🟨 3Sum (#15)

🔗 Problem Link

• LeetCode - 3Sum
• NeetCode - 3Sum

📋 Problem Statement

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.

💡 Examples

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.

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]]

🔑 Key Insights & Approach

Core Observation: Sort array, fix one number, use two pointers for remaining two numbers to find sum of 0.

Why Sort + Two Pointers?

• O(n²) time complexity vs O(n³) brute force
• Easy duplicate handling with sorted array
• Can skip duplicates efficiently

Approaches:

1. Brute force three loops: O(n³)
2. Hash map: O(n²) but complex duplicate handling
3. Sort + Two pointers: O(n²) - optimal

Pattern: "Two Pointers - Fixed + Moving" pattern for sum problems.

🐍 Solution: Python

Approach: Sort + Two Pointers

Time Complexity: O(n²) | Space Complexity: O(1) or O(n) depending on sort

from typing import List

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        result = []
        n = len(nums)

        for i in range(n - 2):
            # Skip duplicates for first number
            if i > 0 and nums[i] == nums[i-1]:
                continue

            # Two pointers for remaining two numbers
            left, right = i + 1, n - 1
            target = -nums[i]

            while left < right:
                current_sum = nums[left] + nums[right]

                if current_sum == target:
                    result.append([nums[i], nums[left], nums[right]])

                    # Skip duplicates for second number
                    while left < right and nums[left] == nums[left+1]:
                        left += 1
                    # Skip duplicates for third number
                    while left < right and nums[right] == nums[right-1]:
                        right -= 1

                    left += 1
                    right -= 1
                elif current_sum < target:
                    left += 1
                else:
                    right -= 1

        return result

🔵 Solution: Golang

Approach: Sort + Two Pointers

Time Complexity: O(n²) | Space Complexity: O(1)

import "sort"

func threeSum(nums []int) [][]int {
    sort.Ints(nums)
    result := [][]int{}
    n := len(nums)

    for i := 0; i < n-2; i++ {
        // Skip duplicates
        if i > 0 && nums[i] == nums[i-1] {
            continue
        }

        left, right := i+1, n-1
        target := -nums[i]

        for left < right {
            currentSum := nums[left] + nums[right]

            if currentSum == target {
                result = append(result, []int{nums[i], nums[left], nums[right]})

                // Skip duplicates
                for left < right && nums[left] == nums[left+1] {
                    left++
                }
                for left < right && nums[right] == nums[right-1] {
                    right--
                }

                left++
                right--
            } else if currentSum < target {
                left++
            } else {
                right--
            }
        }
    }

    return result
}

📚 References

• NeetCode - 3Sum