Three Sum Problem

Given an array of integers, your task is to find all the unique triplets (a, b, c) in the array such that:

• a + b + c = 0 (the sum of the three elements is zero)
• Each triplet should be returned in sorted order and must be unique (i.e., no duplicate triplets in the result)

If no such triplets exist, return an empty list.

Understanding the Problem

The Three Sum problem asks us to find all unique triplets in an array such that the sum of the three numbers is zero. In other words, we need to find all combinations of three numbers (a, b, c) from the array where a + b + c = 0.

It’s important to avoid returning duplicate triplets. For example, [-1, 0, 1] and [0, -1, 1] are considered the same triplet and should appear only once in the output.

We are going to solve this step-by-step, with a beginner-friendly explanation, and handle all edge cases carefully.

Step-by-Step Solution Using an Example

Let’s take the example array: [-1, 0, 1, 2, -1, -4]

Step 1: Sort the Array

We start by sorting the array. This helps in skipping duplicates and using the two-pointer method effectively.

Sorted array: [-4, -1, -1, 0, 1, 2]

Step 2: Fix One Element and Use Two Pointers

We loop through each element (let’s call the index i) and treat it as the first element of a potential triplet.

Then, we use two pointers: left starting from i + 1 and right starting from the end of the array.

We calculate the sum of the elements at positions i, left, and right and act based on the result:

• If the sum is zero: we’ve found a valid triplet. Add it to the result.
• If the sum is less than zero: we need a bigger number, so we move the left pointer forward.
• If the sum is more than zero: we need a smaller number, so we move the right pointer backward.

Step 3: Skip Duplicates

To avoid duplicate triplets:

• Skip repeated values at the i position.
• After finding a valid triplet, skip repeated values at left and right positions.

Example Walkthrough

Let’s go through the sorted array step by step:

• i = 0: value = -4, left = 1 (-1), right = 5 (2) → sum = -3 → move left
• left = 2 (-1), right = 5 (2) → sum = -3 → move left
• left = 3 (0), right = 5 (2) → sum = -2 → move left
• left = 4 (1), right = 5 (2) → sum = -1 → move left
• i = 1: value = -1, left = 2 (-1), right = 5 (2) → sum = 0 → valid triplet [-1, -1, 2]
• Skip duplicate -1 at left, move left → left = 3 (0), right = 4 (1) → sum = 0 → valid triplet [-1, 0, 1]
• i = 2: value = -1 (duplicate), skip
• i = 3: value = 0, left = 4 (1), right = 5 (2) → sum = 3 → move right

Final result: [[-1, -1, 2], [-1, 0, 1]]

Handling Edge Cases

• Less than 3 elements: No triplets possible, return an empty list.
• All zeros: If the array is like [0, 0, 0, 0], return [[0, 0, 0]] only once.
• Empty array: Return an empty list.
• Duplicates: Skip over duplicate values during iteration to prevent redundant triplets.

Why This Approach Works

This method uses sorting + two pointers and avoids brute-force triplet checking. The time complexity is O(n²) compared to O(n³) in brute-force, making it efficient for larger inputs.

1. Given an array arr of integers.
2. Sort the array in ascending order.
3. Traverse the array with index i from 0 to n-2:
4. → If i > 0 and arr[i] == arr[i-1], skip to avoid duplicates.
5. → Initialize two pointers: left = i + 1 and right = n - 1.
6. → While left < right:
7. → → Calculate the sum: arr[i] + arr[left] + arr[right].
8. → → If sum is zero, add triplet to result and move left and right to skip duplicates.
9. → → If sum < 0, move left forward.
10. → → If sum > 0, move right backward.
11. Return the list of unique triplets.

#include <stdio.h>
#include <stdlib.h>

int cmp(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

void threeSum(int* nums, int n) {
    qsort(nums, n, sizeof(int), cmp);
    for (int i = 0; i < n - 2; i++) {
        if (i > 0 && nums[i] == nums[i-1]) continue;
        int left = i + 1, right = n - 1;
        while (left < right) {
            int total = nums[i] + nums[left] + nums[right];
            if (total == 0) {
                printf("[%d, %d, %d]\n", 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 (total < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
}

int main() {
    int arr[] = {-1, 0, 1, 2, -1, -4};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Triplets:\n");
    threeSum(arr, n);
    return 0;
}

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<vector<int>> threeSum(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    vector<vector<int>> result;
    int n = nums.size();

    for (int i = 0; i < n - 2; i++) {
        if (i > 0 && nums[i] == nums[i-1]) continue;
        int left = i + 1, right = n - 1;
        while (left < right) {
            int total = nums[i] + nums[left] + nums[right];
            if (total == 0) {
                result.push_back({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 (total < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
}

int main() {
    vector<int> arr = {-1, 0, 1, 2, -1, -4};
    vector<vector<int>> res = threeSum(arr);
    cout << "Triplets:\n";
    for (auto triplet : res) {
        cout << "[";
        for (auto num : triplet) {
            cout << num << " ";
        }
        cout << "]\n";
    }
    return 0;
}

def three_sum(nums):
    nums.sort()
    result = []
    n = len(nums)

    for i in range(n - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue
        left, right = i + 1, n - 1
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            if total == 0:
                result.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left + 1]:
                    left += 1
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1
                left += 1
                right -= 1
            elif total < 0:
                left += 1
            else:
                right -= 1
    return result

arr = [-1, 0, 1, 2, -1, -4]
print("Triplets:", three_sum(arr))

import java.util.*;

public class ThreeSum {
    public static List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        int n = nums.length;

        for (int i = 0; i < n - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int left = i + 1, right = n - 1;
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                if (total == 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 (total < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] arr = {-1, 0, 1, 2, -1, -4};
        System.out.println("Triplets: " + threeSum(arr));
    }
}

function threeSum(nums) {
    nums.sort((a, b) => a - b);
    const result = [];
    const n = nums.length;

    for (let i = 0; i < n - 2; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) continue;
        let left = i + 1, right = n - 1;
        while (left < right) {
            const total = nums[i] + nums[left] + nums[right];
            if (total === 0) {
                result.push([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 (total < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
}

const arr = [-1, 0, 1, 2, -1, -4];
console.log("Triplets:", threeSum(arr));

package main

import (
    "fmt"
    "sort"
)

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

    for i := 0; i < n-2; i++ {
        if i > 0 && nums[i] == nums[i-1] {
            continue
        }
        left, right := i+1, n-1
        for left < right {
            total := nums[i] + nums[left] + nums[right]
            if total == 0 {
                result = append(result, []int{nums[i], nums[left], nums[right]})
                for left < right && nums[left] == nums[left+1] {
                    left++
                }
                for left < right && nums[right] == nums[right-1] {
                    right--
                }
                left++
                right--
            } else if total < 0 {
                left++
            } else {
                right--
            }
        }
    }
    return result
}

func main() {
    arr := []int{-1, 0, 1, 2, -1, -4}
    fmt.Println("Triplets:", threeSum(arr))
}

fn three_sum(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
    nums.sort();
    let n = nums.len();
    let mut result = Vec::new();

    for i in 0..n-2 {
        if i > 0 && nums[i] == nums[i - 1] {
            continue;
        }
        let (mut left, mut right) = (i + 1, n - 1);
        while left < right {
            let total = nums[i] + nums[left] + nums[right];
            if total == 0 {
                result.push(vec![nums[i], nums[left], nums[right]]);
                while left < right && nums[left] == nums[left + 1] {
                    left += 1;
                }
                while left < right && nums[right] == nums[right - 1] {
                    right -= 1;
                }
                left += 1;
                right -= 1;
            } else if total < 0 {
                left += 1;
            } else {
                right -= 1;
            }
        }
    }
    result
}

fn main() {
    let arr = vec![-1, 0, 1, 2, -1, -4];
    println!("Triplets: {:?}", three_sum(arr));
}

fun threeSum(nums: IntArray): List<List<Int>> {
    nums.sort()
    val result = mutableListOf<List<Int>>()
    val n = nums.size

    for (i in 0 until n - 2) {
        if (i > 0 && nums[i] == nums[i - 1]) continue
        var left = i + 1
        var right = n - 1
        while (left < right) {
            val total = nums[i] + nums[left] + nums[right]
            when {
                total == 0 -> {
                    result.add(listOf(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--
                }
                total < 0 -> left++
                else -> right--
            }
        }
    }
    return result
}

fun main() {
    val arr = intArrayOf(-1, 0, 1, 2, -1, -4)
    println("Triplets: ${threeSum(arr)}")
}

import Foundation

func threeSum(_ nums: [Int]) -> [[Int]] {
    let nums = nums.sorted()
    var result: [[Int]] = []
    let n = nums.count

    for i in 0..<n-2 {
        if i > 0 && nums[i] == nums[i - 1] { continue }
        var left = i + 1, right = n - 1
        while left < right {
            let total = nums[i] + nums[left] + nums[right]
            if total == 0 {
                result.append([nums[i], nums[left], nums[right]])
                while left < right && nums[left] == nums[left + 1] { left += 1 }
                while left < right && nums[right] == nums[right - 1] { right -= 1 }
                left += 1
                right -= 1
            } else if total < 0 {
                left += 1
            } else {
                right -= 1
            }
        }
    }
    return result
}

let arr = [-1, 0, 1, 2, -1, -4]
print("Triplets:", threeSum(arr))

function threeSum(nums: number[]): number[][] {
    nums.sort((a, b) => a - b);
    const result: number[][] = [];
    const n = nums.length;

    for (let i = 0; i < n - 2; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) continue;
        let left = i + 1, right = n - 1;
        while (left < right) {
            const total = nums[i] + nums[left] + nums[right];
            if (total === 0) {
                result.push([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 (total < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
}

const arr: number[] = [-1, 0, 1, 2, -1, -4];
console.log("Triplets:", threeSum(arr));

Time complexity

Space complexity

O(1)

The algorithm uses constant extra space apart from the output list. Sorting is done in-place and only a few variables (i, left, right) are used.