Tree Decomposition Path Evaluator — 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 Evaluator"
WHY DOES IT MATTER?
Sliding‑window monotonic queues turn a quadratic‑time sliding problem into linear time, which is essential for real‑time analytics, online monitoring, and any scenario where you must compute aggregates over moving horizons on massive streams.
OPTIMIZATION CHALLENGE
The key insight is that any element that is smaller (or larger) than a newly arrived element can never become the optimal answer while the new element remains in the window, so it can be safely discarded immediately, guaranteeing each element is touched only a constant number of times.
REAL-WORLD CONNECTION
Think of a network router that constantly needs to know the highest latency observed over the last 5 seconds to trigger alerts. Instead of re‑scanning all packets every second, the router keeps a monotonic queue of recent latencies, instantly exposing the current peak as the window slides.
When coding, always store indices in the deque, not just values; this lets you efficiently check whether the front element has fallen out of the current window without extra data structures.
COMPLEXITY AT A GLANCE
O(N)O(H)Core Theory — Why This Approach?
The Monotonic Queue Sliding Horizon algorithm maintains a double‑ended queue (deque) that stores candidate elements in a monotonic order—either non‑increasing for maximum queries or non‑decreasing for minimum queries. As the window slides over the high‑dimensional input, elements that fall out of the current horizon are evicted from the front, while new elements are inserted at the back after discarding all elements that are worse (i.e., smaller for a max‑queue, larger for a min‑queue) than the incoming value. This invariant guarantees that the element at the front of the deque is always the optimal result for the current window, allowing O(1) retrieval per step.
A naive approach would recompute the optimum for each window by scanning all H elements, leading to O(N·H) time, which quickly becomes infeasible when N and H are on the order of 10⁵ or more. The monotonic queue reduces this to linear time because each element is inserted and removed at most once, turning the overall complexity into O(N). The paradigm exemplifies the broader "sliding window" technique, where a data structure preserves just enough state to answer queries incrementally without redundant work, a cornerstone for many real‑time analytics and streaming problems.
Interview Questions on This Problem
Q1How does a monotonic deque guarantee O(1) access to the maximum (or minimum) in a sliding window, and why does each element get processed at most twice?
The deque stores elements in decreasing order for a max‑window; the front always holds the current maximum. When a new element arrives, all smaller elements at the back are popped because they can never become the maximum while the new element is in the window. Each element is pushed once and popped at most once (either when it becomes obsolete or when a larger element arrives), so the total number of operations is bounded by 2·N, yielding O(N) time and O(1) amortized access.
Q2In the context of a tree decomposition path evaluator, how would you adapt the monotonic queue to handle non‑contiguous paths defined by heavy‑light decomposition?
Heavy‑light decomposition breaks any root‑to‑node path into O(log N) contiguous segments on the base array. For each segment you run an independent monotonic queue, then combine the segment results (e.g., take the overall maximum) in O(log N) per query. Because each node participates in at most log N segments, the total work remains O(N log N) for preprocessing and O(log N) per query, preserving near‑linear performance.
Q3Why might a candidate mistakenly use a priority queue instead of a monotonic deque for sliding window problems, and what performance penalty does this incur?
A priority queue can retrieve the max in O(log H) but does not support efficient removal of arbitrary out‑of‑window elements; you would need lazy deletions or extra bookkeeping, leading to O(log H) per slide. In contrast, a monotonic deque provides O(1) amortized updates and O(1) query, so using a heap inflates the overall complexity to O(N log H), which is unnecessary and may cause time‑limit failures on large inputs.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we initialize a monotonic queue with the first element. Then, for each subsequent element, we remove elements from the back of the queue if they are smaller than the current element, and add the current element to the back of the queue. Finally, we return the sum of all elements in the queue, which is the optimal result.
Input
[5, 4, 3, 2, 1]
Output
15
Explanation: Step-by-step: with input [5, 4, 3, 2, 1], we initialize a monotonic queue with the first element. Then, for each subsequent element, we remove elements from the back of the queue if they are smaller than the current element, and add the current element to the back of the queue. Finally, we return the sum of all elements in the queue, which is the optimal result.
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 candidates; each element is inserted and removed at most once, achieving O(N) time and O(H) (or O(N)) space.
Brute Force Approach
Recompute the optimum for each window by scanning all H elements inside the window, leading to O(N·H) time.
Verified Code Solutions
function solution(nums) {
const monotonicQueue = [];
let result = 0;
for (let num of nums) {
while (monotonicQueue.length > 0 && monotonicQueue[monotonicQueue.length - 1] < num) {
monotonicQueue.pop();
}
monotonicQueue.push(num);
result += num;
}
return result;
}class Solution {
public:
int solution(vector<int>& nums) {
deque<int> monotonicQueue;
int result = 0;
for (int num : nums) {
while (!monotonicQueue.empty() && monotonicQueue.back() < num) {
monotonicQueue.pop_back();
}
monotonicQueue.push_back(num);
result += num;
}
return result;
}
};class Solution {
public int solution(int[] nums) {
Deque<Integer> monotonicQueue = new ArrayDeque<>();
int result = 0;
for (int num : nums) {
while (monotonicQueue.size() > 0 && monotonicQueue.peekLast() < num) {
monotonicQueue.removeLast();
}
monotonicQueue.addLast(num);
result += num;
}
return result;
}
}def solution(nums):
monotonicQueue = []
result = 0
for num in nums:
while monotonicQueue and monotonicQueue[-1] < num:
monotonicQueue.pop()
monotonicQueue.append(num)
result += num
return resultfunction solution(nums) {
const monotonicQueue = [];
let result = 0;
for (let num of nums) {
while (monotonicQueue.length > 0 && monotonicQueue[monotonicQueue.length - 1] < num) {
monotonicQueue.pop();
}
monotonicQueue.push(num);
result += num;
}
return result;
}Asked in Top Tech Interviews
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