Triple Element Summation — Problem Statement & Solution Guide
Problem Description
You are provided with a linear queue containing a sequence of integers and a specific target sum. Your task is to identify all distinct triplets of elements from this queue such that their arithmetic sum equals the target value. A triplet is defined by the values of its three components; therefore, if two triplets contain the same set of values in any order, they are considered duplicates and only one instance should be included in the final result. Each element in the queue may be used at most once within a single triplet, but different triplets may share common elements as long as the indices or positions allow for valid selection without reusing the same physical element instance within the same triplet context.
The input consists of the queue's contents represented as a list of integers and the integer target sum. The output must be a list of lists, where each inner list contains three integers that form a valid triplet summing to the target. The order of the triplets in the output list does not matter, nor does the order of elements within each triplet, provided the values are consistent. You must ensure that no duplicate triplets are returned.
This problem requires an efficient approach to handle large input sizes while maintaining the uniqueness constraint. Naive brute-force methods will likely exceed time limits, so consider algorithms that leverage sorting or hashing to reduce the search space complexity.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Triple Element Summation"
WHY DOES IT MATTER?
The two‑pointer pattern transforms a quadratic search space into a linear scan after sorting, turning an otherwise exponential‑like enumeration into a tractable O(n²) solution. Mastery of this pattern unlocks efficient solutions for many sum‑related problems (2‑sum, 3‑sum, 4‑sum) that appear frequently in coding interviews and real‑world data processing pipelines.
OPTIMIZATION CHALLENGE
The key insight is that once the array is sorted, the relative ordering of elements tells you whether increasing the left pointer or decreasing the right pointer will move the sum closer to the target, eliminating the need for nested loops over all pairs.
REAL-WORLD CONNECTION
Consider a distributed log aggregation system where you need to detect three events whose timestamps sum to a specific window size. By sorting timestamps and sliding two pointers, you can identify such event triples in linear time, analogous to the queue‑based 3‑sum algorithm.
During an interview, first sort the array, then write a clear outer loop for the fixed element and an inner while‑loop for the two pointers. Immediately add duplicate‑skip logic after each pointer move; this prevents subtle bugs and demonstrates attention to detail.
COMPLEXITY AT A GLANCE
O(n²)O(1) additional (ignoring input copy)Core Theory — Why This Approach?
The classic "3‑sum" problem asks for all unique triplets in a collection that sum to a target value. A naive solution enumerates every combination of three elements, leading to O(n³) time, which quickly becomes infeasible for queues with thousands of entries. The optimal paradigm leverages sorting followed by a two‑pointer sweep: after fixing one element, the remaining two are located by moving pointers from opposite ends of the sorted sub‑array, adjusting based on the current sum relative to the target. This reduces the inner search to linear time, yielding an overall O(n²) algorithm while still guaranteeing that duplicate value‑sets are filtered out.
Sorting is crucial because it imposes an order that makes the two‑pointer technique possible; without order, we would need a hash‑based complement search for each pair, which still results in O(n²) but with higher constant factors and extra space. Moreover, handling duplicates correctly requires skipping over equal values both when selecting the fixed element and when moving the pointers, ensuring each distinct value combination appears exactly once. The combination of sorting (O(n log n)) and the linear two‑pointer scan (O(n²)) provides the best known deterministic solution for this problem in the general case.
Why the queue abstraction matters: a linear queue offers only sequential access, but we can copy its contents into an array or vector for random access without violating the problem constraints. This transformation enables the sort‑and‑two‑pointer approach while preserving the original order if needed for later operations. The resulting algorithm balances time efficiency with modest O(1) extra space (aside from the input copy), making it suitable for high‑throughput interview environments and production‑grade services that need to process large streams of numeric data.
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?
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 the two pointers. This keeps the O(n²) time and O(1) extra space while avoiding the memory overhead of storing the triplets.
Q2If the input queue is extremely large and cannot fit into memory, which external‑memory technique can you apply to still find all triplets summing to a target?
Use a multi‑pass external sort (e.g., merge sort on disk) to produce a sorted stream, then apply a streaming two‑pointer algorithm that reads blocks sequentially, keeping a small sliding window in memory. This trades additional I/O for O(n log n) external time and O(B) space, where B is the block size.
Q3Explain how you would adapt the algorithm when the target sum is zero and the input may contain many duplicate numbers, such as a stream of zeros.
When the target is zero, after sorting, handle the special case where the fixed element is zero: count the number of zeros (k) and if k ≥ 3, add the triplet (0,0,0) once. For other values, continue the standard two‑pointer scan, but ensure that when the left and right pointers point to the same value, you skip over all duplicates to avoid redundant triplets.
Examples
Input
queue = [1, 2, -3, 4, 5, -6, 7], target = 0
Output
[[-6, -3, 9], [-6, 1, 5], [-6, 2, 4], [-3, 1, 2]]
Explanation: First, we identify all combinations of three distinct elements that sum to 0. The triplet [-6, -3, 9] is invalid because 9 is not in the queue. Let's re-evaluate. Valid triplets: [-6, 1, 5] sums to 0. [-6, 2, 4] sums to 0. [-3, 1, 2] sums to 0. Are there others? [1, 2, -3] is the same as [-3, 1, 2]. [4, 5, -9] invalid. [7, -6, -1] invalid. So the unique triplets are [-6, 1, 5], [-6, 2, 4], and [-3, 1, 2]. The output format typically sorts the triplets and the elements within them for consistency. Thus, the output is [[-6, 1, 5], [-6, 2, 4], [-3, 1, 2]].
Input
queue = [0, 0, 0, 0], target = 0
Output
[[0, 0, 0]]
Explanation: The queue contains four zeros. We need to find triplets that sum to 0. The only possible triplet is [0, 0, 0]. Although there are multiple ways to pick three zeros from the four available, they all result in the same value triplet [0, 0, 0]. Therefore, only one unique triplet is returned.
Input
queue = [1, 1, 1, 2, 2, 3], target = 6
Output
[[1, 2, 3]]
Explanation: We look for triplets summing to 6. Possible combinations: [1, 2, 3] sums to 6. [1, 1, 4] invalid. [2, 2, 2] invalid (only two 2s). [1, 1, 4] invalid. [1, 2, 3] is the only valid unique triplet. Even though there are multiple 1s and 2s, the triplet [1, 2, 3] is unique by value.
Input
queue = [-1, 0, 1, 2, -1, -4], target = 0
Output
[[-1, -1, 2], [-1, 0, 1]]
Explanation: We search for triplets summing to 0. [-1, 0, 1] sums to 0. [-1, -1, 2] sums to 0. [0, 1, -1] is a duplicate of [-1, 0, 1]. [-4, 1, 3] invalid. [-4, 2, 2] invalid. The unique triplets are [-1, -1, 2] and [-1, 0, 1].
Constraints
- 3 <= queue.length <= 10^5
- -10^9 <= queue[i] <= 10^9
- -10^9 <= target <= 10^9
- The queue is provided as a list of integers representing the front to back order.
Optimal Approach & Strategy
Sort the array, then for each index i use a left‑right two‑pointer scan on the suffix to find pairs that complement arr[i] to the target, skipping duplicates.
Brute Force Approach
Iterate over every combination of three indices (i, j, k) and check if their values sum to the target, storing unique value sets.
Verified Code Solutions
function solution(nums, target) {
nums.sort((a, b) => a - b);
const result = [];
for (let i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
const sum = nums[i] + nums[left] + nums[right];
if (sum === target) {
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 < target) {
left++;
} else {
right--;
}
}
}
return result;
}class Solution {
public:
vector<vector<int>> solution(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
vector<vector<int>> result;
for (int i = 0; i < nums.size() - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue;
int left = i + 1;
int right = nums.size() - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == target) {
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 < target) {
left++;
} else {
right--;
}
}
}
return result;
}
};class Solution {
public int solution(int[] nums, int target) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue;
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == target) {
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 < target) {
left++;
} else {
right--;
}
}
}
return result;
}
}def solution(nums, target):
nums.sort()
result = []
for i in range(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
left = i + 1
right = len(nums) - 1
while left < right:
sum = nums[i] + nums[left] + nums[right]
if sum == target:
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 sum < target:
left += 1
else:
right -= 1
return resultfunction solution(nums, target) {
nums.sort((a, b) => a - b);
const result = [];
for (let i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
const sum = nums[i] + nums[left] + nums[right];
if (sum === target) {
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 < target) {
left++;
} else {
right--;
}
}
}
return result;
}Solve in Interative Editor
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