1386. Shift 2D Grid | DSA Solvebook

📋Problem

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

💡Solution

java

class Solution {
    public List<List<Integer>> shiftGrid(int[][] grid, int k) {
        int m = grid.length;
        int n = grid[0].length;

        int total = m * n;
        k %= total;

        List<List<Integer>> ans = new ArrayList<>();

        for (int i = 0; i < m; i++) {
            List<Integer> row = new ArrayList<>();
            for (int j = 0; j < n; j++)
                row.add(0);
            ans.add(row);
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {

                // index in 1D array (before rotation)
                int oldIndex = i * n + j;

                // index in 1D array (after rotation)
                int newIndex = (oldIndex + k) % total;

                // changing from 1D back to 2D
                int newRow = newIndex / n;
                int newCol = newIndex % n;

                ans.get(newRow).set(newCol, grid[i][j]);
            }
        }

        return ans;
    }
}

📋Problem

💡Solution

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