Tome Voyage Extractor 9 — 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, where K is the maximum allowed value.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Extractor 9"
WHY DOES IT MATTER?
Greedy selection transforms an exponential subset problem into a linear scan, making large inputs tractable.
OPTIMIZATION CHALLENGE
The key is reducing the combinatorial explosion by proving that a sorted order yields an optimal prefix.
REAL-WORLD CONNECTION
It mirrors budget allocation where you fund the cheapest projects first to maximize the number of initiatives.
Always verify the greedy‑choice property and optimal‑substructure before coding; a quick proof saves debugging later.
COMPLEXITY AT A GLANCE
O(n log n)O(1)Core Theory — Why This Approach?
Greedy algorithms make a locally optimal choice at each step with the hope that these choices lead to a globally optimal solution. For the Tome Voyage Extractor problem, the locally optimal decision is to always pick the smallest remaining metric because any larger metric would consume more of the limited K budget and can only reduce the number of elements we can include, thus lowering the total extractor value.
A naive exhaustive search would examine every subset of the sequence, resulting in O(2^n) time, which quickly becomes infeasible for n > 30. By recognizing that the objective is monotonic with respect to the sum of selected metrics and that the constraint K is a simple upper bound, we can sort the array once (O(n log n)) and then greedily accumulate elements until the next addition would breach K, guaranteeing optimality in O(n log n) overall.
Interview Questions on This Problem
Q1Why does selecting the smallest metrics first guarantee the maximum number of elements under the K constraint?
Choosing the smallest values consumes the least budget per element, leaving more capacity for additional picks. Any solution that replaces a smaller metric with a larger one cannot increase the count and may violate K.
Q2What is the time and space complexity of the optimal solution?
The algorithm runs in O(n log n) time due to sorting and uses O(1) extra space beyond the input array. The linear scan after sorting adds only O(n) time.
Q3How would you modify the algorithm if the goal were to maximize the sum of selected metrics without exceeding K?
The same greedy approach works because maximizing the sum under a hard cap is equivalent to filling the knapsack with the smallest items first until no more fit. If ties exist, any order among equal values yields the same sum.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
Output
90
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K=3, we iterate through the array. When we encounter 30, which is greater than K=3, we do not add it to the sum. Therefore, the optimal extractor value is 90.
Input
[]
Output
0
Explanation: Step-by-step: Given an empty input array, we return 0 as there are no elements to process.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort the array and greedily take elements from the start until the next addition would breach K.
Brute Force Approach
Enumerate every possible subset, compute its sum, and keep the best that does not exceed K.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= K) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] <= K) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] <= K) {
sum += nums[i];
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num <= K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= K) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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