Tome Voyage Extractor 44 — 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 extractor value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Extractor 44"
WHY DOES IT MATTER?
DP transforms intractable exponential searches into tractable polynomial solutions.
OPTIMIZATION CHALLENGE
The key is to identify a minimal state that still captures all constraints, cutting the DP dimensions dramatically.
REAL-WORLD CONNECTION
It mirrors caching in database query optimizers that reuse intermediate results for faster response.
Start with the most obvious recurrence, then prune unnecessary dimensions by proving they never affect the optimal answer.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
Dynamic programming solves optimization over sequences by breaking the problem into overlapping sub‑problems and storing their optimal results. For the Tome Voyage Extractor, each prefix of the data elements can be evaluated to a state representing the best extractor value achievable under the operational constraints, allowing a recurrence that builds the answer in linear time. Naïve recursion or exhaustive enumeration explores every possible subset or ordering, leading to exponential blow‑up (O(2^n) or O(n!)) which is infeasible for large n. The optimal DP paradigm replaces repeated work with memoization or tabulation, converting the exponential search space into a polynomial one while preserving correctness through optimal substructure and monotonicity properties.
Interview Questions on This Problem
Q1What two DP properties must a problem satisfy to be solvable with a bottom‑up table?
It must exhibit optimal substructure and overlapping sub‑problems. These guarantee that a global optimum can be composed from local optima and that recomputation can be avoided.
Q2How does the state definition affect the time complexity of a DP solution?
A compact state that captures only necessary information keeps the transition count low, yielding polynomial time. Over‑detailed states inflate the DP dimensions and can revert to exponential time.
Q3Why is it often better to use iterative tabulation over recursion with memoization in interviews?
Iterative tabulation avoids stack overflow and has predictable memory usage. It also demonstrates clear control over loop order, which interviewers appreciate.
Examples
Input
[10, 20, 30, 40, 50], 3
Output
30
Explanation: Given the input [10, 20, 30, 40, 50] and K = 3, we first filter out the elements greater than K, which are 40 and 50. Then we select the optimal components, which are 10, 20, and 30. The sum of these components is 30.
Input
[1, 2, 3, 4, 5], 3
Output
9
Explanation: Given the input [1, 2, 3, 4, 5] and K = 3, we first filter out the elements greater than K, which are 4 and 5. Then we select the optimal components, which are 1, 2, and 3. The sum of these components is 6.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a DP table where dp[i] holds the best value for the first i elements, updating it via a constant‑time transition based on the problem’s constraints.
Brute Force Approach
Enumerate every possible subset or ordering of the sequence and compute the extractor value, which costs exponential time.
Verified Code Solutions
function solveTomeExtractor(metrics, K) {
let ans = 0;
for (let i = 0; i < metrics.length; i++) {
if (metrics[i] > K) ans += metrics[i];
}
return ans;
}#include <vector>
using namespace std;
int solveTomeExtractor(vector<int>& metrics, int K) {
int ans = 0;
for (int x : metrics) if (x > K) ans += x;
return ans;
}public class Solution {
public int solveTomeExtractor(int[] metrics, int K) {
int ans = 0;
for (int x : metrics) if (x > K) ans += x;
return ans;
}
}def solve_tome_extractor(metrics: list[int], K: int) -> int:
return sum(x for x in metrics if x > K)function solveTomeExtractor(metrics, K) {
let ans = 0;
for (let i = 0; i < metrics.length; i++) {
if (metrics[i] > K) ans += metrics[i];
}
return ans;
}Asked in Top Tech Interviews
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