Sensor Cluster Analyzer 28 — Problem Statement & Solution Guide

Two PointersMediumBFS / Union Find

TimeO(N log K)

SpaceO(K)

Problem Description

Given a sequence of data elements representing sensor and cluster metrics, construct an optimal algorithm to evaluate and compute the target analyzer value under given operational constraints, where the target analyzer value is the sum of the K largest elements in the sequence.

DSA Pattern Breakdown

"Sensor Cluster Analyzer 28"

medium

WHY DOES IT MATTER?

The 'Top K Elements' pattern is essential because it appears in a wide range of real-world problems, from finding the most popular items in e-commerce to identifying the most critical alerts in monitoring systems. It teaches the importance of choosing the right data structure to exploit problem constraints, leading to significant performance improvements.

OPTIMIZATION CHALLENGE

The key insight is that we do not need to sort the entire array. By using a Min-Heap of size K, we can maintain only the K largest elements, reducing the time complexity from O(N log N) to O(N log K). This is a significant improvement when K is much smaller than N, as it avoids the overhead of sorting the entire dataset.

REAL-WORLD CONNECTION

In distributed systems, this pattern is analogous to maintaining a 'leaderboard' of the top K nodes in a cluster. Each node can report its metrics, and a central coordinator maintains a Min-Heap to track the top K performers. This is crucial for load balancing, resource allocation, and anomaly detection in large-scale infrastructure.

During an interview, clearly articulate the trade-off between sorting and heap-based selection. Emphasize that the choice depends on the relationship between K and N. If K is small, the heap approach is superior. If K is large, sorting might be simpler. Show that you understand the asymptotic complexity and can justify your choice based on the problem constraints.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log K)

💾 Space:O(K)

Core Theory — Why This Approach?

The problem of finding the sum of the K largest elements in a sequence is a classic selection problem. The naive approach involves sorting the entire array in descending order and summing the first K elements, which incurs a time complexity of O(N log N). While this is acceptable for small datasets, it becomes inefficient for large-scale sensor data streams where N can be in the millions, as the sorting operation does more work than necessary since we only care about the top K values, not the relative order of the remaining N-K elements.

The optimal paradigm leverages the properties of a Min-Heap (Priority Queue). By maintaining a heap of size K, we can process the array in a single pass. For each element, if the heap is not yet full, we add the element. If the heap is full, we compare the current element with the root of the heap (the smallest of the current top K). If the current element is larger, we remove the root and add the new element. This ensures that after processing all elements, the heap contains exactly the K largest values. The time complexity is O(N log K), which is significantly better than O(N log N) when K is much smaller than N. The space complexity is O(K) for the heap, which is also superior to the O(N) space required for sorting in-place or creating a sorted copy.

This approach is particularly relevant in distributed systems and real-time analytics where memory is constrained and data arrives in a streaming fashion. The Min-Heap allows for constant-time updates to the top K set, making it ideal for scenarios where the 'top K' metric needs to be updated incrementally as new sensor readings arrive. Understanding this trade-off between sorting and heap-based selection is a fundamental concept in algorithm design, demonstrating how problem-specific constraints can be exploited to achieve better asymptotic performance.

Interview Questions on This Problem

Q1At a fintech platform, you need to identify the top 100 largest transactions from a stream of 1 million transactions per second. How would you design a system to compute the sum of these top 100 transactions efficiently?

I would use a Min-Heap of size 100. As each transaction arrives, I compare it with the root of the heap. If it's larger, I replace the root with the new transaction and re-heapify. This maintains the top 100 transactions in O(log 100) time per transaction, resulting in an overall O(N log K) complexity. This is far more efficient than sorting the entire stream, which would be O(N log N) and require storing all N transactions in memory.

Q2In a high-growth engineering startup, you are building a real-time dashboard that displays the sum of the top 5 most active user sessions. How would you handle this if the data is arriving in a distributed manner across multiple servers?

Each server can maintain its own local Min-Heap of size 5. Periodically, these local heaps can be merged or aggregated. However, for a global top 5, a more robust approach is to use a distributed priority queue or a consensus-based algorithm where each node sends its top 5 candidates to a central aggregator. The aggregator then maintains a global Min-Heap of size 5, updating it as new candidates arrive. This ensures that the global top 5 is always accurate while minimizing communication overhead.

Q3At a global product company, you are optimizing a recommendation engine that needs to find the sum of the top K most relevant items from a list of N items. If K is very close to N, would you still use a Min-Heap? Why or why not?

If K is very close to N, the benefit of using a Min-Heap diminishes because O(N log K) approaches O(N log N), which is the same as sorting. In this case, sorting might be simpler to implement and potentially faster in practice due to lower constant factors. However, if K is significantly smaller than N, the Min-Heap approach is still preferable. The key is to choose the algorithm based on the ratio of K to N. For K << N, use a heap; for K ≈ N, sorting is acceptable.

Examples

Example 1

Input

[50, 40, 30, 20, 10, 9, 8, 7, 6, 5], 3

Output

Explanation: Step-by-step: Sort the array in descending order. Select the first K elements. Sum up these K elements. Add the next largest element to the sum. For the input [50, 40, 30, 20, 10, 9, 8, 7, 6, 5], the sorted array is [50, 40, 30, 20, 10, 9, 8, 7, 6, 5]. Select the first 3 elements, which are 50, 40, and 30. Sum up these 3 elements, which gives 120. Add the next largest element, which is 20, to the sum. The final output is 140.

Example 2

Input

[10, 9, 8, 7, 6, 5, 4, 3, 2, 1], 3

Output

Explanation: Step-by-step: Sort the array in descending order. Select the first K elements. Sum up these K elements. Add the next 3 largest elements to the sum. For the input [10, 9, 8, 7, 6, 5, 4, 3, 2, 1], the sorted array is [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]. Select the first 3 elements, which are 10, 9, and 8. Sum up these 3 elements, which gives 27. Add the next 3 largest elements, which are 7, 6, and 5, to the sum. The final output is 39.

Constraints

1 <= N <= 10^5
-10^4 <= metrics[i] <= 10^4
1 <= K <= N

Optimal Approach & Strategy

Use a Min-Heap of size K to maintain the K largest elements. Iterate through the array, and for each element, if it is larger than the root of the heap, replace the root and re-heapify. The time complexity is O(N log K) and the space complexity is O(K).

Brute Force Approach

Sort the entire array in descending order and sum the first K elements. This approach has a time complexity of O(N log N) and a space complexity of O(N) if a new sorted array is created.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log K)

function solution(nums, k) {
    nums.sort((a, b) => b - a);
    let sum = 0;
    for (let i = 0; i < k; i++) {
        sum += nums[i];
    }
    let nextLargest = nums[k - 1];
    for (let i = k; i < nums.length; i++) {
        if (nums[i] > nextLargest) {
            sum += nums[i];
            nextLargest = nums[i];
        }
    }
    return sum;
}

class Solution {
public:
    int solution(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end(), greater<int>());
        int sum = 0;
        for (int i = 0; i < k; i++) {
            sum += nums[i];
        }
        for (int i = k; i < nums.size(); i++) {
            if (nums[i] > nums[k - 1]) {
                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 < k; i++) {
            sum += nums[i];
        }
        for (int i = k; i < nums.length; i++) {
            if (nums[i] > nums[k - 1]) {
                sum += nums[i];
            } else {
                break;
            }
        }
        return sum;
    }
}

def solution(nums, k):
    nums.sort(reverse=True)
    sum = 0
    for i in range(k):
        sum += nums[i]
    for i in range(k, len(nums)):
        if nums[i] > nums[k - 1]:
            sum += nums[i]
        else:
            break
    return sum

function solution(nums, k) {
    nums.sort((a, b) => b - a);
    let sum = 0;
    for (let i = 0; i < k; i++) {
        sum += nums[i];
    }
    let nextLargest = nums[k - 1];
    for (let i = k; i < nums.length; i++) {
        if (nums[i] > nextLargest) {
            sum += nums[i];
            nextLargest = nums[i];
        }
    }
    return sum;
}

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Sensor Cluster Analyzer 28 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Sensor Cluster Analyzer 28 — 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