JavaScript

``````const twoSum = (nums, target) => {
const map = {};

for (let i = 0; i < nums.length; i++) {
const another = target - nums[i];

if (another in map) {
return [map[another], i];
}

map[nums[i]] = i;
}

return null;
};
``````
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