Problem #15: 3Sum

Intuition & Strategy Breakdown

Sort the array and iterate through each number as a fixed target. Use two pointers (left and right) to find pairs that sum to -target, skipping duplicates to avoid repeated triplets.

Python Solution

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

Java Solution

class Solution {
    public List> threeSum(int[] nums) {
        Arrays.sort(nums);
        List> res = new ArrayList<>();
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int l = i + 1, r = nums.length - 1;
            while (l < r) {
                int sum = nums[i] + nums[l] + nums[r];
                if (sum < 0) l++;
                else if (sum > 0) r--;
                else {
                    res.add(Arrays.asList(nums[i], nums[l], nums[r]));
                    l++;
                    while (l < r && nums[l] == nums[l - 1]) l++;
                }
            }
        }
        return res;
    }
}