15. 3Sum

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15. 3Sum

Description

Difficulty: Medium

Related Topics: Array, Two Pointers, Sorting

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:

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Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

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Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

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Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

  • 3 <= nums.length <= 3000
  • -105 <= nums[i] <= 105

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    vector<vector<int>> res;

    vector<vector<int>> threeSum(vector<int>& nums) {
        ranges::sort(nums);
        int n = nums.size();
        for (int i = 0; i < nums.size() - 2; i++) {
            if (nums[i] + nums[n - 2] + nums[n - 1] < 0) continue;
            if (nums[i] + nums[i + 1] + nums[i + 2] > 0) break;
            twoSum(i, -nums[i], nums);
            while (i + 1 < nums.size() && nums[i] == nums[i + 1]) {
                i++;
            }
        }
        return res;
    }

    void twoSum(int index, int target, vector<int>& nums) {
        int left = index + 1, right = nums.size() - 1;
        while (left < right) {
            int sum = nums[left] + nums[right];
            if (sum == target) {
                res.push_back({-target, nums[left], nums[right]});
                while (left + 1 < nums.size() && nums[left] == nums[left + 1]) left++;
                while (right - 1 >= 0 && nums[right] == nums[right - 1]) right--;
                left++; right--;
            } else if (sum > target) {
                right--;
            } else {
                left++;
            }
        }
    }
};