1.two-sum.cpp

#include <bits/stdc++.h>

using namespace std;

// @lc code=start
class Solution
{
public:
    vector<int> twoSum(vector<int> &nums, int target)
    {

        map<int, int> mp;

        for (int i = 0; i < nums.size(); i++)
        {
            int num = nums[i];
            int comp = target - num;
            if (mp.contains(comp))
            {
                return {mp[comp], i};
            }
            mp[num] = i;
        }

        return {-1, -1};
    }
};
// @lc code=end

/*
 * @lc app=leetcode id=1 lang=cpp
 *
 * [1] Two Sum
 *
 * https://leetcode.com/problems/two-sum/description/
 *
 * algorithms
 * Easy (56.21%)
 * Likes:    66054
 * Dislikes: 2458
 * Total Accepted:    19.9M
 * Total Submissions: 35M
 * Testcase Example:  '[2,7,11,15]\n9'
 *
 * 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[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
 * Only one valid answer exists.
 *
 *
 *
 * Follow-up: Can you come up with an algorithm that is less than O(n^2) time
 * complexity?
 */