3Sum

Problem

Given an integer array nums, return all unique triplets [a, b, c] such that a + b + c = 0. The answer must not contain duplicate triplets.

Constraints

• 3 ≤ nums.length ≤ 3000
• −10⁵ ≤ nums[i] ≤ 10⁵

Approach — Two Pointers

This is a Two Pointers problem. The idea: walk two indices through the data together (or toward each other), turning an O(n²) pair search into one linear pass. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(n) time, O(1) extra space.

Solution code

Python

class Solution:
    def threeSum(self, nums):
        nums.sort()
        res = []
        n = len(nums)
        for i in range(n):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            l, r = i + 1, n - 1
            while l < r:
                s = nums[i] + nums[l] + nums[r]
                if s < 0:
                    l += 1
                elif s > 0:
                    r -= 1
                else:
                    res.append([nums[i], nums[l], nums[r]])
                    l += 1; r -= 1
                    while l < r and nums[l] == nums[l-1]:
                        l += 1
                    while l < r and nums[r] == nums[r+1]:
                        r -= 1
        return res

Java

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate anchors so the same triplet isn't recorded twice.
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int left = i + 1;
            int right = nums.length - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum == 0) {
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    left++;
                    right--;
                } else if (sum < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return result;
    }
}

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.