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In this post, you will find the solution for the **Two Sum** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Two Sum– LeetCode Problem

Two Sum– LeetCode Problem

**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 = 9Output:[0,1]Output: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.**

Two Sum– LeetCode Solutions

class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> numToIndex; for (int i = 0; i < nums.size(); ++i) { if (numToIndex.count(target - nums[i])) return {numToIndex[target - nums[i]], i}; numToIndex[nums[i]] = i; } throw; } };

class Solution { public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> numToIndex = new HashMap<>(); for (int i = 0; i < nums.length; ++i) { if (numToIndex.containsKey(target - nums[i])) return new int[] {numToIndex.get(target - nums[i]), i}; numToIndex.put(nums[i], i); } throw new IllegalArgumentException(); } }

class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: numToIndex = {} for i, num in enumerate(nums): if target - num in numToIndex: return numToIndex[target - num], i numToIndex[num] = i