Monotonic Envelope Protocol 2 — Problem Statement & Solution Guide

Two PointersHard3Sum Zero Target

TimeO(n^2)

SpaceO(1) additional (ignoring output storage)

Problem Description

You are tasked with analyzing a sequence of integer values representing system load metrics. The objective is to identify all unique triplets of indices (i, j, k) such that i < j < k and the sum of the corresponding values equals exactly zero. This problem requires an efficient approach to avoid redundant computations and ensure that no duplicate triplets are reported in the final output.

Given an array of integers, return a list of all unique triplets [nums[i], nums[j], nums[k]] that sum to zero. The solution must handle large input sizes efficiently, leveraging sorting and two-pointer techniques to achieve optimal time complexity. The output should be sorted in lexicographical order to ensure consistency and ease of verification.

The challenge lies in managing the constraints of large datasets while maintaining precision in identifying valid triplets. You must ensure that the algorithm avoids O(n^3) brute-force approaches and instead utilizes a more sophisticated strategy to meet the performance requirements.

DSA Pattern Breakdown

"Monotonic Envelope Protocol 2"

hard

WHY DOES IT MATTER?

The two‑pointer pattern on a sorted array converts a combinatorial explosion into a linear scan, a cornerstone technique for many sum‑related problems. Mastery of this pattern unlocks efficient solutions for 2‑Sum, 3‑Sum, and k‑Sum variants, which appear frequently in coding interviews and real‑world data processing.

OPTIMIZATION CHALLENGE

The key insight is that after sorting, the relative order of elements guarantees that moving the left pointer increases the sum and moving the right pointer decreases it. This monotonicity eliminates the need for nested loops over the inner pair, collapsing O(n³) to O(n²).

REAL-WORLD CONNECTION

Think of a load balancer that must pair high‑load and low‑load servers to achieve a target capacity. By sorting server loads, the balancer can pair the heaviest with the lightest, adjusting pointers until the combined load meets the target, mirroring the two‑pointer adjustment in 3‑Sum.

During an interview, fix the first element and immediately skip duplicates before entering the two‑pointer loop; this prevents subtle bugs where the same triplet is reported multiple times and keeps the code clean.

COMPLEXITY AT A GLANCE

⏱ Time:O(n^2)

💾 Space:O(1) additional (ignoring output storage)

Core Theory — Why This Approach?

The classic 3‑Sum problem asks for all unique index triples (i, j, k) with i < j < k such that nums[i] + nums[j] + nums[k] = 0. A naïve solution enumerates every combination, leading to O(n³) time, which quickly becomes infeasible for n > 10⁴. The breakthrough comes from sorting the array first, which imposes a monotonic order and enables the two‑pointer technique: for each fixed element, we slide two pointers from the ends of the remaining sub‑array toward each other, adjusting based on the current sum. This reduces the inner search to linear time, yielding an overall O(n²) algorithm while also making duplicate elimination straightforward by skipping equal values during iteration.

Sorting also transforms the problem into a structured search space where the sum of the leftmost and rightmost elements provides deterministic guidance: if the sum is too low we must increase the left pointer, if too high we decrease the right pointer. This deterministic movement eliminates the exponential blow‑up of the brute‑force approach and guarantees that each potential triplet is examined at most once. The final solution therefore balances time efficiency with simplicity, making it the optimal paradigm for large inputs.

Interview Questions on This Problem

Q1How would you modify the 3‑Sum solution to return the count of unique triplets instead of the triplets themselves, and what is the impact on space complexity?

After sorting, use the same two‑pointer loop but increment a counter each time a valid triplet is found, still skipping duplicates. Since we no longer store the triplets, the auxiliary space drops to O(1) beyond the input array.

Q2Explain how you would adapt the algorithm to handle the variant where the target sum is an arbitrary integer T instead of zero.

Replace the zero check with (sum == T) inside the two‑pointer loop. The rest of the logic—sorting, duplicate skipping, pointer movement—remains unchanged, preserving O(n²) time and O(1) extra space.

Q3In a distributed system processing massive streams of numbers, how could you approximate the 3‑Sum count using limited memory?

One approach is to maintain a bounded hash‑based sketch of value frequencies and sample windows of the stream, then apply a probabilistic two‑pointer scan on the sampled data. This trades exactness for O(1) memory per node while providing an estimate with bounded error.

Examples

Example 1

Input

nums = [-1, 0, 1, 2, -1, -4]

Output

[[-1, -1, 2], [-1, 0, 1]]

Explanation: First, sort the array to [-4, -1, -1, 0, 1, 2]. Iterate through each element as the first element of the triplet. For -4, no valid triplet is found. For the first -1, use two pointers to find pairs that sum to 1, yielding [-1, 0, 1]. For the second -1, skip it to avoid duplicates. For 0, no valid triplet is found. The final output is [[-1, -1, 2], [-1, 0, 1]].

Example 2

Input

nums = [0, 0, 0, 0]

Output

[[0, 0, 0]]

Explanation: Sort the array to [0, 0, 0, 0]. Iterate through each element. For the first 0, use two pointers to find pairs that sum to 0, yielding [0, 0, 0]. Skip subsequent 0s to avoid duplicates. The final output is [[0, 0, 0]].

Example 3

Input

nums = [1, 2, -3, 4, -5, 6]

Output

[[-5, 1, 4], [-5, 2, 3], [-3, 1, 2]]

Explanation: Sort the array to [-5, -3, 1, 2, 4, 6]. Iterate through each element. For -5, find pairs that sum to 5, yielding [-5, 1, 4] and [-5, 2, 3]. For -3, find pairs that sum to 3, yielding [-3, 1, 2]. The final output is [[-5, 1, 4], [-5, 2, 3], [-3, 1, 2]].

Constraints

3 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
The answer must be sorted in lexicographical order.
All triplets must be unique.

Optimal Approach & Strategy

Sort the array, fix i, then apply a two‑pointer scan on the suffix to find pairs that sum to -nums[i], while skipping duplicates to ensure uniqueness.

Brute Force Approach

Enumerate every i, j, k triple with three nested loops and check if their sum is zero; record unique triplets using a set.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n^2)

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(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 sum = nums[i] + nums[left] + nums[right];
            if (sum === 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 (sum < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
};

class Solution {
public:
    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 sum = nums[i] + nums[left] + nums[right];
                if (sum == 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 (sum < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return result;
    }
};

class Solution {
    public 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 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;
    }
}

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        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

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(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 sum = nums[i] + nums[left] + nums[right];
            if (sum === 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 (sum < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
};

Asked in Top Tech Interviews

MetaAmazon

Solve in Interative Editor

Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.

Monotonic Envelope Protocol 2 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Monotonic Envelope Protocol 2 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor