0001.cpp

// Source: https://leetcode.com/problems/two-sum
// Title: Two Sum
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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?
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

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