15_3sum.cpp

/*
LeetCode 15 - 3Sum
Difficulty: Medium

Problem:
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that
i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.

Example 1:
    Input: nums = [-1,0,1,2,-1,-4]
    Output: [[-1,-1,2],[-1,0,1]]

Example 2:
    Input: nums = [0,1,1]
    Output: []

Example 3:
    Input: nums = [0,0,0]
    Output: [[0,0,0]]

Constraints:
    3 <= nums.length <= 3000
    -10^5 <= nums[i] <= 10^5

Time Complexity: O(n^2)
Space Complexity: O(1)  (excluding output)
*/

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> result;

        for (int i = 0; i < nums.size() - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue; // Skip duplicate values for the first element
            }

            int left = i + 1, right = nums.size() - 1;
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                if (total == 0) {
                    result.push_back({nums[i], nums[left], nums[right]});
                    left++;
                    right--;
                    while (left < right && nums[left] == nums[left - 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right + 1]) {
                        right--;
                    }
                } else if (total < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }

        return result;
    }
};

int main() {
    Solution solution;

    vector<int> nums1 = {-1, 0, 1, 2, -1, -4};
    vector<vector<int>> result1 = solution.threeSum(nums1);
    cout << "[";
    for (int i = 0; i < result1.size(); i++) {
        cout << "[";
        for (int j = 0; j < result1[i].size(); j++) {
            cout << result1[i][j];
            if (j < result1[i].size() - 1) cout << ", ";
        }
        cout << "]";
        if (i < result1.size() - 1) cout << ", ";
    }
    cout << "]" << endl; // [[-1, -1, 2], [-1, 0, 1]]

    vector<int> nums2 = {0, 1, 1};
    vector<vector<int>> result2 = solution.threeSum(nums2);
    cout << "[]" << endl; // []

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

    return 0;
}