15. 3Sum Medium

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.

Approach

Input: An integer array nums, elements can be positive or negative.

Output: Find all unique triplets [a, b, c] that satisfy a + b + c == 0.

This is a classic problem of sorting + opposite-direction two pointers.

We can sort the array first, then fix the first number nums[i], and use two pointers in the range i+1 to n-1:

1. Define two pointers: left = i + 1 and right = len(nums) - 1, pointing to candidate positions for the second and third numbers respectively;

2. For each nums[i]:

   • If the current number is the same as the previous one (nums[i] == nums[i - 1]), it means we might generate duplicate triplets, so we skip it;
   • Use two pointers to iterate through the range nums[i+1:right], calculate the sum of the three numbers;
   • If the sum is less than 0, move the left pointer to the right (left += 1);
   • If the sum is greater than 0, move the right pointer to the left (right -= 1);
   • If the sum equals 0, add it to the result set, and skip the duplicate nums[left] and nums[right] elements;

3. When the left and right pointers meet, the processing for the current i ends, and we proceed to the next iteration in the loop.

Essentially, by fixing a number nums[i], we are finding the sum of two remaining numbers that equals -nums[i].

By using sorting + deduplication + two pointers, the time complexity of this solution is O(n^2). It is the optimal standard approach for solving the "3Sum" problem.

Implementation

python

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()  # Sort the array first to easily use two pointers later

        ans = []     # Store the final results
        n = len(nums)

        for i in range(n - 2):  # Iterate over the first number, leave room for j and k
            x = nums[i]

            # Deduplicate: if the current value is the same as the previous one, skip (to avoid duplicate triplets)
            if i > 0 and nums[i] == nums[i - 1]:
                continue

            # Pruning 1: if the current minimum sum of 3 numbers > 0, no solution is possible later, exit directly
            if x + nums[i + 1] + nums[i + 2] > 0:
                break

            # Pruning 2: if the current maximum sum of 3 numbers < 0, it's impossible to sum to 0, skip current i
            if x + nums[-1] + nums[-2] < 0:
                continue

            # Find combinations where the sum of three numbers is 0 using two pointers from i+1 to the end
            j = i + 1
            k = n - 1
            while j < k:
                s = x + nums[j] + nums[k]  # Current sum of three numbers

                if s < 0:
                    j += 1  # Sum is too small, move left pointer to increase sum
                elif s > 0:
                    k -= 1  # Sum is too large, move right pointer to decrease sum
                else:
                    ans.append([x, nums[j], nums[k]])  # Found a valid combination

                    # Skip duplicate second numbers
                    j += 1
                    while j < k and nums[j] == nums[j - 1]:
                        j += 1

                    # Skip duplicate third numbers
                    k -= 1
                    while k > j and nums[k] == nums[k + 1]:
                        k -= 1

        return ans

javascript

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
const threeSum = function(nums) {
    nums.sort((a, b) => (a - b));
    
    const ans = [];

    for (let i = 0; i < nums.length - 2; i++) {
        const x = nums[i];

        if (i > 0 && nums[i] == nums[i - 1])
            continue
        
        if ((x + nums[i + 1] + nums[i + 2]) > 0)
            break
        
        if ((x + nums.at(-1) + nums.at(-2)) < 0)
            continue
        
        let j = i + 1;
        let k = nums.length - 1;

        while (j < k) {
            const s = x + nums[j] + nums[k];

            if (s > 0) {
                k--;
            } else if (s < 0) {
                j++;
            } else {
                ans.push([x, nums[j], nums[k]]);

                j++;
                while (j < k && nums[j] == nums[j - 1]) {
                    j++;
                }

                k--;
                while (j < k && nums[k] == nums[k + 1]) {
                    k--;
                }
            }
        }
    }
    return ans;
};

Complexity Analysis

• Time Complexity: O(n^2)
• Space Complexity: O(1)

Links

15. 3Sum (English)15. 三数之和 (Chinese)