1_two_sum.cpp

/*
LeetCode 1 - Two Sum
Difficulty: Easy

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.

Example 1:
    Input: nums = [2,7,11,15], target = 9
    Output: [0,1]
    Explanation: 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

Time Complexity: O(n)
Space Complexity: O(n)
*/

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> seen;  // value -> index
        for (int i = 0; i < nums.size(); i++) {
            int complement = target - nums[i];
            if (seen.find(complement) != seen.end()) {
                return {seen[complement], i};
            }
            seen[nums[i]] = i;
        }
        return {};
    }
};

int main() {
    Solution solution;

    vector<int> nums1 = {2, 7, 11, 15};
    vector<int> result1 = solution.twoSum(nums1, 9);
    cout << "[" << result1[0] << ", " << result1[1] << "]" << endl;  // [0, 1]

    vector<int> nums2 = {3, 2, 4};
    vector<int> result2 = solution.twoSum(nums2, 6);
    cout << "[" << result2[0] << ", " << result2[1] << "]" << endl;  // [1, 2]

    vector<int> nums3 = {3, 3};
    vector<int> result3 = solution.twoSum(nums3, 6);
    cout << "[" << result3[0] << ", " << result3[1] << "]" << endl;  // [0, 1]

    return 0;
}