Skip to content

Two Sum⚓︎

Link

Description⚓︎

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 <= 104
  • -10^9 <= nums[i] <= 10^9
  • -10^9 <= target <= 10^9
  • Only one valid answer exists.

Solution⚓︎

Double Pointer⚓︎

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<pair<int, int>> sumPos;

        for (int i = 0; i < nums.size(); i++) {
            sumPos.push_back(make_pair(nums[i], i));
        }

        sort(sumPos.begin(), sumPos.end());
        for (int i = 0, j = nums.size() - 1; i < nums.size(); i++) {
            while (j >= 0 && sumPos[i].first + sumPos[j].first > target) j--;
            if (sumPos[i].first + sumPos[j].first == target) {
                vector<int> res = {sumPos[i].second, sumPos[j].second};
                return res;
            }
        }
        return {};
    }
};

Hash Map⚓︎

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

Time and space complexity: \(O(n)\)