Optimal Grid Path Protocol 8 — Problem Statement & Solution Guide
Problem Description
Given an integer array nums, find every distinct triplet of indices (i, j, k) with i < j < k such that nums[i] + nums[j] + nums[k] equals zero. Each triplet should be reported as a list of its three values sorted in non‑decreasing order, and the overall result must not contain duplicate triplets. The order of the triplets in the output is irrelevant. The array may contain repeated values, but each unique combination of values that sums to zero should appear only once in the answer. A typical solution sorts the array and then uses a two‑pointer technique to achieve an O(n²) time complexity.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Grid Path Protocol 8"
WHY DOES IT MATTER?
The two-pointer pattern transforms a cubic search into a quadratic one by exploiting sorted order, which is critical for large datasets where O(n^3) would be infeasible. It also naturally handles duplicate avoidance without extra data structures, keeping space usage minimal.
OPTIMIZATION CHALLENGE
The central optimization is recognizing that once the array is sorted, the sum of two moving pointers can be compared to a target in constant time, allowing the inner loop to run in linear time per outer iteration. This reduces the overall complexity from O(n^3) to O(n^2).
REAL-WORLD CONNECTION
Think of a warehouse inventory system where items are sorted by weight. To find three items that balance a shipment to zero net weight, you can pick the lightest item and then use a balancing scale (two pointers) to find the other two that together counterbalance it, rather than checking every possible combination of items.
When explaining this in an interview, emphasize the importance of early duplicate checks for the fixed element and the pointers, and demonstrate how skipping duplicates keeps the algorithm efficient and the output clean.
COMPLEXITY AT A GLANCE
O(n^2)O(1)Core Theory — Why This Approach?
The optimal solution for the three-sum problem relies on sorting the input array and then applying a two-pointer technique for each fixed element. By sorting, we can guarantee that any two numbers that sum to a target will appear in a predictable order, allowing us to move pointers inward or outward based on the current sum. This reduces the naive O(n^3) brute-force enumeration of all triplets to an O(n^2) algorithm, which is essential for handling arrays with thousands of elements within practical time limits.
The key insight is that after sorting, for a chosen first element nums[i], the remaining two elements must satisfy nums[j] + nums[k] = -nums[i]. Because the subarray nums[i+1..n-1] is sorted, we can start with j = i+1 and k = n-1 and adjust them in a single pass: if the sum is too small, increment j; if too large, decrement k; if it matches, record the triplet and skip duplicates. This single-pass scan for each i yields a quadratic time complexity while maintaining linear additional space (excluding the output).
Interview Questions on This Problem
Q1How does the two-pointer approach guarantee that all unique triplets are found without missing any combinations?
After sorting, fixing the first element and scanning the remaining subarray with two pointers ensures that every pair that sums to the required value is examined exactly once. Skipping over duplicate values for both the fixed element and the pointers prevents repeated triplets, guaranteeing completeness and uniqueness.
Q2What would be the impact on time complexity if we used a hash set to store pairs instead of sorting?
Using a hash set to store pairs would still require O(n^2) time to generate all pairs, but it would add O(n^2) space overhead and lose the ability to skip duplicates efficiently. Sorting provides O(n log n) preprocessing and O(1) extra space for the two-pointer scan, making it more space-efficient.
Q3In a distributed system, how could the three-sum problem be parallelized while preserving the two-pointer logic?
One can partition the sorted array into segments, assign each segment to a worker to fix the first element, and perform the two-pointer scan locally. After local results are collected, a merge step removes cross-segment duplicates. This approach keeps the core two-pointer logic intact while leveraging parallelism for the outer loop.
Examples
Input
[ -1, 0, 1, 2, -1, -4 ]
Output
[ [ -1, -1, 2 ], [ -1, 0, 1 ] ]
Explanation: Sorting gives [-4, -1, -1, 0, 1, 2]. For each fixed left element, a right pointer starts at the end. The pairs that sum to the negative of the left element are found: (-1, 2) and (0, 1). These are the only unique triplets that sum to zero.
Input
[ 0, 0, 0, 0 ]
Output
[ [ 0, 0, 0 ] ]
Explanation: All elements are zero. The only triplet that sums to zero is (0, 0, 0). Duplicates are removed, so the result contains a single triplet.
Input
[ 3, -2, 1, 0, -1, 2, -3, 4 ]
Output
[ [ -3, -1, 4 ], [ -3, 0, 3 ], [ -3, 1, 2 ], [ -2, -1, 3 ], [ -2, 0, 2 ], [ -1, 0, 1 ] ]
Explanation: After sorting: [-3, -2, -1, 0, 1, 2, 3, 4]. Using two pointers for each left element yields the six unique combinations that sum to zero, as listed.
Input
[ 1, 2, 3, 4, 5 ]
Output
[]
Explanation: No three numbers in the array can sum to zero, so the output is an empty list.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The number of distinct triplets returned will not exceed 10000
- The algorithm must run in O(n^2) time and O(1) additional space (excluding the output list)
- The input array may contain duplicate values
Optimal Approach & Strategy
Sort the array and for each index i, use two pointers j and k to find pairs summing to -nums[i] in linear time. Skip duplicates for i, j, and k to ensure unique triplets, achieving O(n^2) time and O(1) extra space.
Brute Force Approach
Check every possible combination of three indices using three nested loops, which takes O(n^3) time. Store each found triplet in a set to avoid duplicates, but this approach is too slow for large inputs.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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