Tree Decomposition Path Evaluator 4 — 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 4"
WHY DOES IT MATTER?
Sliding‑window monotonic structures turn quadratic‑time window scans into linear‑time streams, which is essential for real‑time analytics, high‑frequency trading, and large‑scale telemetry where latency and throughput are critical.
OPTIMIZATION CHALLENGE
The key insight is that an element that is larger (for a min‑queue) than a later element can never become the minimum for any future window, so it can be removed preemptively. This eliminates the need to revisit stale elements, collapsing the nested loop into a single pass.
REAL-WORLD CONNECTION
Think of a moving surveillance camera that only keeps the brightest pixel in its view. As the camera pans, older pixels fall out of view and newer ones enter; the camera's hardware can discard dimmer pixels instantly, analogous to the deque discarding dominated values.
When coding, always store indices alongside values in the deque; this makes out‑of‑range checks trivial and prevents subtle bugs when duplicate values appear.
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 data‑structure technique that maintains elements of a sliding window in a strictly monotonic order (either non‑increasing or non‑decreasing). By preserving this order, the extremal value (minimum or maximum) of the current window is always at the front of the deque, allowing O(1) retrieval. When the window slides forward, elements that fall out of the range are popped from the front, and newly arriving elements are inserted after discarding all elements that are worse (larger for a min‑queue, smaller for a max‑queue) than the newcomer. This guarantees each element is pushed and popped at most once, yielding linear overall time.
A naive approach would recompute the optimum for each window by scanning all K elements, leading to O(N·K) time, which quickly becomes infeasible for large N (up to 10^6 or more) and high‑dimensional data. The monotonic queue eliminates redundant comparisons by exploiting the fact that once an element is dominated by a newer one, it can never become the optimum for any future window. This insight transforms the problem into a streaming one where the algorithm processes each datum exactly once, achieving optimal O(N) time and O(K) auxiliary space.
The paradigm fits the broader class of sliding‑window optimization problems, such as "minimum in each subarray of size K", "maximum profit with a limited look‑ahead", or more complex tree‑decomposition path evaluations where the window corresponds to a path length constraint. By abstracting the window as a horizon that moves across the dataset, the monotonic queue provides a universal, cache‑friendly solution that scales to massive inputs without sacrificing correctness.
Interview Questions on This Problem
Q1How does a monotonic deque guarantee O(N) time for sliding‑window min/max problems?
Each element is inserted exactly once and removed at most once. When inserting, we pop from the back all elements that are worse than the new one, ensuring monotonicity. When the window slides, we pop from the front if the front element is out of range. Since each element participates in at most two operations, the total work is linear.
Q2In a tree‑decomposition path evaluation, why might you replace a recursive DP with a monotonic queue over a BFS order?
Recursive DP on a tree can lead to O(N·K) when the DP state depends on a sliding range of ancestors. By linearizing the tree via Euler tour or BFS order, the ancestor window becomes a contiguous segment, allowing a monotonic queue to maintain the optimal DP value for the current depth range, thus reducing the complexity to O(N).
Q3What edge cases must you handle when the window size K equals 1 or N in a monotonic queue solution?
When K=1, every element is its own window, so the deque should never hold more than one element; you must still pop the front before inserting the next. When K=N, the window never slides, so you must avoid popping elements due to out‑of‑range checks and simply return the global optimum after processing all elements.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we initialize the queue with the first three elements [1, 2, 3]. The sum of elements in the queue is 1 + 2 + 3 = 6. We then slide the window to the right by removing the leftmost element 1 and adding the next element 4. The new sum of elements in the queue is 2 + 3 + 4 = 9. We continue this process until the end of the array, giving the final sum of elements in the queue as 15.
Input
[1, 1, 1, 1, 1]
Output
5
Explanation: Step-by-step: with input [1, 1, 1, 1, 1], we initialize the queue with the first three elements [1, 1, 1]. The sum of elements in the queue is 1 + 1 + 1 = 3. We then slide the window to the right by removing the leftmost element 1 and adding the next element 1. The new sum of elements in the queue is 1 + 1 + 1 = 3. We continue this process until the end of the array, giving the final sum of elements in the queue as 5.
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 optimal element of the current window, updating it in O(1) per slide, resulting in O(N) total time.
Brute Force Approach
Recompute the optimum for every window by scanning all K elements inside the window, leading to O(N·K) time.
Verified Code Solutions
function solution(nums) {
let queue = [];
let sum = 0;
for (let i = 0; i < nums.length; i++) {
while (queue.length > 0 && queue[0] < nums[i]) {
sum -= queue.shift();
}
queue.push(nums[i]);
sum += nums[i];
if (i >= nums.length - 3) {
return sum;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
vector<int> queue;
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
while (!queue.empty() && queue[0] < nums[i]) {
sum -= queue[0];
queue.erase(queue.begin());
}
queue.push_back(nums[i]);
sum += nums[i];
if (i >= nums.size() - 3) {
return sum;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int[] queue = new int[nums.length];
int sum = 0;
int front = 0;
int rear = 0;
for (int i = 0; i < nums.length; i++) {
while (front < rear && queue[front] < nums[i]) {
sum -= queue[front];
front++;
}
queue[rear++] = nums[i];
sum += nums[i];
if (i >= nums.length - 3) {
return sum;
}
}
return sum;
}
}def solution(nums):
queue = []
sum = 0
for i in range(len(nums)):
while queue and queue[0] < nums[i]:
sum -= queue.pop(0)
queue.append(nums[i])
sum += nums[i]
if i >= len(nums) - 3:
return sum
return sumfunction solution(nums) {
let queue = [];
let sum = 0;
for (let i = 0; i < nums.length; i++) {
while (queue.length > 0 && queue[0] < nums[i]) {
sum -= queue.shift();
}
queue.push(nums[i]);
sum += nums[i];
if (i >= nums.length - 3) {
return sum;
}
}
return sum;
}Asked in Top Tech Interviews
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