Triple Sum Zero — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Two Pointers and solve the Triple Sum Zero problem optimally.
O(n²)O(1)Problem Description
Given an integer array nums, determine how many distinct unordered triplets (i, j, k) with i<j<k exist such that nums[i]+nums[j]+nums[k]=0. Two triplets are considered the same if they contain the same three values, regardless of their positions in the array. Return the count of such unique triplets. The algorithm must run in O(n²) time or better and use only O(1) additional memory apart from the input storage.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Triple Sum Zero"
WHY DOES IT MATTER?
The two‑pointer pattern turns a combinatorial explosion into a linear scan after sorting, making O(n²) the best achievable bound for 3‑Sum without additional constraints. Mastery of this pattern unlocks efficient solutions for many related problems like 4‑Sum, closest‑sum, and range‑pair queries.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that once the array is sorted, the sum of the smallest and largest remaining elements provides a monotonic direction for pointer movement, allowing us to discard large swaths of impossible pairs in constant time per iteration.
REAL-WORLD CONNECTION
Think of a sorted list of timestamps from distributed logs; finding three events that exactly cancel each other's latency is analogous to the three‑sum zero search. The two‑pointer sweep mirrors a sliding window that expands or contracts based on cumulative latency, a technique used in real‑time monitoring systems.
During an interview, first state the O(n³) baseline, then immediately propose sorting + two‑pointer as the O(n²) improvement, and walk through duplicate‑skipping logic with a concrete example. This shows both algorithmic insight and attention to edge‑case correctness.
COMPLEXITY AT A GLANCE
O(n²)O(1)Core Theory — Why This Approach?
The "Triple Sum Zero" problem asks for the number of distinct unordered triplets (i, j, k) with i < j < k such that nums[i] + nums[j] + nums[k] = 0. A straightforward solution enumerates every combination of three indices using three nested loops, which incurs O(n³) time and quickly becomes infeasible for n in the tens of thousands. Moreover, this brute‑force method does not address duplicate values; without careful de‑duplication it would over‑count triplets that share the same three numbers but appear in different positions.
The optimal paradigm leverages two classic algorithmic ideas: sorting and the two‑pointer technique. By sorting the array in O(n log n) time, we impose an order that lets us treat the problem as a variant of the classic "2‑Sum" sub‑problem for each fixed first element. For each index i, we set two pointers, left = i + 1 and right = n – 1, and move them inward based on the sum of nums[i] + nums[left] + nums[right]. If the sum is too low we increment left, if it is too high we decrement right, and when the sum equals zero we record a unique triplet and skip over any duplicate values on both sides. This reduces the inner search to O(n) for each i, yielding an overall O(n²) time algorithm while using only O(1) extra space beyond the input array.
The key to achieving O(1) auxiliary space lies in performing all duplicate elimination in‑place after sorting, and never allocating additional data structures such as hash sets. The two‑pointer sweep is deterministic and cache‑friendly, making it not only theoretically optimal but also practically fast on large datasets. This pattern—sort then two‑pointer scan—is a cornerstone for many "k‑Sum" problems and exemplifies how ordering can transform an exponential search into a quadratic one.
Interview Questions on This Problem
Q1How would you adapt the two‑pointer solution to return the actual list of unique triplets instead of just the count, and what impact does that have on space complexity?
After sorting, we still fix each element and use two pointers, but each time we find a zero‑sum we push the triplet into a result list. To keep uniqueness we skip duplicates for the fixed element and for both pointers. This changes auxiliary space from O(1) to O(k), where k is the number of distinct triplets, which in the worst case can be O(n²) but is usually much smaller.
Q2In a fintech platform, you often need to detect three‑way arbitrage opportunities where the sum of three price differentials equals zero. How does the "Triple Sum Zero" algorithm map to that scenario, and what practical considerations would you add?
Each price differential can be treated as an integer (or scaled integer) and the arbitrage detection becomes exactly a three‑sum zero problem. The sorted two‑pointer approach works, but in production we must handle streaming data, floating‑point precision, and possibly a sliding window, so we might maintain a balanced BST or hash‑based index to support incremental updates while preserving near‑linear performance.
Q3Why is it safe to skip duplicate values while moving the left and right pointers, and what bug can arise if you forget to do this when counting distinct triplets?
Skipping duplicates ensures that each unique combination of values is counted only once, because after sorting identical values are contiguous. If you forget to skip them, the algorithm will count the same numeric triplet multiple times, inflating the answer and violating the problem's definition of unordered distinct triplets.
Examples
Input
[-1,0,1,2,-1,-4]
Output
2
Explanation: After sorting the array becomes [-4,-1,-1,0,1,2]. Scanning with two pointers yields the zero‑sum combinations (-1,-1,2) and (-1,0,1). No other value triples sum to zero, so the answer is 2.
Input
[0,0,0,0]
Output
1
Explanation: Sorting gives [0,0,0,0]. The only value triple that sums to zero is (0,0,0). Although there are four choose three index selections, they all correspond to the same value set, so the unique count is 1.
Input
[3,-2,1,-5,4]
Output
1
Explanation: Sorted array: [-5,-2,1,3,4]. The two‑pointer scan discovers the single zero‑sum triple (-5,1,4). No other combination of three values adds to zero, therefore the result is 1.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The answer fits in a 32‑bit signed integer
Optimal Approach & Strategy
Sort the array, then for each i use a left/right two‑pointer scan to find pairs that sum to -nums[i], skipping duplicates; this runs in O(n²) time with O(1) extra space.
Brute Force Approach
Iterate over all i < j < k with three nested loops and check if nums[i] + nums[j] + nums[k] == 0, counting each distinct value set; this runs in O(n³) time.
Code Solutions
function countTriplets(nums) {
nums.sort((a, b) => a - b);
const n = nums.length;
let count = 0;
for (let i = 0; i < n - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue; // skip duplicate first element
let left = i + 1;
let right = n - 1;
while (left < right) {
const sum = nums[i] + nums[left] + nums[right];
if (sum === 0) {
count++;
const curL = nums[left];
const curR = nums[right];
while (left < right && nums[left] === curL) left++;
while (left < right && nums[right] === curR) right--;
} else if (sum < 0) {
left++;
} else {
right--;
}
}
}
return count;
}
// Driver
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
if (input.length > 0) {
const n = input[0];
const nums = input.slice(1, 1 + n);
console.log(countTriplets(nums));
}#include <bits/stdc++.h>
using namespace std;
int countTriplets(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
int count = 0;
for (int i = 0; i < n - 2; ++i) {
if (i > 0 && nums[i] == nums[i-1]) continue; // skip duplicate first element
int left = i + 1, right = n - 1;
while (left < right) {
long sum = (long)nums[i] + nums[left] + nums[right];
if (sum == 0) {
++count;
int curL = nums[left], curR = nums[right];
while (left < right && nums[left] == curL) ++left;
while (left < right && nums[right] == curR) --right;
} else if (sum < 0) {
++left;
} else {
--right;
}
}
}
return count;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
if(!(cin >> n)) return 0;
vector<int> nums(n);
for(int i = 0; i < n; ++i) cin >> nums[i];
cout << countTriplets(nums) << "\n";
return 0;
}import java.io.*;
import java.util.*;
public class Main {
public static int countTriplets(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
int count = 0;
for (int i = 0; i < n - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue; // skip duplicate first element
int left = i + 1;
int right = n - 1;
while (left < right) {
long sum = (long) nums[i] + nums[left] + nums[right];
if (sum == 0) {
count++;
int curL = nums[left];
int curR = nums[right];
while (left < right && nums[left] == curL) left++;
while (left < right && nums[right] == curR) right--;
} else if (sum < 0) {
left++;
} else {
right--;
}
}
}
return count;
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
if (line == null || line.isEmpty()) return;
int n = Integer.parseInt(line.trim());
int[] nums = new int[n];
if (n > 0) {
StringTokenizer st = new StringTokenizer(br.readLine());
for (int i = 0; i < n; i++) {
nums[i] = Integer.parseInt(st.nextToken());
}
}
System.out.println(countTriplets(nums));
}
}def count_triplets(nums):
nums.sort()
n = len(nums)
count = 0
i = 0
while i < n - 2:
if i > 0 and nums[i] == nums[i - 1]:
i += 1
continue
left, right = i + 1, n - 1
while left < right:
s = nums[i] + nums[left] + nums[right]
if s == 0:
count += 1
curL, curR = nums[left], nums[right]
while left < right and nums[left] == curL:
left += 1
while left < right and nums[right] == curR:
right -= 1
elif s < 0:
left += 1
else:
right -= 1
i += 1
return count
if __name__ == "__main__":
import sys
data = sys.stdin.read().strip().split()
if data:
n = int(data[0])
nums = list(map(int, data[1:1+n]))
print(count_triplets(nums))function countTriplets(nums) {
nums.sort((a, b) => a - b);
const n = nums.length;
let count = 0;
for (let i = 0; i < n - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue; // skip duplicate first element
let left = i + 1;
let right = n - 1;
while (left < right) {
const sum = nums[i] + nums[left] + nums[right];
if (sum === 0) {
count++;
const curL = nums[left];
const curR = nums[right];
while (left < right && nums[left] === curL) left++;
while (left < right && nums[right] === curR) right--;
} else if (sum < 0) {
left++;
} else {
right--;
}
}
}
return count;
}
// Driver
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
if (input.length > 0) {
const n = input[0];
const nums = input.slice(1, 1 + n);
console.log(countTriplets(nums));
}Asked in Top Tech Interviews
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