Two Sum
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# LeetCode #1: Two Sum (C/C++).

easy

source: https://leetcode.com/problems/two-sum
C/C++ Solution to LeetCode problem 1. Two Sum.

## 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.

## Examples

### 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.

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

## Solution

Two solutions using hashmap (in reality the second one is an optimization of the first one).

• Add all values to a hash table as key and value a vector of the indexes where the value is located.
• Iterate through the hash table, we need to find a value x = target - current_key.
• if the x key exist, then we get the index for that value and for the current_key.
• if not, we move the the next key.

The second solution performs the search of x, if not found then the current_key it is added to the hash table.

### Solution 1:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, vector<int>> hasht; for(int i=0; i<nums.size(); i+=1) hasht[nums[i]].push_back(i); for(auto const& [key, value]: hasht) { int need = target - key; auto a = hasht.find(need); if(a != hasht.end()) { if (key == a->first) { if (value.size() == 1) continue; return {value, value}; } else return {value, a->second}; } } return {}; } }; 

### Solution 2:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> hasht; for(int i = 0; i < nums.size(); i++){ if(hasht.find(target - nums[i]) == hasht.end()) hasht[nums[i]] = i; else return {hasht[target - nums[i]], i}; } return {}; } };