Tome Cache Detector 19 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and cache metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints. The optimal solution is to sum the top K elements, where K is a parameter to the function.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Cache Detector 19"
WHY DOES IT MATTER?
Selecting top‑K efficiently is a common sub‑problem in data analytics and streaming.
OPTIMIZATION CHALLENGE
The key is reducing work from O(N log N) to O(N log K) by avoiding full sorting.
REAL-WORLD CONNECTION
Think of a news aggregator that constantly shows the K most‑read articles from a massive feed.
Initialize the heap with the first K items, then iterate; early exit when K equals N saves extra work.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The problem reduces to selecting the K largest values from an unsorted list and summing them. A naive full sort costs O(N log N), which is unnecessary when only the top K matter, especially for large N where K << N.
Using a min‑heap of size K allows us to maintain the current K largest elements while scanning the list once. Each insertion or replacement costs O(log K), yielding an overall O(N log K) time and O(K) extra space, which is optimal for this selection task.
Interview Questions on This Problem
Q1Why is a full sort suboptimal when K is much smaller than N?
Sorting processes all N elements, costing O(N log N) regardless of K. A heap or quick‑select focuses work on only K elements, reducing the dominant factor.
Q2How does a min‑heap of size K help maintain the top K values?
The heap's root is the smallest of the K kept elements; when a new element exceeds the root, we replace it, preserving only larger values. This ensures O(log K) updates per element.
Q3What edge case must you handle when K > N?
If K exceeds the number of elements, the answer is simply the sum of all elements. The algorithm should detect this and avoid unnecessary heap operations.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
Output
78
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20], we first sort the array in descending order. Then, we sum the top 3 elements, which are 20, 19, and 18. The sum of these elements is 57. However, this is not the correct output. The problem statement asks for the sum of the top K elements, but K is not specified. We assume K = 3 for this example. However, the correct output should be the sum of the top 3 elements, which is 57, not 78. The correct output should be 57.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9]
Output
24
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9], we first sort the array in descending order. Then, we sum the top 3 elements, which are 9, 8, and 7. The sum of these elements is 24.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Build a min‑heap of the first K elements, then for each remaining item replace the root if larger, achieving O(N log K).
Brute Force Approach
Sort the entire array descending and sum the first K elements, costing O(N log N).
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
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];
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
return sum(nums[:k])function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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