Tree Decomposition Path Validator — Problem Statement & Solution Guide
Problem Description
Given a high-dimensional input dataset or state graph of length $N$, calculate the optimal result using the **Monotonic Queue Sliding Horizon** algorithm.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tree Decomposition Path Validator"
WHY DOES IT MATTER?
Monotonic queue patterns turn otherwise quadratic sliding‑window problems into linear ones, which is essential for real‑time analytics, streaming validation, and any scenario where a sliding extremum must be queried millions of times per second.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that each element only influences future windows until it is eclipsed by a larger (or smaller) element, allowing us to discard it early and keep the deque size bounded by the window length.
REAL-WORLD CONNECTION
Think of a network router that constantly tracks the peak bandwidth usage over the last minute to trigger throttling; the router uses a sliding horizon and a monotonic structure to update the peak instantly as new packets arrive and old ones expire.
When coding, always store indices—not just values—in the deque; this lets you efficiently drop elements that slide out of the window without extra look‑ups.
COMPLEXITY AT A GLANCE
O(N)O(K)Core Theory — Why This Approach?
The Monotonic Queue Sliding Horizon (also known as monotonic deque) is a powerful technique for maintaining the extremal (minimum or maximum) element of a moving window in O(1) amortized time per update. The core idea is to store indices (or values) in a double‑ended queue such that the queue’s elements are monotonic—either non‑increasing for a maximum query or non‑decreasing for a minimum query. When the window slides forward, elements that fall out of the window are removed from the front, and any new element that would break the monotonic order is pruned from the back, guaranteeing that the front always holds the current optimal value. Naïve approaches recompute the extremum for each window by scanning all K elements, leading to O(N·K) time, which is infeasible for large N (up to 10⁶ or more) and high‑dimensional data streams. By contrast, the monotonic queue reduces the total work to linear time because each element is inserted and removed at most once.
In the context of a "Tree Decomposition Path Validator," the problem can be modeled as a linearized traversal of a tree (e.g., Euler tour or heavy‑light decomposition) where each position in the array represents a node’s state. The validation rule often requires checking that a certain metric (like depth, weight, or a custom predicate) stays within a sliding bound along any root‑to‑leaf path. Applying the monotonic queue to this linear representation enables constant‑time verification of the bound for every possible sub‑path of length ≤ K, turning an otherwise exponential verification into a deterministic O(N) solution. The optimal paradigm therefore combines tree linearization, window sliding, and monotonic data structures to achieve both speed and low memory overhead.
Interview Questions on This Problem
Q1How would you validate that every sub‑path of length K in a tree (after heavy‑light decomposition) satisfies a maximum‑value constraint in O(N) time?
Linearize the tree using heavy‑light decomposition to obtain an array of node values. Then slide a window of size K across the array while maintaining a decreasing monotonic deque that stores indices of potential maximums. The front of the deque always holds the current window's maximum, allowing O(1) checks per position and overall O(N) time.
Q2Explain why a simple for‑loop that recomputes the max for each window leads to TLE on N = 10⁶ and K = 10⁵, and how the monotonic queue avoids this.
Recomputing the max for each window costs O(K) per slide, resulting in O(N·K) ≈ 10¹¹ operations, which exceeds typical time limits. The monotonic queue inserts each element once and removes it at most once, so the total number of deque operations is O(N). This amortized constant‑time per slide eliminates the quadratic blow‑up.
Q3A fintech platform needs to monitor the highest transaction amount over the last 30 seconds in a high‑throughput stream. Which data structure would you choose and why?
A monotonic deque is ideal because it provides O(1) amortized updates and queries for the maximum while automatically discarding stale transactions that fall outside the 30‑second window, ensuring both speed and bounded memory usage.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5], we first initialize a monotonic queue with the first element 1. Then, we slide the horizon to the right, adding elements 2 and 3 to the queue. The maximum sum of the queue is 6. Next, we slide the horizon to the right, adding element 4 to the queue. The maximum sum of the queue is 10. Finally, we slide the horizon to the right, adding element 5 to the queue. The maximum sum of the queue is 15. Therefore, the optimal result using the Monotonic Queue Sliding Horizon algorithm is 15.
Input
[5, 4, 3, 2, 1]
Output
15
Explanation: Step-by-step: Given the input array [5, 4, 3, 2, 1], we first initialize a monotonic queue with the first element 5. Then, we slide the horizon to the right, adding elements 4 and 3 to the queue. The maximum sum of the queue is 12. Next, we slide the horizon to the right, adding element 2 to the queue. The maximum sum of the queue is 14. Finally, we slide the horizon to the right, adding element 1 to the queue. The maximum sum of the queue is 15. Therefore, the optimal result using the Monotonic Queue Sliding Horizon algorithm is 15.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N log N) or O(N log^2 N)
- Space Complexity: O(N)
Optimal Approach & Strategy
Use a monotonic deque to maintain the window's maximum in amortized O(1) per slide, achieving O(N) total time and O(K) space.
Brute Force Approach
For each possible window, scan all K elements to compute the maximum, resulting in O(N·K) time.
Verified Code Solutions
function solution(nums) {
const n = nums.length;
let maxSum = nums[0];
let queue = [nums[0]];
for (let i = 1; i < n; i++) {
while (queue.length > 0 && nums[i] > queue[queue.length - 1]) {
queue.pop();
}
queue.push(nums[i]);
maxSum = Math.max(maxSum, queue.reduce((a, b) => a + b, 0));
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
int max_sum = nums[0];
deque<int> queue;
queue.push_back(nums[0]);
for (int i = 1; i < n; i++) {
while (!queue.empty() && nums[i] > queue.back()) {
queue.pop_back();
}
queue.push_back(nums[i]);
max_sum = max(max_sum, accumulate(queue.begin(), queue.end(), 0));
}
return max_sum;
}
}class Solution {
public int solution(int[] nums) {
int n = nums.length;
int maxSum = nums[0];
Deque<Integer> queue = new ArrayDeque<>();
queue.add(nums[0]);
for (int i = 1; i < n; i++) {
while (!queue.isEmpty() && nums[i] > queue.peekLast()) {
queue.removeLast();
}
queue.add(nums[i]);
maxSum = Math.max(maxSum, queue.stream().mapToInt(Integer::intValue).sum());
}
return maxSum;
}
}def solution(nums):
n = len(nums)
max_sum = nums[0]
queue = [nums[0]]
for i in range(1, n):
while queue and nums[i] > queue[-1]:
queue.pop()
queue.append(nums[i])
max_sum = max(max_sum, sum(queue))
return max_sumfunction solution(nums) {
const n = nums.length;
let maxSum = nums[0];
let queue = [nums[0]];
for (let i = 1; i < n; i++) {
while (queue.length > 0 && nums[i] > queue[queue.length - 1]) {
queue.pop();
}
queue.push(nums[i]);
maxSum = Math.max(maxSum, queue.reduce((a, b) => a + b, 0));
}
return maxSum;
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