Pipeline Vector Aligner 38 — Problem Statement & Solution Guide
Problem Description
In a high-throughput data processing system, a sequence of integer metrics is generated representing the alignment scores of vector components within a pipeline. The system requires the computation of a specific 'aligner value' for each position in the sequence. For a given index i, the aligner value is defined as the sum of all preceding elements (indices j where j < i) that are strictly greater than the element at index i. If no such preceding elements exist, the aligner value for that position is 0.
Your task is to implement an efficient algorithm that processes the input sequence and returns an array where each element at index i corresponds to the computed aligner value for that position. The solution must handle large input sizes efficiently, avoiding the O(n^2) complexity of a brute-force approach by leveraging appropriate data structures or algorithmic patterns to maintain the necessary state of preceding elements.
Input: An array of integers metrics representing the sequence of alignment scores.
Output: An array of integers result of the same length, where result[i] is the sum of all metrics[j] such that j < i and metrics[j] > metrics[i].
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Vector Aligner 38"
WHY DOES IT MATTER?
Efficient cumulative aggregation turns quadratic work into linear, a core performance win in data‑intensive pipelines.
OPTIMIZATION CHALLENGE
The key is to avoid recomputing the same partial sums by storing them once and reusing them instantly.
REAL-WORLD CONNECTION
Streaming analytics often need the running total of metrics to compute moving averages or thresholds in real time.
Initialize a single long‑type accumulator and update it in‑place; avoid extra arrays unless you need random access later.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The naive solution iterates over each index i and sums all elements before i, leading to O(n²) time, which quickly becomes infeasible for large n (e.g., 10⁵ or more). The optimal paradigm leverages the prefix‑sum technique: maintain a running total while scanning the array once, updating the aligner value for i in O(1) and achieving overall O(n) time. Although the problem is categorized under the Heap pattern, the underlying operation is a cumulative aggregation; a heap would only be useful for variants that require ordering (e.g., sum of k‑largest previous elements). By recognizing that the required value is simply the sum of all earlier entries, we can replace any complex data structure with a single accumulator, dramatically reducing both time and space overhead.
Interview Questions on This Problem
Q1Why does the double‑loop solution exceed time limits for n = 10⁵?
It performs O(n²) additions, resulting in ~10¹⁰ operations, which far exceeds typical 1‑2 second limits.
Q2How does a prefix‑sum array enable O(1) query for the aligner value?
Each entry stores the sum of all elements up to that index, so the aligner value at i is just prefix[i‑1].
Q3When would a heap be a better choice than a simple accumulator for a similar problem?
If the query required the sum of the k‑largest (or smallest) previous elements, a max‑ or min‑heap maintains those candidates in O(log k) per insertion.
Examples
Input
metrics = [5, 2, 8, 1, 4]
Output
[0, 5, 0, 15, 15]
Explanation: Index 0: No preceding elements, result is 0. Index 1: Preceding [5]. 5 > 2, sum = 5. Index 2: Preceding [5, 2]. Neither 5 nor 2 is > 8, sum = 0. Index 3: Preceding [5, 2, 8]. 5 > 1, 2 > 1, 8 > 1. Sum = 5 + 2 + 8 = 15. Index 4: Preceding [5, 2, 8, 1]. 5 > 4, 2 is not > 4, 8 > 4, 1 is not > 4. Sum = 5 + 8 = 13. Wait, let me re-verify the logic. The problem asks for sum of preceding elements strictly greater. For index 4 (value 4), preceding are 5, 2, 8, 1. 5>4 (yes), 2>4 (no), 8>4 (yes), 1>4 (no). Sum = 5+8=13. My previous output said 15. Let me correct the example output and explanation to be mathematically accurate. Corrected Output: [0, 5, 0, 15, 13]
Input
metrics = [10, 10, 10, 10]
Output
[0, 0, 0, 0]
Explanation: Index 0: No preceding, 0. Index 1: Preceding [10]. 10 is not strictly greater than 10. Sum = 0. Index 2: Preceding [10, 10]. Neither is strictly greater. Sum = 0. Index 3: Preceding [10, 10, 10]. None are strictly greater. Sum = 0.
Input
metrics = [3, 1, 2]
Output
[0, 3, 3]
Explanation: Index 0: No preceding, 0. Index 1: Preceding [3]. 3 > 1. Sum = 3. Index 2: Preceding [3, 1]. 3 > 2 (yes), 1 > 2 (no). Sum = 3.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- The sum of all metrics[i] can exceed 32-bit integer range, so use 64-bit integers for accumulation.
Optimal Approach & Strategy
Single pass with a running accumulator; answer[i] = accumulator before adding arr[i].
Brute Force Approach
Nested loops: for each i, loop j = 0..i‑1 and accumulate arr[j].
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, k):
nums.sort()
sum = 0
for num in nums:
if num > k:
sum += num
else:
break
return sumfunction solution(nums, k) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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