What is the time complexity of Cyclically Rotating a Grid?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Cyclically Rotating a Grid?

The space complexity is O(m+n).

What data structures are used to solve Cyclically Rotating a Grid?

This problem uses Array, Matrix, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Cyclically Rotating a Grid?

Insight 1: The key is to process each layer separately. Identify the boundaries of each layer iteratively. Insight 2: Reduce the number of rotations 'k' by taking the modulo with the number of elements in the current layer (k % num_elements). This avoids unnecessary full rotations and improves efficiency.

Which companies ask LeetCode 1914. Cyclically Rotating a Grid in interviews?

This problem is frequently asked at Applied Intuition.

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Cyclically Rotating a Grid - LeetCode 1914 Solution

Cyclically Rotating a Grid - Complete Solution Guide

Cyclically Rotating a Grid is LeetCode problem 1914, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an m x n integer matrix grid , where m and n are both even integers, and an integer k . The matrix is composed of several layers, which is shown in the below image, where each color is its own layer: A cyclic rotation of the matrix is done by cyclically rotating each layer in the matrix. To cyclically rotate a layer once, each element in the layer will take the place of the adjacent element in the counter-clockwise direction. An example rotation is shown below: Return the matrix

Detailed Explanation

The problem asks you to rotate the layers of a given m x n integer matrix 'grid' cyclically 'k' times in the counter-clockwise direction. The matrix is composed of concentric rectangular layers, and each layer needs to be rotated independently. The inputs are the 'grid' matrix (where m and n are even) and an integer 'k' representing the number of rotations. The output is the modified 'grid' after performing 'k' rotations on each layer.

Solution Approach

The solution iterates through each layer of the grid from the outermost to the innermost layer. For each layer, it extracts the elements into a temporary list. Then, it rotates the elements in the list by k positions (using the modulo operator to handle large k values). Finally, it places the rotated elements back into the corresponding positions in the grid.

Step-by-Step Algorithm

Step 1: Initialize the boundaries of the grid (top, bottom, left, right).
Step 2: Iterate while the inner loop is valid (top < bottom and left < right). This loop processes each layer.
Step 3: Extract the elements of the current layer into a temporary list in counter-clockwise order: top row, right column, bottom row, left column.
Step 4: Calculate the effective number of rotations using k % num_elements, where num_elements is the number of elements in the current layer.
Step 5: Rotate the temporary list by the effective number of rotations.
Step 6: Place the rotated elements back into the grid in the same counter-clockwise order.
Step 7: Update the boundaries (top++, bottom--, left++, right--) to process the next inner layer.
Step 8: Return the modified grid.

Key Insights

Insight 1: The key is to process each layer separately. Identify the boundaries of each layer iteratively.
Insight 2: Reduce the number of rotations 'k' by taking the modulo with the number of elements in the current layer (k % num_elements). This avoids unnecessary full rotations and improves efficiency.
Insight 3: Store the elements of a layer in a temporary data structure (like a list or vector) to facilitate easy rotation and re-insertion.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m+n)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: Applied Intuition.

Cyclically Rotating a Grid | LeetCode 1914

Medium

Array Matrix Simulation

Time: O(m*n)Space: O(m+n)

Solution Code

Available in:

class Solution {
    public int[][] rotateGrid(int[][] grid, int k) {
        int m = grid.length;
        int n = grid[0].length;
        int top = 0, bottom = m - 1, left = 0, right = n - 1;

        while (top < bottom && left < right) {
            List<Integer> layerElements = new ArrayList<>();

            // Extract top row
            for (int c = left; c < right; c++) {
                layerElements.add(grid[top][c]);
            }
            // Extract right column
            for (int r = top; r < bottom; r++) {
                layerElements.add(grid[r][right]);
            }
            // Extract bottom row
            for (int c = right; c > left; c--) {
                layerElements.add(grid[bottom][c]);
            }
            // Extract left column
            for (int r = bottom; r > top; r--) {
                layerElements.add(grid[r][left]);
            }

            int numElements = layerElements.size();
            int effectiveK = k % numElements;

            List<Integer> rotatedLayer = new ArrayList<>();
            for(int i = effectiveK; i < numElements; i++){
                rotatedLayer.add(layerElements.get(i));
            }
            for(int i = 0; i < effectiveK; i++){
                rotatedLayer.add(layerElements.get(i));
            }

            int idx = 0;
            // Place back top row
            for (int c = left; c < right; c++) {
                grid[top][c] = rotatedLayer.get(idx);
                idx++;
            }
            // Place back right column
            for (int r = top; r < bottom; r++) {
                grid[r][right] = rotatedLayer.get(idx);
                idx++;
            }
            // Place back bottom row
            for (int c = right; c > left; c--) {
                grid[bottom][c] = rotatedLayer.get(idx);
                idx++;
            }
            // Place back left column
            for (int r = bottom; r > top; r--) {
                grid[r][left] = rotatedLayer.get(idx);
                idx++;
            }

            top++;
            bottom--;
            left++;
            right--;
        }

        return grid;
    }
}

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Problem Info

Description

Problem 1914: Cyclically Rotating a Grid - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

Applied Intuition

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1914

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize the boundaries of the grid (top, bottom, left, right).
2
Step 2: Iterate while the inner loop is valid (top < bottom and left < right). This loop processes each layer.
3
Step 3: Extract the elements of the current layer into a temporary list in counter-clockwise order: top row, right column, bottom row, left column.
4
Step 4: Calculate the effective number of rotations using k % num_elements, where num_elements is the number of elements in the current layer.
5
Step 5: Rotate the temporary list by the effective number of rotations.
6
Step 6: Place the rotated elements back into the grid in the same counter-clockwise order.
7
Step 7: Update the boundaries (top++, bottom--, left++, right--) to process the next inner layer.
8
Step 8: Return the modified grid.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(m+n)

📝

Why?

The time complexity is O(m*n) because we visit each element in the grid once to extract, rotate, and place the elements. Although there is a loop within a loop to extract elements from the borders, each element is visited only once during the extraction and again during the placing process. Space complexity is O(m+n) because, in the worst-case scenario (outer layer), the number of elements in a layer is proportional to the sum of the number of rows and columns. Therefore, the space required to store the temporary list is O(m+n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

In-place rotation

Instead of extracting the layer elements into a separate list, one could attempt to perform the rotation directly within the grid using a series of swaps. This eliminates the O(m+n) space complexity but makes the implementation more complex and error-prone.

Time: O(m*n)Space: O(1)

Trade-offs: This approach saves space but makes the code more difficult to understand, debug, and maintain. It's only beneficial if memory usage is extremely restricted.

✨

Key Insights

1Insight 1: The key is to process each layer separately. Identify the boundaries of each layer iteratively.
2Insight 2: Reduce the number of rotations 'k' by taking the modulo with the number of elements in the current layer (k % num_elements). This avoids unnecessary full rotations and improves efficiency.
3Insight 3: Store the elements of a layer in a temporary data structure (like a list or vector) to facilitate easy rotation and re-insertion.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: k is larger than the number of elements in a layer. The modulo operator (%) handles this case by only performing the effective rotations.
Edge case 2: Single element layer (when top == bottom and left == right). The while loop condition (top < bottom && left < right) prevents this layer from being processed, effectively skipping it, which is the correct behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using the modulo operator (%) to calculate the effective number of rotations. This can lead to unnecessary iterations and potential time limit exceeded errors, especially when k is very large.

✗Common mistake 2: Incorrectly extracting or placing the elements of a layer, leading to incorrect rotations or missing elements. Carefully double-check the loop bounds and indexing when handling the top row, right column, bottom row, and left column.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Image processing. Rotating layers of an image or portions of an image can be useful for various effects and transformations.

🏦Real-world scenario 2: Data encryption. Cyclically shifting data within a matrix or data structure could be used as a basic encryption technique (although this example is not robust and would require significantly more sophisticated implementations for secure encryption).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Cyclically Rotating a Grid"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether matrix techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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