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.

// 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) {
        queue<int> q;
        int M = A.size(), N = A[0].size();
        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;
    }
};

Java

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;
    }
}

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