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3 Sum⚓︎

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Description⚓︎

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]]

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:

  • Input: nums = [0,1,1]
  • Output: []

Explanation: The only possible triplet does not sum up to 0.

Example 3:

  • Input: nums = [0,0,0]
  • Output: [[0,0,0]]

Explanation: The only possible triplet sums up to 0.

Constraints:

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

Solution⚓︎

Double Pointer Starter Way⚓︎

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class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > 0) return res;

            // wrong! e.g. [-1, -1, 2]
            /*
            if (nums[i] == nums[i + 1]) {
                continue;
            }
            */
            //the correct way to remove duplicates
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int left = i + 1, right = nums.size() - 1;
            while (left < right) {
                if (nums[i] + nums[left] + nums[right] > 0) right--;
                else if (nums[i] + nums[left] + nums[right] < 0) left++;
                else {
                    res.push_back(vector<int>{nums[i], nums[left], nums[right]});
                    // remove duplicates
                    while (right > left && nums[right] == nums[right - 1]) right--;
                    while (right > left && nums[left] == nums[left + 1]) left++;
                    right--;
                    left++;
                }
            }
        }
        return res;
    }
};

Time complexity: \(O(n^2)\)

Double Pointer Simpler Way⚓︎

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

        for (int i = 0; i < nums.size(); i++) {
            if (i && nums[i] == nums[i - 1]) continue;
            for (int j = i + 1, k = nums.size() - 1; j < k; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;
                while (j < k - 1 && nums[i] + nums[j] + nums[k - 1] >= 0) k--;
                if (nums[i] + nums[j] + nums[k] == 0)
                    res.push_back({nums[i], nums[j], nums[k]});
            }
        }
        return res;
    }
};

Note: Why are we checking for nums[i] + nums[j] + nums[k-1] in the while loop, but nums[i] + nums[j] + nums[k] when adding the result? This is because in the while loop, if a k has been found that makes the sum of the three elements less than 0, then the next k (i.e., k-1) must make the sum of the three elements smaller (because nums[k-1] <= nums[k]). So, if nums[i] + nums[j] + nums[k-1] is still greater than or equal to 0, then we need to keep decreasing k. Then at the end of the while loop, since we may have missed the k that makes the sum of the three elements equal to 0, we finally check if nums[i] + nums[j] + nums[k] == 0 or not.

If we do not use k - 1 in that section, we can rewrite the code:

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

        for (int i = 0; i < nums.size(); i++) {
            if (i && nums[i] == nums[i - 1]) continue;
            for (int j = i + 1, k = nums.size() - 1; j < k; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;
                while (j < k && nums[i] + nums[j] + nums[k] > 0) k--;
                // we need to check if j < k here!
                if (j < k && nums[i] + nums[j] + nums[k] == 0)
                    res.push_back({nums[i], nums[j], nums[k]});
            }
        }
        return res;
    }
};