Leetcode: Two Sum

Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

Idea: One trivial solution is to for each element, loop through all remaining elements and check if their sum is equal to the target. The time complexity is O(n^2), where n is the size of array. To seek the efficiency, we need the extra space for this problem. The idea of this problem is using a hashtable (unordered_map) to store unordered_map<target-numbers[i], i+1>, where the key is the difference of the target and the current element and the corresponding value of hashtable is the index (i+1). Therefore, we will find the two elements in one loop.

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        std::unordered_map<int, int> hash;
        std::vector<int> sol;
        for (int i=0; i < numbers.size(); ++i)
        {
            auto &found = hash.find(numbers[i]); 
            if(found != hash.end())
            {
                sol.push(found->second);
                sol.push(i+1);
                return sol;
            }
            else
            {
                hash[target - numbers[i]] = i+1;
            }
        }
    }
};