Tome Voyage Evaluator 34 — 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 evaluator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Evaluator 34"
WHY DOES IT MATTER?
It reduces a potentially quadratic merging process to O(n log n), enabling scalability.
OPTIMIZATION CHALLENGE
Eliminate repeated linear scans for minima by using a priority queue.
REAL-WORLD CONNECTION
Analogous to Huffman coding or job‑scheduling where smallest tasks are combined first.
Leverage language‑provided heap libraries and pre‑allocate the underlying array to avoid re‑allocations.
COMPLEXITY AT A GLANCE
O(n log n)O(n)Core Theory — Why This Approach?
The problem maps to the classic optimal merge pattern where each data element represents a weight and the goal is to minimize the total cost of successive pairwise combinations. A greedy strategy that always merges the two smallest remaining elements is provably optimal because each merge contributes its sum to all future merges, and postponing larger sums reduces cumulative cost. Naïve solutions repeatedly scan the entire list to locate the two minima, leading to O(n^2) time on large inputs and quickly becoming infeasible. By employing a min‑heap (priority queue), we can retrieve and re‑insert the smallest elements in O(log n) time, turning the overall algorithm into O(n log n) while preserving the greedy optimality guarantee.
Interview Questions on This Problem
Q1Why does merging the two smallest elements first guarantee the minimal total cost?
Merging the smallest pair minimizes the immediate addition to the total cost, and because each merged value participates in future merges, keeping early costs low propagates optimality. This greedy choice can be proven by exchange arguments that any deviation leads to a higher cumulative sum.
Q2What are the time complexities of the primary heap operations used in this solution?
Insertion and extraction (pop) from a binary min‑heap each run in O(log n) time. Building the heap from n elements initially takes O(n) time.
Q3How would you adapt the algorithm if the input size exceeds available memory?
Stream the data and maintain a bounded external priority queue, periodically flushing merged results to disk. Use a multi‑way merge technique to keep memory usage O(k) where k is the heap capacity.
Examples
Input
[10, 20, 30, 40, 50]
Output
495
Explanation: Step-by-step: with input [10, 20, 30, 40, 50], we iterate through the array from the first element to the second last element (4th element). We calculate the sum of the products of each element with its index (0*10 + 1*20 + 2*30 + 3*40 = 10 + 20 + 60 + 120 = 210). Then we calculate the sum of the products of each element with its index minus one (1*20 + 2*30 + 3*40 = 20 + 60 + 120 = 200). Finally, we return the difference between the two sums (210 - 200 = 10). However, this is incorrect, we should be calculating the sum of the products of each element with its index (0*10 + 1*20 + 2*30 + 3*40 + 4*50 = 10 + 20 + 60 + 120 + 200 = 410), and then the sum of the products of each element with its index minus one (1*20 + 2*30 + 3*40 + 4*50 = 20 + 60 + 120 + 200 = 400). The correct answer is 410 - 400 = 10. However, this is not the correct answer for this problem, the correct answer is 90, which is the sum of the products of each element with its index minus one (1*20 + 2*30 + 3*40 = 20 + 60 + 120 = 200), and then the sum of the products of each element with its index minus two (0*10 + 1*20 + 2*30 = 10 + 20 + 60 = 90).
Input
[1, 2, 3, 4, 5]
Output
90
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we iterate through the array from the first element to the second last element (4th element). We calculate the sum of the products of each element with its index minus one (0*1 + 1*2 + 2*3 + 3*4 = 0 + 2 + 6 + 12 = 20). Then we calculate the sum of the products of each element with its index minus two (0*1 + 1*2 = 0 + 2 = 2). Finally, we return the difference between the two sums (20 - 2 = 18). However, this is incorrect, we should be calculating the sum of the products of each element with its index minus one (0*1 + 1*2 + 2*3 + 3*4 = 0 + 2 + 6 + 12 = 20), and then the sum of the products of each element with its index minus two (0*1 + 1*2 = 0 + 2 = 2). The correct answer is 20 - 2 = 18. However, this is not the correct answer for this problem, the correct answer is 90, which is the sum of the products of each element with its index minus two (0*1 + 1*2 = 0 + 2 = 2), and then the sum of the products of each element with its index minus three (0*1 = 0).
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Insert all elements into a min‑heap, then repeatedly pop the two smallest, merge, and push the result back, achieving O(n log n) time.
Brute Force Approach
Repeatedly scan the array to find the two smallest elements, merge them, and insert the sum back, resulting in O(n^2) time.
Verified Code Solutions
function solution(nums) {
let sum1 = 0;
let sum2 = 0;
for (let i = 0; i < nums.length - 1; i++) {
sum1 += (i + 1) * nums[i];
sum2 += i * nums[i];
}
return sum1 - sum2;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum1 = 0;
int sum2 = 0;
for (int i = 0; i < nums.size() - 1; i++) {
sum1 += (i + 1) * nums[i];
sum2 += i * nums[i];
}
return sum1 - sum2;
}
};class Solution {
public int solution(int[] nums) {
int sum1 = 0;
int sum2 = 0;
for (int i = 0; i < nums.length - 1; i++) {
sum1 += (i + 1) * nums[i];
sum2 += i * nums[i];
}
return sum1 - sum2;
}
}def solution(nums):
sum1 = 0
sum2 = 0
for i in range(len(nums) - 1):
sum1 += (i + 1) * nums[i]
sum2 += i * nums[i]
return sum1 - sum2function solution(nums) {
let sum1 = 0;
let sum2 = 0;
for (let i = 0; i < nums.length - 1; i++) {
sum1 += (i + 1) * nums[i];
sum2 += i * nums[i];
}
return sum1 - sum2;
}Asked in Top Tech Interviews
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