Centroid Tree Metric Analyzer — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Stack and solve the Centroid Tree Metric Analyzer 3 problem optimally.
O(N)O(K)Problem Description
You are given an integer array nums of length N and a positive integer K (K ≤ N). For every contiguous subarray (window) of length K, determine the maximum element inside that window. Return the sum of all these maximum values. The required time complexity is O(N) and the intended solution uses a monotonic decreasing deque (also known as a monotonic queue) to maintain candidates for the window maximum while sliding the window from left to right.
Input: The first line contains two integers N and K. The second line contains N space‑separated integers representing nums.
Output: A single integer – the sum of the maximum values of all N‑K+1 windows of size K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Centroid Tree Metric Analyzer"
WHY DOES IT MATTER?
Monotonic queues are a cornerstone of sliding‑window problems because they provide constant‑time access to extremal values while preserving linear overall work. Mastery of this pattern unlocks efficient solutions for a wide class of real‑time analytics and streaming algorithms.
OPTIMIZATION CHALLENGE
The key insight is that any element smaller than a newly arrived larger element can never become the maximum of any future window, so it can be discarded immediately. This aggressive pruning reduces both time and space from O(N·K) to O(N).
REAL-WORLD CONNECTION
Think of a network traffic monitor that continuously reports the peak bandwidth over the last minute. As new packets arrive, older measurements expire, and the monitor must instantly know the current peak without rescanning the entire minute’s data—exactly what a monotonic deque does.
When coding, always store indices—not just values—in the deque. This makes it trivial to drop elements that slide out of the window and avoids subtle bugs when duplicate values appear.
COMPLEXITY AT A GLANCE
O(N)O(K)Core Theory — Why This Approach?
The sliding‑window maximum problem asks for the largest element in every contiguous subarray of length K. A naïve scan of each window costs O(K) per window, leading to O(N·K) overall, which is prohibitive when N and K approach 10^6. The optimal paradigm leverages a monotonic decreasing deque (also called a monotonic queue) that stores indices of elements in decreasing order of their values. As the window slides, the deque discards indices that fall out of the current window and removes from the back any indices whose values are smaller than the incoming element, guaranteeing that the front of the deque always holds the index of the current window’s maximum. This structure enables each array element to be inserted and removed at most once, delivering a linear O(N) runtime while using O(K) auxiliary space.
The monotonic deque is a classic example of a sliding‑window data structure that transforms a seemingly quadratic problem into linear time by exploiting the order of operations. By maintaining a strict decreasing order, the algorithm avoids redundant comparisons: when a larger element arrives, all smaller elements behind it can never become a future maximum, so they are safely evicted. This insight is the heart of many real‑time analytics, such as computing moving peaks in sensor streams or financial tick data, where latency constraints demand O(N) solutions.
Interview Questions on This Problem
Q1How would you modify the monotonic deque solution to also return the indices of the maximum elements for each window?
Store indices in the deque instead of values; the front always holds the index of the current maximum. When recording results, simply read deque[0] for each window. The same push‑pop logic applies, ensuring O(N) time.
Q2Can you adapt the sliding‑window maximum algorithm to compute the sum of minimums of all windows of size K? Explain the changes.
Yes. Use a monotonic increasing deque (values in ascending order). When a new element arrives, pop from the back while it is smaller than the incoming element, ensuring the front holds the minimum. The rest of the logic—removing out‑of‑range indices and adding the front value to the sum—remains identical.
Q3Why does the monotonic queue guarantee O(N) time even though each window triggers push and pop operations?
Each array element is pushed exactly once and can be popped at most once (either when it falls out of the window or when a larger element evicts it). Hence the total number of deque operations across the entire scan is bounded by 2N, yielding linear time.
Examples
Input
8 3 1 3 -1 -3 5 3 6 7
Output
29
Explanation: The windows of size 3 are: 1) [1,3,-1] → max = 3 2) [3,-1,-3] → max = 3 3) [-1,-3,5] → max = 5 4) [-3,5,3] → max = 5 5) [5,3,6] → max = 6 6) [3,6,7] → max = 7 Sum = 3+3+5+5+6+7 = 29.
Input
5 2 2 2 2 2 2
Output
8
Explanation: All windows of length 2 contain only the value 2, so each maximum is 2. There are 4 windows, therefore the sum is 2×4 = 8.
Input
6 4 9 -1 3 7 2 5
Output
23
Explanation: The three windows of size 4 are: 1) [9,-1,3,7] → max = 9 2) [-1,3,7,2] → max = 7 3) [3,7,2,5] → max = 7 Sum = 9+7+7 = 23.
Constraints
- 1 <= N <= 2*10^5
- 1 <= K <= N
- -10^9 <= nums[i] <= 10^9
- The answer fits in a signed 64‑bit integer.
Optimal Approach & Strategy
Maintain a monotonic decreasing deque of indices; update it as the window slides, reading the front as the current maximum. This yields O(N) time and O(K) space.
Brute Force Approach
Iterate over every window of size K, scan the K elements to find the maximum, and add it to the sum; this costs O(N·K) time.
Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let maxSum = nums[0];
let currentSum = nums[0];
let queue = [nums[0]];
for (let i = 1; i < nums.length; i++) {
while (queue.length > 0 && queue[0] < nums[i]) {
currentSum -= queue.shift();
}
currentSum += nums[i];
queue.push(nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int maxSum = nums[0];
int currentSum = nums[0];
deque<int> queue;
queue.push_back(nums[0]);
for (int i = 1; i < nums.size(); i++) {
while (!queue.empty() && queue.front() < nums[i]) {
currentSum -= queue.front();
queue.pop_front();
}
currentSum += nums[i];
queue.push_back(nums[i]);
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int maxSum = nums[0];
int currentSum = nums[0];
Deque<Integer> queue = new ArrayDeque<>();
queue.add(nums[0]);
for (int i = 1; i < nums.length; i++) {
while (!queue.isEmpty() && queue.peek() < nums[i]) {
currentSum -= queue.pollFirst();
}
currentSum += nums[i];
queue.add(nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}def solution(nums):
if not nums:
return 0
max_sum = nums[0]
current_sum = nums[0]
queue = [nums[0]]
for i in range(1, len(nums)):
while queue and queue[0] < nums[i]:
current_sum -= queue.pop(0)
current_sum += nums[i]
queue.append(nums[i])
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums) {
if (nums.length === 0) return 0;
let maxSum = nums[0];
let currentSum = nums[0];
let queue = [nums[0]];
for (let i = 1; i < nums.length; i++) {
while (queue.length > 0 && queue[0] < nums[i]) {
currentSum -= queue.shift();
}
currentSum += nums[i];
queue.push(nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
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