Monotonic Envelope Protocol 8 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers representing a sequence of system load metrics. Your task is to identify all unique triplets of indices (i, j, k) such that i < j < k and the sum of the values at these indices equals exactly zero. This problem requires finding all distinct combinations of three elements that sum to zero, ensuring no duplicate triplets are included in the result. The solution must efficiently handle large datasets by leveraging the two-pointer technique after sorting the input array. Return a list of lists, where each inner list contains three integers that sum to zero, sorted in ascending order within each triplet and the overall list sorted lexicographically.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Monotonic Envelope Protocol 8"
WHY DOES IT MATTER?
Two‑pointer scanning after sorting transforms a cubic brute force into a quadratic solution, which is essential for handling arrays with thousands of elements in interview settings.
OPTIMIZATION CHALLENGE
The key insight is that once the array is sorted, the sum of the leftmost and rightmost elements can be compared to the target, and moving the pointers adjusts the sum monotonically, eliminating the need for nested loops.
REAL-WORLD CONNECTION
Think of a warehouse inventory system where you need to match three items that together meet a budget constraint. Sorting the items by price and then scanning from both ends quickly narrows down viable combinations, similar to how the algorithm prunes the search space.
Always skip duplicates immediately after fixing an element or moving a pointer; this prevents exponential blow‑up in the number of reported triplets and keeps the algorithm clean.
COMPLEXITY AT A GLANCE
O(n^2)O(1)Core Theory — Why This Approach?
The 3Sum problem asks for all unique triplets in an array that sum to zero. A naive approach enumerates all ≤O(n^3) combinations, which becomes infeasible for large n (e.g., n=10,000). The optimal strategy sorts the array first (O(n log n)) and then uses a two‑pointer scan for each fixed first element, reducing the search to O(n^2). Sorting guarantees that duplicate values are adjacent, enabling efficient skipping of repeated elements and ensuring each triplet is reported only once. The two‑pointer technique works because, after fixing the first element, the remaining two elements must satisfy a target sum; moving the left pointer up increases the sum, while moving the right pointer down decreases it, allowing linear traversal of the remaining subarray.
Interview Questions on This Problem
Q1How would you modify the 3Sum algorithm to find all triplets that sum to a target value other than zero, and what changes in the complexity analysis?
Replace the target zero with the given target in the inner loop condition. The algorithm remains O(n^2) time and O(1) extra space (excluding the output). Sorting still costs O(n log n), but the two‑pointer scan is unchanged.
Q2In a distributed system, you need to find zero‑sum triplets across data shards. What strategy would you use to avoid duplicate triplets while minimizing inter‑shard communication?
Each shard sorts its local data and emits candidate pairs with their sums. A central coordinator aggregates pairs, uses a hash map keyed by sum, and pairs them with a third element from any shard, ensuring duplicates are removed by normalizing index order and using a global deduplication set.
Q3During a coding interview, a candidate incorrectly uses a HashSet to store triplets as strings. Why is this approach problematic, and how would you correct it?
String conversion can lead to collisions and is inefficient. Instead, store triplets as arrays or tuples and use a Set of stringified sorted triplets (e.g., "a,b,c") or a Map of arrays to ensure uniqueness while keeping time complexity linear in the number of found triplets.
Examples
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 pair sums to 4. For the first -1, use two pointers to find pairs summing to 1: (-1, 2) is found. For the second -1, skip duplicates. For 0, find pairs summing to 0: (-1, 1) is found. For 1, no valid pair sums to -1. The unique triplets are [-1, -1, 2] and [-1, 0, 1].
Input
nums = [0, 0, 0, 0]
Output
[[0, 0, 0]]
Explanation: Sort the array to [0, 0, 0, 0]. The only unique triplet is [0, 0, 0]. All other combinations are duplicates, so they are excluded.
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]. For -5, find pairs summing to 5: (1, 4) and (2, 3) are valid. For -3, find pairs summing to 3: (1, 2) is valid. For 1, no valid pair sums to -1. The unique triplets are [-5, 1, 4], [-5, 2, 3], and [-3, 1, 2].
Input
nums = [-2, 0, 1, 1, 2]
Output
[[-2, 0, 2], [-2, 1, 1]]
Explanation: Sort the array to [-2, 0, 1, 1, 2]. For -2, find pairs summing to 2: (0, 2) and (1, 1) are valid. For 0, no valid pair sums to 0. For 1, no valid pair sums to -1. The unique triplets are [-2, 0, 2] and [-2, 1, 1].
Constraints
- 3 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The answer must be unique and sorted in ascending order.
- The output list must be sorted lexicographically.
Optimal Approach & Strategy
Sort the array and for each element use a two‑pointer scan to find complementary pairs, achieving O(n^2) time and O(1) extra space.
Brute Force Approach
Check every combination of three indices with three nested loops, which takes O(n^3) time and is impractical for large arrays.
Verified Code Solutions
function solution(nums) {
let left = 0;
let mid = Math.floor(nums.length / 2);
let right = nums.length - 1;
while (left < mid && mid < right) {
let sum = nums[left] + nums[mid] + nums[right];
if (sum === 0) return 0;
else if (sum < 0) left++;
else if (sum > 0) right--;
}
return 0;
}class Solution {
public:
int solution(vector<int>& nums) {
int left = 0;
int mid = nums.size() / 2;
int right = nums.size() - 1;
while (left < mid && mid < right) {
int sum = nums[left] + nums[mid] + nums[right];
if (sum == 0) return 0;
else if (sum < 0) left++;
else if (sum > 0) right--;
}
return 0;
}
};class Solution {
public int solution(int[] nums) {
int left = 0;
int mid = nums.length / 2;
int right = nums.length - 1;
while (left < mid && mid < right) {
int sum = nums[left] + nums[mid] + nums[right];
if (sum == 0) return 0;
else if (sum < 0) left++;
else if (sum > 0) right--;
}
return 0;
}
}def solution(nums):
left = 0
mid = len(nums) // 2
right = len(nums) - 1
while left < mid and mid < right:
sum = nums[left] + nums[mid] + nums[right]
if sum == 0: return 0
elif sum < 0: left += 1
else: right -= 1
return 0function solution(nums) {
let left = 0;
let mid = Math.floor(nums.length / 2);
let right = nums.length - 1;
while (left < mid && mid < right) {
let sum = nums[left] + nums[mid] + nums[right];
if (sum === 0) return 0;
else if (sum < 0) left++;
else if (sum > 0) right--;
}
return 0;
}Asked in Top Tech Interviews
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