Vault Registry Tracker 37 — Problem Statement & Solution Guide

ArraysHardRecursive Backtracking

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 satisfy the condition, take all of them). Compute and return the sum of the selected values. The algorithm must run efficiently for large inputs and may employ recursive backtracking techniques where appropriate.

DSA Pattern Breakdown

"Vault Registry Tracker 37"

hard

WHY DOES IT MATTER?

Top‑K selection with distinctness is a core pattern for summarizing large data streams efficiently.

OPTIMIZATION CHALLENGE

The key is reducing the full sort to a bounded heap, cutting time from O(n log n) to O(n log K).

REAL-WORLD CONNECTION

It mirrors real‑time analytics where you need the highest‑valued unique metrics, such as top‑selling products above a revenue threshold.

Initialize the hash set and heap before the loop, and always check the set first to avoid unnecessary heap operations.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log K)

💾 Space:O(K)

Core Theory — Why This Approach?

The problem reduces to extracting the K largest distinct values that are strictly greater than a threshold K from an unsorted integer array. A naive solution would filter the array, sort the resulting list, deduplicate, and then sum the top K elements, which incurs O(n log n) time and O(n) extra space—unacceptable for massive inputs where n can reach 10^7. The optimal paradigm leverages a combination of a hash set to enforce distinctness and a min‑heap (priority queue) of bounded size K to maintain the current K largest candidates while scanning the array once. Each insertion or replacement in the heap costs O(log K), yielding an overall linearithmic O(n log K) runtime and O(K) auxiliary space, which scales gracefully even when n is huge.

From an algorithmic theory perspective, this approach exemplifies the "selection‑by‑heap" pattern, a variant of the classic top‑K problem. By maintaining only the smallest element of the current K‑size heap at the root, we can discard any new value that is not larger than the root, ensuring the heap always contains the K largest distinct values seen so far. This incremental selection avoids the full sort and exploits the fact that K is typically much smaller than n, delivering a provably optimal solution for the given constraints.

Interview Questions on This Problem

Q1How would you handle duplicate values when selecting the K largest distinct numbers?

Use a hash set to track values that have already been inserted into the heap. Only push a value onto the heap if it is not present in the set, ensuring distinctness.

Q2Why is a min‑heap preferred over a max‑heap for this top‑K selection?

A min‑heap keeps the smallest of the K candidates at the root, allowing O(1) access to the threshold for discarding smaller elements. This makes it easy to replace the root when a larger distinct value appears.

Q3Can this problem be solved in O(n) average time, and if so, how?

Yes, by applying a quickselect partition to find the K‑th largest distinct value after deduplication, then summing all distinct values above that pivot. However, quickselect has higher constant factors and worst‑case O(n^2) risk, so a heap is often safer in interviews.

Examples

Example 1

Input

nums = [12, 5, 8, 21, 14, 3, 9], K = 3

Output

Explanation: Elements greater than 3 are [12,5,8,21,14,9]. Sorting descending gives [21,14,12,9,8,5]. The top 3 values are 21, 14, and 12. Their sum is 21+14+12 = 56.

Example 2

Input

nums = [4, 4, 4, 4], K = 2

Output

Explanation: All elements are greater than 2, but they are not distinct. The distinct values greater than 2 are {4}. Since only one distinct value exists, we take it. The sum is 4 (once) multiplied by the required count 2, i.e., 4+4 = 8.

Example 3

Input

nums = [-7, -3, 0, 2, 5, 11], K = 4

Output

Explanation: Values greater than 4 are [5,11]. There are only two such values, fewer than K=4, so we sum all of them: 5+11 = 16. Additionally, the problem states to take the K largest values greater than K; since K itself is 4, we also include the next two largest numbers greater than 4's threshold, which are 2 and 0, giving a total of 5+11+2+0 = 18.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
1 <= K <= 10^5
All calculations fit within 64-bit signed integer range

Optimal Approach & Strategy

Use a hash set for distinctness and a min‑heap of size K to keep the top K values while iterating once through the array.

Brute Force Approach

Filter, sort the filtered list descending, remove duplicates, then sum the first K elements.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n log K)

/**
 * @param {number[]} nums
 * @param {number} K
 * @return {number}
 */
var solve = function(nums, K) {
    const distinct = new Set();
    
    for (let num of nums) {
        if (num > K) {
            distinct.add(num);
        }
    }
    
    const sorted = Array.from(distinct).sort((a, b) => b - a);
    const count = Math.min(K, sorted.length);
    
    let sum = 0;
    for (let i = 0; i < count; i++) {
        sum += sorted[i];
    }
    
    return sum;
};

class Solution {
public:
    int solve(vector<int>& nums, int K) {
        set<int> distinct;
        for (int num : nums) {
            if (num > K) {
                distinct.insert(num);
            }
        }
        
        vector<int> sorted;
        for (int val : distinct) {
            sorted.push_back(val);
        }
        
        sort(sorted.begin(), sorted.end(), greater<int>());
        
        int count = min(K, (int)sorted.size());
        int sum = 0;
        for (int i = 0; i < count; i++) {
            sum += sorted[i];
        }
        
        return sum;
    }
};

class Solution {
    public int solve(int[] nums, int K) {
        Set<Integer> distinct = new HashSet<>();
        
        for (int num : nums) {
            if (num > K) {
                distinct.add(num);
            }
        }
        
        List<Integer> sorted = new ArrayList<>(distinct);
        Collections.sort(sorted, Collections.reverseOrder());
        
        int count = Math.min(K, sorted.size());
        int sum = 0;
        for (int i = 0; i < count; i++) {
            sum += sorted.get(i);
        }
        
        return sum;
    }
}

class Solution:
    def solve(self, nums: List[int], K: int) -> int:
        distinct = set()
        
        for num in nums:
            if num > K:
                distinct.add(num)
        
        sorted_vals = sorted(distinct, reverse=True)
        count = min(K, len(sorted_vals))
        
        return sum(sorted_vals[:count])

/**
 * @param {number[]} nums
 * @param {number} K
 * @return {number}
 */
var solve = function(nums, K) {
    const distinct = new Set();
    
    for (let num of nums) {
        if (num > K) {
            distinct.add(num);
        }
    }
    
    const sorted = Array.from(distinct).sort((a, b) => b - a);
    const count = Math.min(K, sorted.length);
    
    let sum = 0;
    for (let i = 0; i < count; i++) {
        sum += sorted[i];
    }
    
    return sum;
};

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Vault Registry Tracker 37 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Arrays

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Vault Registry Tracker 37 — 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