Pipeline Grid Aligner 50 — Problem Statement & Solution Guide

ArraysMediumFrequency Hash Map

TimeO(n)

SpaceO(1)

Problem Description

You are tasked with processing a stream of sensor readings from an industrial pipeline grid. The system provides an array metrics containing integer values representing pressure and flow rates, along with a threshold integer K. Your objective is to compute the 'Aligner Value', defined as the sum of all elements in metrics that are strictly greater than K. Elements equal to or less than K are considered within tolerance and are excluded from the calculation. If no elements exceed the threshold, the Aligner Value is 0.

The input consists of an array of integers metrics and an integer K. You must return a single integer representing the computed Aligner Value. The solution should efficiently process the array to determine the sum without unnecessary overhead, leveraging the properties of linear scans or frequency maps where applicable to handle large datasets within time limits.

DSA Pattern Breakdown

"Pipeline Grid Aligner 50"

medium

WHY DOES IT MATTER?

This pattern is essential because it forms the basis for many data filtering and aggregation tasks. Understanding how to efficiently filter and sum elements based on a condition is a fundamental skill in data processing and algorithm design.

OPTIMIZATION CHALLENGE

The key insight is that you only need to inspect each element once, and you can accumulate the sum in a single pass. This avoids the need for additional data structures or multiple passes over the data.

REAL-WORLD CONNECTION

In real-world engineering, this pattern is used in sensor data processing, where you need to filter out noise or irrelevant data points based on a threshold. For example, in an industrial pipeline, you might want to sum up all pressure readings that exceed a certain threshold to identify potential issues.

In an interview, emphasize the simplicity and efficiency of the linear scan approach. Highlight that you are making a single pass over the data, which is optimal for this problem. Also, mention potential optimizations for sorted arrays or distributed systems to show depth of understanding.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem of computing the sum of elements strictly greater than a threshold K in an array is a fundamental linear scan operation. While it appears trivial, it serves as a building block for more complex filtering and aggregation tasks in data processing pipelines. The naive approach involves iterating through the array once, checking each element against K, and accumulating the sum if the condition is met. This approach is optimal in terms of time complexity, as every element must be inspected at least once to determine if it contributes to the sum, resulting in O(n) time complexity.

Interview Questions on This Problem

Q1How would you modify this solution if the array was sorted in ascending order?

If the array is sorted, you can use binary search to find the first element greater than K. Then, you can sum the elements from that index to the end of the array. This reduces the time complexity to O(log n + m), where m is the number of elements greater than K, which can be more efficient if m is small.

Q2What if the array is extremely large and distributed across multiple machines? How would you handle this?

In a distributed system, you can partition the array across multiple machines. Each machine computes the sum of elements greater than K in its partition. Then, you aggregate the partial sums from all machines to get the final result. This approach leverages parallelism to reduce the overall computation time.

Q3How would you handle floating-point precision issues if the metrics were floating-point numbers instead of integers?

For floating-point numbers, you need to be cautious with comparisons due to precision issues. Instead of using strict greater-than (>) for floating-point numbers, you might use a small epsilon value to account for floating-point errors. For example, you could check if an element is greater than K + epsilon, where epsilon is a small positive number.

Examples

Example 1

Input

metrics = [12, 5, 20, 8, 15], K = 10

Output

Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 20 > 10 (add 20), 8 <= 10 (skip), 15 > 10 (add 15). Sum = 12 + 20 + 15 = 47. Wait, let me re-calculate. 12+20+15 = 47. Let's adjust the example to be cleaner. Let's use metrics = [12, 5, 20, 8, 15], K = 10. 12>10, 20>10, 15>10. Sum = 12+20+15 = 47. Let's try another set. metrics = [3, 7, 12, 1, 9], K = 5. 7>5, 12>5, 9>5. Sum = 7+12+9 = 28. Let's use this one. Input: metrics = [3, 7, 12, 1, 9], K = 5. Output: 28. Explanation: 3<=5 (skip), 7>5 (add 7), 12>5 (add 12), 1<=5 (skip), 9>5 (add 9). Total = 7+12+9 = 28.

Example 2

Input

metrics = [10, 10, 10, 10], K = 10

Output

Explanation: All elements are equal to K (10). Since the condition is strictly greater than K, no elements are added to the sum. The result is 0.

Example 3

Input

metrics = [-5, -2, 0, 3, 7], K = 1

Output

Explanation: Check each element against K=1: -5 <= 1 (skip), -2 <= 1 (skip), 0 <= 1 (skip), 3 > 1 (add 3), 7 > 1 (add 7). Sum = 3 + 7 = 10.

Example 4

Input

metrics = [100, 200, 300], K = 50

Output

Explanation: All elements (100, 200, 300) are greater than 50. Sum = 100 + 200 + 300 = 600.

Constraints

1 <= metrics.length <= 10^5
-10^9 <= metrics[i] <= 10^9
-10^9 <= K <= 10^9
The sum of all elements greater than K is guaranteed to fit within a 64-bit integer.

Optimal Approach & Strategy

Use a single pass through the array, accumulating the sum of elements greater than K. If the array is sorted, use binary search to find the first element greater than K and sum the remaining elements.

Brute Force Approach

Iterate through the array, check each element against K, and accumulate the sum if the element is greater than K. This approach is already optimal for unsorted arrays.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, K) {
   let sum = 0;
   for (let num of nums) {
       if (num > K) {
           sum += num;
       }
   }
   return sum;
}

class Solution {
public:
   int solution(vector<int> nums, int K) {
       int sum = 0;
       for (int num : nums) {
           if (num > K) {
               sum += num;
           }
       }
       return sum;
   }
};

class Solution {
   public int solution(int[] nums, int K) {
       int sum = 0;
       for (int num : nums) {
           if (num > K) {
               sum += num;
           }
       }
       return sum;
   }
}

def solution(nums, K):
   sum = 0
   for num in nums:
       if num > K:
           sum += num
   return sum

function solution(nums, K) {
   let sum = 0;
   for (let num of nums) {
       if (num > K) {
           sum += num;
       }
   }
   return sum;
}

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Pipeline Grid Aligner 50 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Arrays

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Pipeline Grid Aligner 50 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor