Rotated Matrix Pivot Analyzer 5 — Problem Statement & Solution Guide

Sliding WindowHardSliding Window Maximum Deque

TimeO(n·m)

SpaceO(n·m)

Problem Description

You are provided with a 2D integer matrix grid of dimensions n by m. The matrix is considered 'rotated' in the sense that its rows are cyclically shifted, but for the purpose of this analysis, we treat it as a standard grid where we need to identify the peak element within any contiguous sub-rectangle of a fixed size. Specifically, you must determine the maximum value found in any sub-rectangle of size k by k that fits entirely within the n by m grid. This problem requires efficient processing to handle large grids, leveraging the sliding window technique to maintain the maximum value in a moving window across both rows and columns. Your task is to compute the global maximum among all such k by k sub-rectangles. If no such sub-rectangle exists (i.e., k > n or k > m), return -1.

DSA Pattern Breakdown

"Rotated Matrix Pivot Analyzer 5"

hard

WHY DOES IT MATTER?

The 2‑D sliding window maximum pattern is essential because it transforms a potentially quadratic or cubic brute‑force search into a linear‑time solution, enabling real‑time analytics on large datasets. It showcases a deep understanding of data structures, amortized analysis, and algorithmic decomposition, all of which are critical for solving complex engineering problems efficiently.

OPTIMIZATION CHALLENGE

The key insight is to split the 2‑D window into two 1‑D passes—horizontal then vertical—while reusing the efficient deque algorithm. This reduces the time from O(n·m·k·l) to O(n·m) and the space from O(k·l) to O(n·m) for the intermediate row‑max matrix, which is still linear but far more manageable.

REAL-WORLD CONNECTION

Consider a distributed monitoring system that aggregates CPU usage across thousands of servers. To detect hotspots, the system needs the maximum usage in every 5‑minute sliding window across a 2‑D grid of server racks. Using the deque‑based sliding window algorithm allows the system to update the maximum in constant time per new data point, ensuring low latency and high throughput.

When implementing, pre‑allocate the deques and reuse them across rows to avoid repeated memory allocations. Also, be careful with index calculations when transitioning from the row‑max matrix to the final column pass; off‑by‑one errors are common and can silently corrupt the result.

COMPLEXITY AT A GLANCE

⏱ Time:O(n·m)

💾 Space:O(n·m)

Core Theory — Why This Approach?

The core of this problem is the 2‑D sliding window maximum. A naive solution would iterate over every possible sub‑rectangle of the required size and scan all its elements, leading to a time complexity of O(n·m·k·l) where k×l is the window size. This quickly becomes infeasible for large grids. The optimal approach extends the 1‑D deque technique to two dimensions: first compute the maximum for every horizontal window of width k in each row using a monotonic deque, producing an intermediate matrix of row‑maxima. Then, treat each column of this intermediate matrix as a 1‑D array and apply the same deque technique to obtain the maximum for every vertical window of height l. The final result is the maximum over all k×l sub‑rectangles, achieved in O(n·m) time and O(n·m) auxiliary space, which is linear in the input size.

The key insight is that the maximum of a k×l window can be decomposed into a horizontal maximum followed by a vertical maximum. Because the maximum operation is associative and idempotent, we can safely split the window into two passes without missing any candidate. This decomposition allows us to reuse the efficient deque algorithm twice, rather than recomputing from scratch for each window.

In practice, this pattern is a staple for problems involving range queries on grids, such as image processing, heat‑map analysis, or real‑time monitoring dashboards. It demonstrates mastery of amortized analysis, data structure design, and algorithmic optimization—skills that are highly prized in technical interviews.

Interview Questions on This Problem

Q1How would you modify the 2‑D sliding window maximum algorithm if the grid is rotated by 90 degrees and you need to find the maximum in every k×l sub‑rectangle in the rotated view?

Rotating the grid by 90 degrees is equivalent to transposing and then reversing rows or columns. Instead of physically rotating, you can adjust the indices: treat the original grid’s columns as rows and vice versa. Then apply the same two‑pass deque algorithm on the transposed dimensions, ensuring you use the correct window sizes (k for the new rows, l for the new columns). This avoids extra memory and keeps the time complexity linear.

Q2A fintech platform needs to detect the highest transaction amount in any 7×7 block of a 10,000×10,000 transaction matrix. What is the most efficient way to answer this query in real time?

Preprocess the matrix using the 2‑D sliding window maximum algorithm to compute the maximum for every 7×7 block once, storing the results in a lookup table of size (n‑6)×(m‑6). Subsequent queries can then be answered in O(1) by indexing into this table. If memory is constrained, you can compute the maximum on demand using the deque approach, which still runs in O(n·m) for a single query but is acceptable if queries are infrequent.

Q3During a coding interview, you are asked to explain why a deque is used instead of a priority queue for the sliding window maximum. What points would you highlight?

A deque maintains elements in decreasing order and allows O(1) removal of the front element when it exits the window. In contrast, a priority queue would require O(log k) per removal and cannot efficiently discard elements that are no longer in the window unless you store indices and perform lazy deletions, which adds overhead. The deque’s amortized O(1) operations make it the optimal choice for sliding window problems.

Examples

Example 1

Input

grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]], k = 2

Output

Explanation: The grid is 3x3 and k=2. The possible 2x2 sub-rectangles are: top-left [[1,2],[4,5]] max=5; top-right [[2,3],[5,6]] max=6; bottom-left [[4,5],[7,8]] max=8; bottom-right [[5,6],[8,9]] max=9. The global maximum is 9.

Example 2

Input

grid = [[10, 20], [30, 40]], k = 2

Output

Explanation: The grid is 2x2 and k=2. There is only one 2x2 sub-rectangle, which is the entire grid. The maximum element in this sub-rectangle is 40.

Example 3

Input

grid = [[5, 1, 3], [2, 4, 6], [7, 8, 9]], k = 3

Output

Explanation: The grid is 3x3 and k=3. There is only one 3x3 sub-rectangle, which is the entire grid. The maximum element in this sub-rectangle is 9.

Example 4

Input

grid = [[1, 2], [3, 4]], k = 3

Output

-1

Explanation: The grid is 2x2 but k=3. Since k > n and k > m, no 3x3 sub-rectangle can fit within the grid. Therefore, the function returns -1.

Constraints

1 <= n, m <= 1000
1 <= k <= 1000
-10^9 <= grid[i][j] <= 10^9
The total number of elements in the grid is at most 10^6

Optimal Approach & Strategy

Compute horizontal sliding maxima with a deque for each row, then compute vertical sliding maxima on the intermediate matrix with another deque. This yields O(n·m) time and O(n·m) space.

Brute Force Approach

Loop over every possible k×l sub‑rectangle, scan all its elements, and keep track of the maximum. This takes O(n·m·k·l) time and is impractical for large grids.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n·m)

function solution(matrix) {
   let max = -Infinity;
   for (let i = 0; i < matrix.length; i++) {
       for (let j = 0; j < matrix[i].length; j++) {
           max = Math.max(max, matrix[i][j]);
       }
   }
   return max;
}

class Solution {
   public:
   int solution(vector<vector<int>>& matrix) {
       int max = INT_MIN;
       for (int i = 0; i < matrix.size(); i++) {
           for (int j = 0; j < matrix[i].size(); j++) {
               max = max(max, matrix[i][j]);
           }
       }
       return max;
   }
};

class Solution {
   public int solution(int[][] matrix) {
       int max = Integer.MIN_VALUE;
       for (int i = 0; i < matrix.length; i++) {
           for (int j = 0; j < matrix[i].length; j++) {
               max = Math.max(max, matrix[i][j]);
           }
       }
       return max;
   }
}

def solution(matrix):
   max = float('-inf')
   for i in range(len(matrix)):
       for j in range(len(matrix[i])):
           max = max(max, matrix[i][j])
   return max

function solution(matrix) {
   let max = -Infinity;
   for (let i = 0; i < matrix.length; i++) {
       for (let j = 0; j < matrix[i].length; j++) {
           max = Math.max(max, matrix[i][j]);
       }
   }
   return max;
}

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Rotated Matrix Pivot Analyzer 5 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Sliding Window

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Rotated Matrix Pivot Analyzer 5 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor