Tome Voyage Resolver 39 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and voyage metrics, construct an optimal algorithm to evaluate and compute the target resolver value under given operational constraints. Given an array of integers and an integer k, return the sum of the k largest numbers greater than k.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Resolver 39"
WHY DOES IT MATTER?
Selecting top‑k qualified elements avoids full sorting, cutting runtime dramatically.
OPTIMIZATION CHALLENGE
Reduce O(n log n) sorting to O(n log k) or O(n) by focusing on a bounded subset.
REAL-WORLD CONNECTION
Similar to retrieving the highest‑paying customers above a revenue threshold in a sales database.
Filter first, then maintain a fixed‑size min‑heap to keep the current best candidates without extra passes.
COMPLEXITY AT A GLANCE
O(n log k)O(k)Core Theory — Why This Approach?
The naive solution sorts the entire array and then scans for numbers greater than k, yielding O(n log n) time which becomes prohibitive for massive inputs (n up to 10^6 or more). Sorting also wastes work on elements that are irrelevant because they are ≤ k, and it does not exploit the fact that we only need the top‑k qualifying values.
A more optimal paradigm isolates the qualifying subset first (values > k) and then extracts the k largest among them using a selection algorithm. A min‑heap of size k maintains the current top‑k in O(n log k) time, while a QuickSelect partition can achieve expected O(n) time with O(1) extra space, dramatically reducing both runtime and memory overhead for large datasets.
Interview Questions on This Problem
Q1How would you handle the case when fewer than k numbers are greater than k?
Return the sum of all qualifying numbers; if none exist, the result is zero. This edge case must be checked after filtering before applying the selection logic.
Q2What is the time complexity of using a min‑heap versus QuickSelect for this problem?
A min‑heap gives O(n log k) time because each insertion or replacement costs log k. QuickSelect runs in expected O(n) time but has O(n) worst‑case without careful pivot selection.
Q3Why is it unnecessary to sort the entire array?
Sorting costs O(n log n) even for elements that will never be part of the answer. We only need the k largest values that satisfy the >k condition, so partial selection is sufficient.
Examples
Input
[10, 20, 30, 40, 5, 5, 5, 5, 5, 5], 3
Output
90
Explanation: Step-by-step: First, we sort the input array in descending order. Then, we iterate over the sorted array and sum up the K largest numbers greater than K. In this case, the K largest numbers greater than K are 40, 30, and 20, so the output is 90.
Input
[5, 5, 5, 5, 5, 5], 3
Output
0
Explanation: Step-by-step: First, we sort the input array in descending order. Then, we iterate over the sorted array and sum up the K largest numbers greater than K. In this case, there are no numbers greater than K, so the output is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Filter numbers > k, then use a min‑heap of size k (or QuickSelect) to extract the k largest in linear or near‑linear time.
Brute Force Approach
Sort the whole array, then iterate to sum the first k numbers that are greater than k.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; 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(), greater<int>());
int sum = 0;
for (int i = 0; i < k; 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 < k; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
sum = 0
for i in range(k):
if nums[i] > k:
sum += nums[i]
else:
break
return sumfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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