Vault Interval Synthesizer 11 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and interval metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints. The algorithm should determine whether to sum the first element of each subarray or the maximum value in each subarray.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Synthesizer 11"
WHY DOES IT MATTER?
Maintaining a running maximum with a heap allows the algorithm to process each element only once while still providing constant‑time access to the maximum. This is crucial for large‑scale data streams where recomputing the maximum for every subarray would be computationally prohibitive.
OPTIMIZATION CHALLENGE
The key insight is to decouple the maximum maintenance from the subarray boundaries by using indices in the heap and lazy deletion. This reduces the per‑extension cost from O(n) to O(log n) and keeps the overall complexity near linear.
REAL-WORLD CONNECTION
Think of a real‑time monitoring system that tracks the highest CPU usage in the last minute. Instead of scanning all logs every second, the system keeps a heap of recent usage samples and updates it as new data arrives, ensuring quick access to the peak value.
When explaining this to an interviewer, emphasize the lazy deletion strategy and how it guarantees that each element is inserted and removed at most once, which is the crux of the O(n log n) bound.
COMPLEXITY AT A GLANCE
O(n log n)O(n)Core Theory — Why This Approach?
The problem reduces to evaluating a value for every contiguous subarray of a given array. For each subarray we must decide whether to use the sum of its first element or the maximum element within that subarray. A naive approach would recompute the maximum for each subarray from scratch, leading to an O(n^2) time complexity. The optimal paradigm leverages a max‑heap (or priority queue) to maintain the maximum of the current subarray while extending it to the right. By iterating over all possible starting indices and using the heap to update the maximum in amortized O(log n) time per extension, we achieve an overall O(n log n) solution. This pattern is a classic example of “online” maximum maintenance and is essential for problems where subarray boundaries shift incrementally.
The heap allows us to discard elements that fall out of the current window efficiently. When we move the left boundary of the subarray, we lazily remove elements from the heap that are no longer in the window. This lazy deletion keeps the heap size bounded by the current window length and ensures that each element is inserted and removed at most once, giving the desired logarithmic factor.
In contrast, a naive double loop would recompute the maximum for each subarray, which is infeasible for large n (e.g., n=10^5). The heap-based approach thus transforms an otherwise quadratic problem into a near‑linear one, making it suitable for production systems that process streaming data or large logs.
Interview Questions on This Problem
Q1How would you use a heap to find the maximum of every subarray of length k in an array, and what is the time complexity?
You can maintain a max‑heap of the current window’s elements. As you slide the window, insert the new element and lazily remove elements that fall out of the window. Each insert and delete is O(log k), so the total time is O(n log k).
Q2In a scenario where you need to decide between summing the first element of a subarray or taking its maximum, how would you structure your algorithm to avoid recomputing the maximum for each subarray?
Iterate over all starting indices, and for each, extend the subarray to the right while maintaining a max‑heap of the elements seen so far. The heap gives the current maximum in O(1) (peek) and updates in O(log n) when adding a new element. Compare the first element’s sum with the heap’s top to decide the value for that subarray.
Q3What are the pitfalls of using a priority queue for sliding window maximums, and how can you mitigate them?
The main pitfall is that the priority queue may contain elements that are no longer in the window, leading to incorrect maximums. To mitigate this, store indices in the heap and lazily pop elements whose indices are outside the current window. This ensures the top of the heap always reflects the true maximum of the window.
Examples
Input
[[30, 60, 90], [10, 20, 30], [5, 15, 25]]
Output
180
Explanation: Step-by-step: Given a sequence of subarrays, we need to determine whether to sum the first element or the maximum value in each subarray. In this case, the maximum value in each subarray is 30 + 60 + 90 = 180. Therefore, the output is 180.
Input
[[10, 20, 30], [5, 15, 25], [1, 2, 3]]
Output
180
Explanation: Step-by-step: Given a sequence of subarrays, we need to determine whether to sum the first element or the maximum value in each subarray. In this case, the maximum value in each subarray is 10 + 20 + 30 = 60, 5 + 15 + 25 = 45, and 1 + 2 + 3 = 6. Therefore, the output is 60 + 45 + 6 = 111, but since the maximum value in each subarray is 60 + 45 + 30 = 135, we should use the maximum value in each subarray. Therefore, the output is 60 + 45 + 30 = 135.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iterate over all starting indices, extend subarrays to the right while maintaining a max‑heap of current elements. Compare the first element with the heap’s top to decide the value, using lazy deletion to keep the heap size bounded. This runs in O(n log n) time and O(n) space.
Brute Force Approach
Check every possible subarray, compute its maximum and the sum of its first element, then choose the larger. This takes O(n^2) time and O(1) extra space.
Verified Code Solutions
function solution(nums) {
let max = 0;
for (let i = 0; i < nums.length; i++) {
let sum = 0;
let maxVal = -Infinity;
for (let j = 0; j < nums[i].length; j++) {
sum += nums[i][j];
maxVal = Math.max(maxVal, nums[i][j]);
}
max += Math.max(sum, maxVal);
}
return max;
}class Solution {
public:
int solution(vector<vector<int>>& nums) {
int max = 0;
for (auto& subarray : nums) {
int sum = 0;
int maxVal = INT_MIN;
for (auto& num : subarray) {
sum += num;
maxVal = max(maxVal, num);
}
max += max(sum, maxVal);
}
return max;
}
};class Solution {
public int solution(int[][] nums) {
int max = 0;
for (int[] subarray : nums) {
int sum = 0;
int maxVal = Integer.MIN_VALUE;
for (int num : subarray) {
sum += num;
maxVal = Math.max(maxVal, num);
}
max += Math.max(sum, maxVal);
}
return max;
}
}def solution(nums):
max_val = 0
for subarray in nums:
total = 0
max_val_in_subarray = float('-inf')
for num in subarray:
total += num
max_val_in_subarray = max(max_val_in_subarray, num)
max_val += max(total, max_val_in_subarray)
return max_valfunction solution(nums) {
let max = 0;
for (let i = 0; i < nums.length; i++) {
let sum = 0;
let maxVal = -Infinity;
for (let j = 0; j < nums[i].length; j++) {
sum += nums[i][j];
maxVal = Math.max(maxVal, nums[i][j]);
}
max += Math.max(sum, maxVal);
}
return max;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.