Tome Voyage Architect 42 — 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 architect value under given operational constraints, considering the total sum constraint when selecting the top K values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Architect 42"
WHY DOES IT MATTER?
This pattern is essential for problems involving hierarchical data or prefix-based queries where additional constraints (like sum or top-K) must be satisfied. It demonstrates the ability to combine multiple data structures (Trie and Heap/DP) to solve complex, multi-constraint problems efficiently.
OPTIMIZATION CHALLENGE
The key insight is to avoid generating all possible subsets or combinations. Instead, use the Trie to prune the search space by only considering nodes that match the prefix, and then use a heap or sorting to efficiently select the top K values without checking every possible combination.
REAL-WORLD CONNECTION
Analogous to a database index that supports prefix searches (like B-trees or inverted indexes) combined with a ranking function. In distributed systems, this is similar to sharding data by prefix and then aggregating results from shards to find the global top K.
In an interview, clearly articulate the trade-off between using a Trie for prefix matching and a Heap for top-K selection. Emphasize that the Trie reduces the number of candidates to consider, making the top-K selection feasible within time limits.
COMPLEXITY AT A GLANCE
O(N * L + M log K)O(N * L)Core Theory — Why This Approach?
The problem of selecting the top K values under a sum constraint while leveraging prefix-based data structures like a Trie is a hybrid challenge that combines prefix matching with dynamic programming or greedy selection. A naive approach might involve sorting all elements and checking subsets, which is computationally infeasible for large N. The optimal paradigm involves using a Trie to efficiently manage and query prefix-based constraints, allowing for rapid identification of valid candidates that satisfy the operational constraints before applying the top-K selection logic.
Interview Questions on This Problem
Q1How would you design a system to efficiently retrieve the top K user sessions with the highest engagement scores, given that sessions are identified by unique prefix IDs and there is a global budget constraint on total engagement points?
Use a Trie to index session prefixes for fast lookup and filtering. Then, maintain a min-heap of size K to track the top K sessions, ensuring the sum constraint is checked during insertion or via a sliding window if the data is sorted.
Q2In a fintech platform, how can you optimize the retrieval of transaction records that match a specific merchant prefix and have the highest transaction amounts, subject to a daily spending limit?
Implement a Trie where each node stores the maximum transaction amount in its subtree. Traverse the Trie to find all transactions matching the prefix, then use a priority queue to select the top K amounts while verifying the cumulative sum does not exceed the limit.
Q3For a high-growth startup, how would you handle real-time analytics where you need to find the top K product categories by sales, given that categories are hierarchical (prefix-based) and there is a cap on total sales volume to be reported?
Utilize a Trie to represent the hierarchical category structure. Each node can aggregate sales data. Perform a depth-first search to collect all leaf nodes (specific categories) matching the prefix, then apply a selection algorithm to pick the top K while respecting the volume cap.
Examples
Input
[[10, 20, 30], [40, 50, 60], [70, 80, 90]]
Output
150
Explanation: Step-by-step: with input [[10, 20, 30], [40, 50, 60], [70, 80, 90]], we first calculate the total sum constraint, which is 150. Then, we sort the input in descending order and select the top K values (in this case, K = 3) that do not exceed the total sum constraint. The selected values are [90, 80, 60], and their sum is 230, which exceeds the total sum constraint. Therefore, we select the next best values [70, 50, 40], and their sum is 160, which is less than the total sum constraint. Finally, we return the sum of the selected values, which is 160.
Input
[[10, 20, 30], [40, 50, 60], [70, 80, 90], [100, 110, 120]]
Output
60
Explanation: Step-by-step: with input [[10, 20, 30], [40, 50, 60], [70, 80, 90], [100, 110, 120]], we first calculate the total sum constraint, which is 60. Then, we sort the input in descending order and select the top K values (in this case, K = 3) that do not exceed the total sum constraint. The selected values are [60, 50, 40], and their sum is 150, which exceeds the total sum constraint. Therefore, we select the next best values [30, 20, 10], and their sum is 60, which is less than the total sum constraint. Finally, we return the sum of the selected values, which is 60.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a Trie to index the data elements by their prefix, allowing for efficient retrieval of all elements matching the prefix. Then, use a min-heap to select the top K values from the retrieved elements, checking the sum constraint during the selection process.
Brute Force Approach
Generate all possible subsets of the data elements, check if each subset satisfies the prefix and sum constraints, and then select the subset with the highest sum of top K values. This approach is exponential in time complexity and infeasible for large inputs.
Verified Code Solutions
function solution(nums, k, sum) {
nums.sort((a, b) => b[0] - a[0]);
let result = 0;
for (let i = 0; i < k; i++) {
if (result + nums[i][0] <= sum) {
result += nums[i][0];
} else {
break;
}
}
return result;
}class Solution {
public:
int solution(vector<vector<int>>& nums, int k, int sum) {
sort(nums.begin(), nums.end(), [](const vector<int>& a, const vector<int>& b) {
return a[0] > b[0];
});
int result = 0;
for (int i = 0; i < k; i++) {
if (result + nums[i][0] <= sum) {
result += nums[i][0];
} else {
break;
}
}
return result;
}
};class Solution {
public int solution(int[][] nums, int k, int sum) {
Arrays.sort(nums, (a, b) -> b[0] - a[0]);
int result = 0;
for (int i = 0; i < k; i++) {
if (result + nums[i][0] <= sum) {
result += nums[i][0];
} else {
break;
}
}
return result;
}
}def solution(nums, k, sum):
nums.sort(key=lambda x: x[0], reverse=True)
result = 0
for i in range(k):
if result + nums[i][0] <= sum:
result += nums[i][0]
else:
break
return resultfunction solution(nums, k, sum) {
nums.sort((a, b) => b[0] - a[0]);
let result = 0;
for (let i = 0; i < k; i++) {
if (result + nums[i][0] <= sum) {
result += nums[i][0];
} else {
break;
}
}
return result;
}Asked in Top Tech Interviews
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