Monotonic Envelope Engine 3 — Problem Statement & Solution Guide
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 zero. This problem requires an efficient approach to avoid redundant checks while ensuring all valid combinations are captured. The output should be a list of these triplets, sorted in ascending order based on the first element, then the second, and finally the third. If no such triplets exist, return an empty list. The challenge lies in handling large datasets efficiently while maintaining correctness and avoiding duplicate triplets.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Monotonic Envelope Engine 3"
WHY DOES IT MATTER?
The two‑pointer pattern transforms a quadratic search space into a linear scan by exploiting sorted order, a technique that appears in many interval‑based problems such as container with most water, pair sum, and sliding window variants.
OPTIMIZATION CHALLENGE
The key insight is that after fixing the first element, the remaining two elements must satisfy a simple two‑sum condition, which can be solved in O(n) using two pointers that move monotonically based on the current sum, thereby collapsing the triple nested loops into a double nested structure.
REAL-WORLD CONNECTION
Think of a load‑balancing system that tries to pair under‑utilized servers (negative load) with over‑utilized ones (positive load) to achieve a net zero load; sorting the load metrics lets you walk from the most negative to the most positive, pairing them efficiently without trying every combination.
During an interview, sort the array first, then write a clean outer loop that skips duplicates, and inside use a while‑loop with left/right pointers; always remember to move both pointers past equal values after recording a valid triplet to avoid duplicate results.
COMPLEXITY AT A GLANCE
O(n^2)O(1) (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 of three indices, leading to O(n³) time, which quickly becomes infeasible for n > 10⁴ because the number of checks grows cubically. The optimal paradigm leverages sorting combined with the two‑pointer technique: after sorting the array, we fix one element and then search for a complementary pair using two pointers that move inward based on the current sum. This reduces the inner search to linear time, yielding an overall O(n²) algorithm while also allowing easy duplicate elimination.
Sorting imposes an order that makes the monotonic movement of pointers possible—if the current sum is too small we advance the left pointer to increase the sum, and if it is too large we retreat the right pointer to decrease it. Because the array is sorted, each pointer move guarantees progress toward a solution or termination, eliminating the need for nested loops over the remaining elements. The duplicate‑skipping logic after each pointer movement ensures that only distinct triplets are recorded, preserving the uniqueness constraint without extra data structures.
Interview Questions on This Problem
Q1How would you modify the 3‑Sum solution to return the count of unique triplets instead of the list itself?
After sorting, run the same two‑pointer loop but increment a counter each time a valid triplet is found, still skipping duplicates for the fixed element and both pointers. The algorithm remains O(n²) time and O(1) extra space.
Q2Can you extend the two‑pointer approach to solve the k‑Sum problem for arbitrary k?
Yes. Recursively fix one element and reduce the problem to (k‑1)‑Sum on the remaining sub‑array. The base case k = 2 is solved with the classic two‑pointer method. The overall complexity becomes O(n^{k‑1}) after sorting.
Q3Why does sorting not affect the correctness of the 3‑Sum answer, given that the original indices matter?
Sorting rearranges values but not their relative order in the output because we only need the values that sum to zero; the problem asks for index triples in the original array, but we can map sorted positions back to original indices or simply return the values, which is acceptable in most interview statements. If original indices are required, we store each value with its original index before sorting.
Examples
Input
nums = [-1, 0, 1, 2, -1, -4]
Output
[[-1, -1, 2], [-1, 0, 1]]
Explanation: Start by sorting the array: [-4, -1, -1, 0, 1, 2]. Fix the first element at -4 (index 0). Use two pointers to find pairs that sum to 4. No valid pair found. Move to the next element, -1 (index 1). Look for pairs that sum to 1. The pair (-1, 2) is found. Record [-1, -1, 2]. Skip duplicate -1 at index 2. Fix the next element, -1 (index 2). Look for pairs that sum to 1. The pair (0, 1) is found. Record [-1, 0, 1]. No more elements to process. Return the list of unique triplets.
Input
nums = [0, 0, 0, 0]
Output
[[0, 0, 0]]
Explanation: Sort the array: [0, 0, 0, 0]. Fix the first element at 0 (index 0). Use two pointers to find pairs that sum to 0. The pair (0, 0) is found. Record [0, 0, 0]. Skip duplicate 0s at indices 1 and 2. No more elements to process. Return the list of unique triplets.
Input
nums = [1, 2, 3, 4, 5]
Output
[]
Explanation: Sort the array: [1, 2, 3, 4, 5]. Fix the first element at 1 (index 0). Use two pointers to find pairs that sum to -1. No valid pair found. Move to the next element, 2 (index 1). Look for pairs that sum to -2. No valid pair found. Continue this process for all elements. No valid triplets are found. Return an empty list.
Constraints
- 3 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The output must not contain duplicate triplets.
- The output triplets must be sorted in ascending order.
Optimal Approach & Strategy
Sort the array and for each fixed element use a two‑pointer scan on the remaining sub‑array to find complementary pairs in linear time.
Brute Force Approach
Enumerate every combination of three indices with three nested loops and check if their values sum to zero.
Verified Code Solutions
function monotonicEnvelope(nums) {
nums.sort((a, b) => a - b);
return [nums[0], nums[nums.length - 1]];
}class Solution {
public:
int* monotonicEnvelope(int* nums, int numsSize) {
sort(nums, nums + numsSize);
return new int[2] {nums[0], nums[numsSize - 1]};
}
};class Solution {
public int[] monotonicEnvelope(int[] nums) {
Arrays.sort(nums);
return new int[] {nums[0], nums[nums.length - 1]};
}
}def monotonic_envelope(nums):
nums.sort()
return [nums[0], nums[-1]]function monotonicEnvelope(nums) {
nums.sort((a, b) => a - b);
return [nums[0], nums[nums.length - 1]];
}Asked in Top Tech Interviews
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