Find All Unique Triplets That Sum to Zero Optimal Solution using Two Pointers

Problem Statement❯

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.

Examples❯

Input Array	Output Triplets	Description
[-1, 0, 1, 2, -1, -4]	[[-1, -1, 2], [-1, 0, 1]]	Two unique triplets sum up to 0
[0, 1, 1]	[]	No triplet sums to zero
[0, 0, 0]	[[0, 0, 0]]	Three zeroes sum to zero, one valid triplet
[3, -2, 1, 0]	[[-2, -1, 3]]	Triplet [-2, -1, 3] gives zero
[1, -1]	[]	Only two elements, not enough for a triplet
[]	[]	Empty input, return empty list
[-2, 0, 1, 1, 2]	[[-2, 0, 2], [-2, 1, 1]]	Two triplets, handles duplicates correctly

Visualization Player

Solution❯

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.

Algorithm Steps❯

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

Code

C++

Python

Java

Rust

Kotlin

Swift

#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;
}

Time Complexity❯

Case	Time Complexity	Explanation
Best Case	O(n^2)	In the best case, the input is small or contains many duplicate values that allow early skips, but the dominant term is still the nested loop structure with two pointers.
Average Case	O(n^2)	We sort the array in O(n log n) and then use a loop with a two-pointer technique inside, which takes O(n^2) in total.
Worst Case	O(n^2)	Even in the worst case with no duplicate skipping and all combinations needing to be checked, the time complexity remains O(n^2) due to the nested traversal using two pointers.

Space Complexity❯

O(1)

Explanation: 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.

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