Formatted question description: https://leetcode.ca/all/1260.html

1260. Shift 2D Grid (Easy)

Given a 2D grid of size m x n and an integer k. You need to shift the grid k times.

In one shift operation:

Element at grid[i][j] moves to grid[i][j + 1].
Element at grid[i][n - 1] moves to grid[i + 1][0].
Element at grid[m - 1][n - 1] moves to grid[0][0].

Return the 2D grid after applying shift operation k times.

Example 1:

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

Example 2:

Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]

Example 3:

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

Constraints:

m == grid.length
n == grid[i].length
1 <= m <= 50
1 <= n <= 50
-1000 <= grid[i][j] <= 1000
0 <= k <= 100

Related Topics:
Array

Solution 1.

class Solution {
    public List<List<Integer>> shiftGrid(int[][] grid, int k) {
        int rows = grid.length, columns = grid[0].length;
        int size = rows * columns;
        k %= size;
        List<List<Integer>> shifted = new ArrayList<List<Integer>>();
        for (int i = 0; i < rows; i++)
            shifted.add(new ArrayList<Integer>());
        int index = (size - k) % size;
        for (int i = 0; i < size; i++) {
            int num = grid[index / columns][index % columns];
            shifted.get(i / columns).add(num);
            index = (index + 1) % size;
        }
        return shifted;
    }
}

############

class Solution {
    public List<List<Integer>> shiftGrid(int[][] grid, int k) {
        int m = grid.length, n = grid[0].length;
        k %= (m * n);
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < m; ++i) {
            List<Integer> t = new ArrayList<>();
            for (int j = 0; j < n; ++j) {
                t.add(0);
            }
            ans.add(t);
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int t = (i * n + j + k) % (m * n);
                ans.get(t / n).set(t % n, grid[i][j]);
            }
        }
        return ans;
    }
}

// OJ: https://leetcode.com/problems/shift-2d-grid/
// Time: O(MN)
// Space: O(K)
class Solution {
public:
    vector<vector<int>> shiftGrid(vector<vector<int>>& A, int k) {
        int M = A.size(), N = A[0].size();
        k %= M * N;
        if (k == 0) return A;
        queue<int> q;
        for (int i = 0; i < M * N + k; ++i) {
            int j = i % (M * N), x = j / N, y = j % N;
            if (i < k) q.push(A[x][y]);
            else {
                q.push(A[x][y]);
                A[x][y] = q.front();
                q.pop();
            }
        }
        return A;
    }
};

class Solution:
    def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        k %= m * n
        t = [grid[i][j] for i in range(m) for j in range(n)]
        t = t[-k:] + t[:-k]
        for i in range(m):
            for j in range(n):
                grid[i][j] = t[i * n + j]
        return grid

############

# 1260. Shift 2D Grid
# https://leetcode.com/problems/shift-2d-grid/

class Solution:
    def shiftGrid(self, grid: List[List[int]], k: int):
        col, nums = len(grid[0]), sum(grid, [])
        k = k % len(nums)
        nums = nums[-k:] + nums[:-k]
        return [nums[i:i+col] for i in range(0, len(nums), col)]

    def shiftGrid(self, grid: List[List[int]], k: int):
        rows, cols = len(grid), len(grid[0])
        
        for _ in range(k):
            res = [[0]*cols for _ in range(rows)]
            for i in range(rows):
                if i + 1 < rows and cols > 0:
                    res[i+1][0] = grid[i][cols-1]
                    
                for j in range(cols):
                    if j + 1 < cols:
                        res[i][j+1] = grid[i][j]

            if rows > 0 and cols > 0:
                res[0][0] = grid[rows-1][cols-1]
            
            grid = res
            
        return grid

func shiftGrid(grid [][]int, k int) [][]int {
	m, n := len(grid), len(grid[0])
	ans := make([][]int, m)
	for i := range ans {
		ans[i] = make([]int, n)
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			t := (i*n + j + k) % (m * n)
			ans[t/n][t%n] = grid[i][j]
		}
	}
	return ans
}

function shiftGrid(grid: number[][], k: number): number[][] {
    const m = grid.length,
        n = grid[0].length;
    const size = m * n;
    k = k % size;
    if (k == 0 || size <= 1) return grid;
    let arr = grid.flat();
    if (size <= 1) return grid;
    let ans = Array.from({ length: m }, v => new Array(n));
    for (let i = 0, j = size - k; i < size; i++) {
        ans[Math.floor(i / n)][i % n] = arr[j];
        j = j == size - 1 ? 0 : j + 1;
    }
    return ans;
}

1260 - Shift 2D Grid

1260. Shift 2D Grid (Easy)

Solution 1.

All Problems

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1260. Shift 2D Grid (Easy)

Solution 1.

All Problems

All Solutions