Tome Cache Consolidator 25 — 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 consolidator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Cache Consolidator 25"
WHY DOES IT MATTER?
Efficient overlap handling turns a potentially quadratic string‑merge into a linear‑ish solution.
OPTIMIZATION CHALLENGE
Pre‑computing overlaps and updating only changed pairs cuts the combinatorial explosion.
REAL-WORLD CONNECTION
Similar to cache deduplication where identical data blocks are merged to save storage.
Cache the overlap values in a matrix and use a priority queue; after each merge, only recompute overlaps involving the new string.
COMPLEXITY AT A GLANCE
O(N·L log N)O(N·L)Core Theory — Why This Approach?
The problem reduces to repeatedly merging strings with maximal suffix‑prefix overlap, a classic string‑consolidation paradigm. Naïve solutions recompute overlaps for every pair after each merge, leading to O(N^2·L) time (where N is the number of strings and L is average length) and quickly explode for large inputs. The optimal approach pre‑computes all pairwise overlaps using linear‑time pattern matching (KMP, Z‑algorithm, or rolling hash) in O(N·L) total time, stores them in a max‑heap, and greedily merges the pair with the greatest overlap, updating only the affected overlaps. This greedy‑heap strategy guarantees the same result as the exhaustive search for the given constraints because each merge strictly reduces the total length and never creates a larger overlap later, yielding an overall O(N·L log N) algorithm that is effectively linear for typical interview limits.
Interview Questions on This Problem
Q1Why does recomputing overlaps after each merge cause a time‑limit exceed in the naïve solution?
Each recomputation scans whole strings, turning a single merge into O(N·L) work; repeated N‑1 merges become O(N^2·L).
Q2How does the Z‑algorithm help compute suffix‑prefix overlaps in linear time?
By concatenating pattern + ‘#’ + text and building the Z‑array, we obtain longest prefix matches for every position in O(total length).
Q3What invariant does the max‑heap maintain during the greedy merging process?
It always holds the current pair with the maximum overlap, ensuring each step yields the greatest possible length reduction.
Examples
Input
[10, 20, 30, 40, 50], 3
Output
90
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K=3, we first sort the array in descending order: [50, 40, 30, 20, 10]. Then, we iterate through the array and add the numbers that are greater than or equal to K (40 and 50) to get the consolidator value: 40 + 50 = 90.
Input
[10, 20, 30, 40, 50], 0
Output
0
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K=0, we first sort the array in descending order: [50, 40, 30, 20, 10]. Then, we iterate through the array and add the numbers that are greater than or equal to K (none) to get the consolidator value: 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Pre‑compute overlaps with KMP/Z, store them in a max‑heap, and update only the overlaps that involve the newly formed string after each merge.
Brute Force Approach
Repeatedly compare every pair, merge the best, and recompute all overlaps from scratch after each merge.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let consolidatorValue = 0;
for (let num of nums) {
if (num >= k) {
consolidatorValue += num;
}
}
return consolidatorValue;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.rbegin(), nums.rend());
int consolidatorValue = 0;
for (int num : nums) {
if (num >= k) {
consolidatorValue += num;
}
}
return consolidatorValue;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int consolidatorValue = 0;
for (int num : nums) {
if (num >= k) {
consolidatorValue += num;
}
}
return consolidatorValue;
}
}def solution(nums, k):
nums.sort(reverse=True)
consolidator_value = 0
for num in nums:
if num >= k:
consolidator_value += num
return consolidator_valuefunction solution(nums, k) {
nums.sort((a, b) => b - a);
let consolidatorValue = 0;
for (let num of nums) {
if (num >= k) {
consolidatorValue += num;
}
}
return consolidatorValue;
}Asked in Top Tech Interviews
Solve in Interative Editor
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