LeetCode #1914 — MEDIUM

Cyclically Rotating a Grid

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 after applying k cyclic rotations to it.

Example 1:

Input: grid = [[40,10],[30,20]], k = 1
Output: [[10,20],[40,30]]
Explanation: The figures above represent the grid at every state.

Example 2:

Input: grid = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]], k = 2
Output: [[3,4,8,12],[2,11,10,16],[1,7,6,15],[5,9,13,14]]
Explanation: The figures above represent the grid at every state.

Constraints:

m == grid.length
n == grid[i].length
2 <= m, n <= 50
Both m and n are even integers.
1 <= grid[i][j] <=5000
1 <= k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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 after applying k cyclic rotations to it.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[40,10],[30,20]]
1

Example 2

[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]
2

Step 02

Core Insight

What unlocks the optimal approach

First, you need to consider each layer separately as an array.
Just cycle this array and then re-assign it.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1914: Cyclically Rotating a Grid
class Solution {
    private int m;
    private int n;
    private int[][] grid;

    public int[][] rotateGrid(int[][] grid, int k) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        for (int p = 0; p < Math.min(m, n) / 2; ++p) {
            rotate(p, k);
        }
        return grid;
    }

    private void rotate(int p, int k) {
        List<Integer> nums = new ArrayList<>();
        for (int j = p; j < n - p - 1; ++j) {
            nums.add(grid[p][j]);
        }
        for (int i = p; i < m - p - 1; ++i) {
            nums.add(grid[i][n - p - 1]);
        }
        for (int j = n - p - 1; j > p; --j) {
            nums.add(grid[m - p - 1][j]);
        }
        for (int i = m - p - 1; i > p; --i) {
            nums.add(grid[i][p]);
        }
        int l = nums.size();
        k %= l;
        if (k == 0) {
            return;
        }
        for (int j = p; j < n - p - 1; ++j) {
            grid[p][j] = nums.get(k++ % l);
        }
        for (int i = p; i < m - p - 1; ++i) {
            grid[i][n - p - 1] = nums.get(k++ % l);
        }
        for (int j = n - p - 1; j > p; --j) {
            grid[m - p - 1][j] = nums.get(k++ % l);
        }
        for (int i = m - p - 1; i > p; --i) {
            grid[i][p] = nums.get(k++ % l);
        }
    }
}

// Accepted solution for LeetCode #1914: Cyclically Rotating a Grid
func rotateGrid(grid [][]int, k int) [][]int {
	m, n := len(grid), len(grid[0])

	rotate := func(p, k int) {
		nums := []int{}
		for j := p; j < n-p-1; j++ {
			nums = append(nums, grid[p][j])
		}
		for i := p; i < m-p-1; i++ {
			nums = append(nums, grid[i][n-p-1])
		}
		for j := n - p - 1; j > p; j-- {
			nums = append(nums, grid[m-p-1][j])
		}
		for i := m - p - 1; i > p; i-- {
			nums = append(nums, grid[i][p])
		}
		l := len(nums)
		k %= l
		if k == 0 {
			return
		}
		for j := p; j < n-p-1; j++ {
			grid[p][j] = nums[k]
			k = (k + 1) % l
		}
		for i := p; i < m-p-1; i++ {
			grid[i][n-p-1] = nums[k]
			k = (k + 1) % l
		}
		for j := n - p - 1; j > p; j-- {
			grid[m-p-1][j] = nums[k]
			k = (k + 1) % l
		}
		for i := m - p - 1; i > p; i-- {
			grid[i][p] = nums[k]
			k = (k + 1) % l
		}
	}

	for i := 0; i < m/2 && i < n/2; i++ {
		rotate(i, k)
	}
	return grid
}

# Accepted solution for LeetCode #1914: Cyclically Rotating a Grid
class Solution:
    def rotateGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
        def rotate(p: int, k: int):
            nums = []
            for j in range(p, n - p - 1):
                nums.append(grid[p][j])
            for i in range(p, m - p - 1):
                nums.append(grid[i][n - p - 1])
            for j in range(n - p - 1, p, -1):
                nums.append(grid[m - p - 1][j])
            for i in range(m - p - 1, p, -1):
                nums.append(grid[i][p])
            k %= len(nums)
            if k == 0:
                return
            nums = nums[k:] + nums[:k]
            k = 0
            for j in range(p, n - p - 1):
                grid[p][j] = nums[k]
                k += 1
            for i in range(p, m - p - 1):
                grid[i][n - p - 1] = nums[k]
                k += 1
            for j in range(n - p - 1, p, -1):
                grid[m - p - 1][j] = nums[k]
                k += 1
            for i in range(m - p - 1, p, -1):
                grid[i][p] = nums[k]
                k += 1

        m, n = len(grid), len(grid[0])
        for p in range(min(m, n) >> 1):
            rotate(p, k)
        return grid

// Accepted solution for LeetCode #1914: Cyclically Rotating a Grid
/**
 * [1914] Cyclically Rotating a Grid
 *
 * 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:
 *
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/06/10/ringofgrid.png" style="width: 231px; height: 258px;" />
 *
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/06/22/explanation_grid.jpg" style="width: 500px; height: 268px;" />
 * Return the matrix after applying k cyclic rotations to it.
 *
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/06/19/rod2.png" style="width: 421px; height: 191px;" />
 *
 * Input: grid = [[40,10],[30,20]], k = 1
 * Output: [[10,20],[40,30]]
 * Explanation: The figures above represent the grid at every state.
 *
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/06/10/ringofgrid5.png" style="width: 231px; height: 262px;" /> <img alt="" src="https://assets.leetcode.com/uploads/2021/06/10/ringofgrid6.png" style="width: 231px; height: 262px;" /> <img alt="" src="https://assets.leetcode.com/uploads/2021/06/10/ringofgrid7.png" style="width: 231px; height: 262px;" />
 *
 *
 * Input: grid = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]], k = 2
 * Output: [[3,4,8,12],[2,11,10,16],[1,7,6,15],[5,9,13,14]]
 * Explanation: The figures above represent the grid at every state.
 *
 *
 *  
 * Constraints:
 *
 *
 * 	m == grid.length
 * 	n == grid[i].length
 * 	2 <= m, n <= 50
 * 	Both m and n are even integers.
 * 	1 <= grid[i][j] <=^ 5000
 * 	1 <= k <= 10^9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/cyclically-rotating-a-grid/
// discuss: https://leetcode.com/problems/cyclically-rotating-a-grid/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn rotate_grid(grid: Vec<Vec<i32>>, k: i32) -> Vec<Vec<i32>> {
        let k = k as usize;
        let mut grid = grid;

        let m = grid.len();
        let n = grid[0].len();

        let num_rings = m.min(n) / 2;

        for i in 0..num_rings {
            let num_rotations = k % (2 * (m + n - 4 * i) - 4);
            for _ in 0..num_rotations {
                for j in i..n - i - 1 {
                    grid[i].swap(j, j + 1);
                }

                for j in i..m - i - 1 {
                    let tmp = grid[j][n - i - 1];
                    grid[j][n - i - 1] = grid[j + 1][n - i - 1];
                    grid[j + 1][n - i - 1] = tmp;
                }

                let mut j = n - i - 1;

                while j > i {
                    grid[m - i - 1].swap(j, j - 1);
                    j -= 1;
                }

                let mut j = m - i - 1;

                while j > i + 1 {
                    let tmp = grid[j][i];
                    grid[j][i] = grid[j - 1][i];
                    grid[j - 1][i] = tmp;
                    j -= 1;
                }
            }
        }

        grid
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1914_example_1() {
        let grid = vec![vec![40, 10], vec![30, 20]];
        let k = 1;

        let result = vec![vec![10, 20], vec![40, 30]];

        assert_eq!(Solution::rotate_grid(grid, k), result);
    }

    #[test]
    fn test_1914_example_2() {
        let grid = vec![
            vec![1, 2, 3, 4],
            vec![5, 6, 7, 8],
            vec![9, 10, 11, 12],
            vec![13, 14, 15, 16],
        ];
        let k = 2;

        let result = vec![
            vec![3, 4, 8, 12],
            vec![2, 11, 10, 16],
            vec![1, 7, 6, 15],
            vec![5, 9, 13, 14],
        ];

        assert_eq!(Solution::rotate_grid(grid, k), result);
    }
}

// Accepted solution for LeetCode #1914: Cyclically Rotating a Grid
function rotateGrid(grid: number[][], k: number): number[][] {
    const m = grid.length;
    const n = grid[0].length;
    const rotate = (p: number, k: number) => {
        const nums: number[] = [];
        for (let j = p; j < n - p - 1; ++j) {
            nums.push(grid[p][j]);
        }
        for (let i = p; i < m - p - 1; ++i) {
            nums.push(grid[i][n - p - 1]);
        }
        for (let j = n - p - 1; j > p; --j) {
            nums.push(grid[m - p - 1][j]);
        }
        for (let i = m - p - 1; i > p; --i) {
            nums.push(grid[i][p]);
        }
        const l = nums.length;
        k %= l;
        if (k === 0) {
            return;
        }
        for (let j = p; j < n - p - 1; ++j) {
            grid[p][j] = nums[k++ % l];
        }
        for (let i = p; i < m - p - 1; ++i) {
            grid[i][n - p - 1] = nums[k++ % l];
        }
        for (let j = n - p - 1; j > p; --j) {
            grid[m - p - 1][j] = nums[k++ % l];
        }
        for (let i = m - p - 1; i > p; --i) {
            grid[i][p] = nums[k++ % l];
        }
    };
    for (let p = 0; p < Math.min(m, n) >> 1; ++p) {
        rotate(p, k);
    }
    return grid;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.