Tree Decomposition Path Evaluator 3 — 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. The input should include the array and the sliding window size k.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tree Decomposition Path Evaluator 3"
WHY DOES IT MATTER?
Sliding window problems appear in time‑series analysis, real‑time monitoring, and streaming data pipelines. An O(N·k) solution cannot keep up with high‑throughput streams, whereas the monotonic queue delivers constant‑amortized work per element, enabling near‑real‑time responsiveness.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that an element dominated by a newer, larger (or smaller) element will never be the answer for any future window, allowing us to prune it immediately. This insight reduces both time and auxiliary space from O(N·k) to O(N).
REAL-WORLD CONNECTION
Think of a conveyor belt with sensors that need to report the highest temperature seen in the last k items. As the belt moves, older sensors fall off and newer ones join; the monotonic queue acts like a smart filter that only keeps the hottest relevant readings, discarding any cooler ones that can never become the hottest again.
When coding, store indices—not just values—in the deque. This lets you efficiently check whether the front element has slid out of the current window, preventing stale data from corrupting the result.
COMPLEXITY AT A GLANCE
O(N)O(k)Core Theory — Why This Approach?
The Monotonic Queue Sliding Horizon algorithm is a classic technique for answering range queries (such as minimum, maximum, or custom aggregate) over a moving window in linear time. The core idea is to maintain a double‑ended queue (deque) that stores indices of elements in a way that their corresponding values are monotonic (either non‑increasing for maximum queries or non‑decreasing for minimum queries). As 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 of the deque always holds the optimal value for the current window.
A naive solution would recompute the answer 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 typical window sizes. The monotonic queue eliminates this redundancy by ensuring each array element is inserted and removed at most once, achieving O(N) total work. This optimal paradigm leverages the fact that once an element is dominated by a newer, better candidate, it can never become the answer for any future window, allowing us to discard it permanently.
In the context of a high‑dimensional dataset or state graph, the sliding horizon often represents a temporal or spatial locality constraint, and the monotonic queue provides a deterministic, cache‑friendly way to propagate optimal sub‑results without the overhead of segment trees or priority queues. The result is a simple, linear‑time solution that scales gracefully with both N and k.
Interview Questions on This Problem
Q1How does a monotonic queue guarantee O(N) time for sliding window maximum/minimum, and why can't a standard queue achieve the same?
A monotonic queue maintains elements in a sorted order by value, discarding any element that is worse than a newer one because it can never become the optimum for any future window. Each element is pushed and popped at most once, giving O(N) total operations. A standard queue lacks this pruning, so it would need to examine all k elements for each slide, resulting in O(N·k).
Q2Explain how you would adapt the monotonic queue to compute the sum of the top‑two maximum values within each sliding window.
Store pairs (value, count) in the deque while preserving non‑increasing order. When inserting a new element, pop from the back while its value exceeds the back's value. To retrieve the top two, look at the first two entries in the deque (or the first entry and its count if it appears multiple times). Adjust counts when elements leave the window to keep the structure accurate.
Q3A fintech platform needs to compute the rolling maximum profit over a sliding window of transaction timestamps. What edge cases must you handle when implementing the monotonic queue, and how would you test them?
Key edge cases include windows larger than the array (returning the global optimum), windows of size 1 (each element is its own answer), duplicate values (ensuring indices are stored to differentiate occurrences), and negative or zero values (algorithm works unchanged but tests must verify correctness). Testing involves arrays with increasing, decreasing, random, and constant values, as well as varying k values including 1, N, and N/2.
Examples
Input
N = 5, k = 3, [1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] of length N = 5 and a sliding window size k = 3, we calculate the sum of the elements within the window. The optimal result is 15, which is the sum of the elements from index 1 to 3 (2 + 3 + 4 + 5).
Input
N = 10, k = 5, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
35
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] of length N = 10 and a sliding window size k = 5, we calculate the sum of the elements within the window. The optimal result is 35, which is the sum of the elements from index 2 to 6 (3 + 4 + 5 + 6 + 7).
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
Maintain a monotonic deque that inserts each element once and removes it at most once, ensuring the front always holds the current window's maximum.
Brute Force Approach
For each window, scan all k elements to find the maximum, repeating this for every possible window start.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length || k <= 0) return 0;
if (nums.length === 0 || nums.length === 1) return nums[0];
let windowSum = 0;
for (let i = 0; i < k; i++) {
windowSum += nums[i];
}
let maxSum = windowSum;
for (let i = k; i < nums.length; i++) {
windowSum = windowSum - nums[i - k] + nums[i];
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k > nums.size() || k <= 0) return 0;
if (nums.size() == 0 || nums.size() == 1) return nums[0];
int windowSum = 0;
for (int i = 0; i < k; i++) {
windowSum += nums[i];
}
int maxSum = windowSum;
for (int i = k; i < nums.size(); i++) {
windowSum = windowSum - nums[i - k] + nums[i];
maxSum = max(maxSum, windowSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length || k <= 0) return 0;
if (nums.length == 0 || nums.length == 1) return nums[0];
int windowSum = 0;
for (int i = 0; i < k; i++) {
windowSum += nums[i];
}
int maxSum = windowSum;
for (int i = k; i < nums.length; i++) {
windowSum = windowSum - nums[i - k] + nums[i];
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}
}def solution(nums, k):
if k > len(nums) or k <= 0:
return 0
if len(nums) == 0 or len(nums) == 1:
return nums[0]
window_sum = 0
for i in range(k):
window_sum += nums[i]
max_sum = window_sum
for i in range(k, len(nums)):
window_sum = window_sum - nums[i - k] + nums[i]
max_sum = max(max_sum, window_sum)
return max_sumfunction solution(nums, k) {
if (k > nums.length || k <= 0) return 0;
if (nums.length === 0 || nums.length === 1) return nums[0];
let windowSum = 0;
for (let i = 0; i < k; i++) {
windowSum += nums[i];
}
let maxSum = windowSum;
for (let i = k; i < nums.length; i++) {
windowSum = windowSum - nums[i - k] + nums[i];
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}Asked in Top Tech Interviews
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