Sensor Packet Resolver 23 — Problem Statement & Solution Guide

Binary TreesMediumMonotonic Stack

TimeO(n log K)

SpaceO(K)

Problem Description

You are given an integer array nums and an integer K. Consider only those elements of nums that are strictly greater than K. From this filtered set, select the K largest distinct values (if fewer than K values exist, select all of them). Return the sum of the selected values. The algorithm must run in O(n log K) time or better, where n is the length of nums.

DSA Pattern Breakdown

"Sensor Packet Resolver 23"

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WHY DOES IT MATTER?

The 'Top K Elements' pattern is fundamental in data processing, recommendation systems, and real-time analytics. It allows efficient selection of extreme values without sorting the entire dataset, which is crucial for performance in large-scale systems.

OPTIMIZATION CHALLENGE

The key insight is to limit the size of the data structure to K. By using a min-heap of size K, we ensure that only the K largest elements are kept in memory, and each insertion/deletion operation is O(log K) instead of O(log n).

REAL-WORLD CONNECTION

Imagine a stock trading platform that needs to display the top K most volatile stocks in real-time. Instead of sorting all stocks every millisecond, a min-heap of size K can be updated incrementally as new price data arrives, ensuring low-latency responses.

During the interview, explicitly state the time complexity trade-off. Mention that if K is close to n, sorting might be faster due to lower constant factors, but for K << n, the heap approach is superior. Also, clarify the handling of distinct values by using a set or checking against the heap's minimum.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log K)

💾 Space:O(K)

Core Theory — Why This Approach?

The problem requires selecting the K largest distinct values from a subset of an array, which is a classic 'Top K' problem. A naive approach involving sorting the entire array takes O(n log n) time, which is suboptimal when K is significantly smaller than n. The optimal paradigm utilizes a min-heap (priority queue) of size K. By maintaining a heap that only stores the current K largest candidates, we ensure that any new element larger than the smallest in the heap replaces it, keeping the heap size constant at K. This reduces the time complexity to O(n log K), as each of the n elements is processed in O(log K) time for heap operations.

Interview Questions on This Problem

Q1How would you modify this solution if the array contained duplicates and you needed the K largest distinct values, but the input size was so large it didn't fit in memory?

You would use an external sorting algorithm or a distributed hash map to first deduplicate the values greater than K. Then, you could use a distributed priority queue or a multi-pass approach where you partition the data into buckets based on value ranges, process each bucket to find local top K, and merge the results. The key is to avoid loading the entire dataset into RAM.

Q2Why is a min-heap preferred over a max-heap for finding the K largest elements?

A min-heap allows us to efficiently access the smallest element among the current K largest candidates. If a new element is larger than this minimum, it belongs in the top K, so we can replace the minimum. With a max-heap, we would have to remove the maximum to check if the new element is larger, which is less efficient for maintaining a fixed-size window of the largest elements.

Q3In a real-time sensor network, if K changes dynamically, how would you optimize the data structure to handle frequent updates without rebuilding the heap?

You could use a balanced binary search tree (like a Red-Black Tree) or a Treap to maintain the K largest distinct values. This allows O(log K) insertion and deletion, and O(1) access to the minimum of the top K. Alternatively, if K changes infrequently, you can rebuild the heap only when K changes significantly, amortizing the cost.

Examples

Example 1

Input

nums = [12, 5, 8, 20, 7, 15], K = 3

Output

Explanation: Elements greater than K (3) are [12,5,8,20,7,15]. Sorting these in descending order gives [20,15,12,8,7,5]. The top 3 values are 20, 15, and 12. Their sum is 20+15+12 = 47.

Example 2

Input

nums = [4, 2, 9, 1, 6], K = 2

Output

Explanation: Values greater than K (2) are [4,9,6]. Sorted descending: [9,6,4]. The two largest are 9 and 6. Sum = 9+6 = 15.

Example 3

Input

nums = [3, 3, 3, 3], K = 5

Output

Explanation: All elements (3) are greater than K (5)? No, because 3 ≤ 5, so no element qualifies. Since the filtered set is empty, the sum is 0. However, the problem states to select up to K values; when none exist, the sum defaults to 0. Therefore output is 0.

Constraints

1 <= nums.length <= 100000
-10^9 <= nums[i] <= 10^9
1 <= K <= 10^5

Optimal Approach & Strategy

Use a min-heap of size K to maintain the K largest distinct values. Iterate through the filtered array, and for each element, if it is larger than the heap's minimum, replace the minimum. Sum the elements in the heap at the end. This takes O(n log K) time.

Brute Force Approach

Filter the array to get elements greater than K, sort them in descending order, and sum the first K distinct values. This takes O(n log n) time due to the sorting step.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n log K)

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

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

class Solution {
      public int solution(int[] nums, int K) {
         Arrays.sort(nums);
         int sum = 0;
         for (int i = nums.length - 1; i >= 0; i--) {
            if (nums[i] > K) {
               sum += nums[i];
               K++;
               if (K == nums.length) break;
            }
         }
         return sum;
      }
   }

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

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

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Sensor Packet Resolver 23 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Binary Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Sensor Packet Resolver 23 — 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