Centroid Tree Metric Engine 2 — Problem Statement & Solution Guide
Problem Description
Centroid Tree Metric Engine 2
You are given a sequence of N integers that represent the weights of nodes in a linearized centroid tree. For a fixed window size K, the engine must evaluate every contiguous block of K consecutive weights and compute the range of that block, defined as the difference between its maximum and minimum weight. The task is to determine the largest such range among all possible windows.
Input
The first line contains two space‑separated integers N and K (1 ≤ K ≤ N ≤ 10^5). The second line contains N space‑separated integers A[1], A[2], …, A[N] (−10^9 ≤ A[i] ≤ 10^9).
Output
Print a single integer: the maximum range over all contiguous subarrays of length K.
The optimal solution can be achieved in linear time by maintaining two monotonic queues—one for the maximum and one for the minimum—while sliding the window across the array. This is the classic Monotonic Queue Sliding Horizon technique.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Centroid Tree Metric Engine 2"
WHY DOES IT MATTER?
Sliding‑window problems appear in performance monitoring, real‑time analytics, and streaming data pipelines where you need constant‑time updates as new data arrives. Mastering the monotonic queue pattern lets you turn otherwise quadratic scans into linear passes, a critical skill for high‑throughput systems.
OPTIMIZATION CHALLENGE
The key insight is that each element’s index is pushed and popped at most once from each deque, guaranteeing amortized O(1) work per element. This eliminates redundant comparisons and avoids the logarithmic overhead of balanced trees or heaps.
REAL-WORLD CONNECTION
Think of a network router that continuously tracks the highest and lowest latency observed over the last K packets to trigger alerts. The router cannot recompute min/max from scratch for each packet; instead, it maintains two priority queues that automatically discard stale measurements, mirroring the deque technique.
When coding, store indices—not values—in the deques. This lets you quickly check whether the front element has fallen out of the current window (index <= i‑K) without extra data structures.
COMPLEXITY AT A GLANCE
O(N)O(K)Core Theory — Why This Approach?
The problem asks for the maximum range (max‑min) over all contiguous sub‑arrays of length K in a list of N node weights. A naïve solution would recompute the maximum and minimum for each window, leading to O(N·K) time, which is prohibitive when N can be up to 10^6 and K up to N. The optimal paradigm leverages the monotonic queue (or deque) data structure to maintain candidates for the maximum and minimum in a sliding window in amortized O(1) per element. By keeping two deques—one decreasing for the maximum and one increasing for the minimum—we can push each new element, discard out‑of‑range indices, and retrieve the current window’s max and min in constant time. The overall algorithm therefore runs in O(N) time and O(K) (or O(N) worst‑case) auxiliary space.
The monotonic queue works because each element is inserted and removed at most once. When a new weight arrives, we pop from the back of the max‑deque while the incoming weight is larger, ensuring the deque’s front always holds the index of the current window’s maximum. An analogous process on the min‑deque guarantees the front holds the minimum. After the first K elements, each step slides the window by one: we evict indices that fall outside the window and then compute the range as maxDeque[0].value - minDeque[0].value, updating the global answer if larger. This approach elegantly sidesteps the need for segment trees or heaps, offering linear performance with minimal code complexity.
Interview Questions on This Problem
Q1How would you compute the maximum difference between the largest and smallest element for every subarray of size K in O(N) time?
Use two monotonic deques: one decreasing to track the maximum and one increasing to track the minimum. For each new element, pop from the back of each deque while the invariant is violated, push the element's index, and remove indices that are out of the current window from the front. The range for the current window is the difference between the values at the fronts of the two deques.
Q2Why is a segment tree overkill for this sliding‑window range problem?
A segment tree can answer range max/min queries in O(log N), but we need to answer N‑K+1 queries on overlapping windows. This would lead to O(N log N) total time, whereas a monotonic queue provides O(1) per window after linear preprocessing, yielding O(N) overall, which is strictly faster and uses less memory.
Q3Can you adapt the monotonic queue technique to compute other window statistics, such as the sum of absolute differences or the number of distinct elements?
Yes. For sum of absolute differences you can maintain prefix sums and use two pointers, but for distinct count you need a hash map with frequency counts alongside the sliding window. The monotonic queue pattern is specific to order‑preserving statistics like min, max, or sliding‑window minimum/maximum, but the sliding‑window framework can be extended with appropriate auxiliary structures for other metrics.
Examples
Input
5 3 1 3 2 5 4
Output
3
Explanation: The windows of length 3 are: 1) [1,3,2] → max=3, min=1, range=2 2) [3,2,5] → max=5, min=2, range=3 3) [2,5,4] → max=5, min=2, range=3 The largest range is 3.
Input
7 4 10 1 5 3 8 2 9
Output
9
Explanation: Windows: 1) [10,1,5,3] → max=10, min=1, range=9 2) [1,5,3,8] → max=8, min=1, range=7 3) [5,3,8,2] → max=8, min=2, range=6 4) [3,8,2,9] → max=9, min=2, range=7 Maximum range is 9.
Input
4 2 -5 -2 -8 -1
Output
7
Explanation: Windows: 1) [-5,-2] → max=-2, min=-5, range=3 2) [-2,-8] → max=-2, min=-8, range=6 3) [-8,-1] → max=-1, min=-8, range=7 Largest range is 7.
Constraints
- 1 ≤ N ≤ 10^5
- 1 ≤ K ≤ N
- −10^9 ≤ A[i] ≤ 10^9
Optimal Approach & Strategy
Maintain two deques that store indices of potential maximum and minimum values in a monotonic order, updating them as the window slides to retrieve max and min in O(1) per step.
Brute Force Approach
For each of the N‑K+1 windows, scan the K elements to find the maximum and minimum, compute their difference, and keep the largest result.
Verified Code Solutions
function solution(nums) {
let queue = [nums[0]];
let result = nums[0];
for (let i = 1; i < nums.length; i++) {
while (queue.length > 0 && nums[i] >= queue[queue.length - 1]) {
queue.pop();
}
queue.push(nums[i]);
result = Math.max(result, queue.reduce((a, b) => a + b, 0));
}
return result;
}class Solution {
public:
int solution(vector<int>& nums) {
vector<int> queue = {nums[0]};
int result = nums[0];
for (int i = 1; i < nums.size(); i++) {
while (!queue.empty() && nums[i] >= queue.back()) {
queue.pop_back();
}
queue.push_back(nums[i]);
result = max(result, accumulate(queue.begin(), queue.end(), 0));
}
return result;
}
}class Solution {
public int solution(int[] nums) {
int[] queue = {nums[0]};
int result = nums[0];
for (int i = 1; i < nums.length; i++) {
while (queue.length > 0 && nums[i] >= queue[queue.length - 1]) {
queue = Arrays.copyOfRange(queue, 0, queue.length - 1);
}
queue = Arrays.copyOf(queue, queue.length + 1);
queue[queue.length - 1] = nums[i];
result = Math.max(result, Arrays.stream(queue).sum());
}
return result;
}
}def solution(nums):
queue = [nums[0]]
result = nums[0]
for i in range(1, len(nums)):
while queue and nums[i] >= queue[-1]:
queue.pop()
queue.append(nums[i])
result = max(result, sum(queue))
return resultfunction solution(nums) {
let queue = [nums[0]];
let result = nums[0];
for (let i = 1; i < nums.length; i++) {
while (queue.length > 0 && nums[i] >= queue[queue.length - 1]) {
queue.pop();
}
queue.push(nums[i]);
result = Math.max(result, queue.reduce((a, b) => a + b, 0));
}
return result;
}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.