LeetCode 1260: Shift 2D Grid
LeetCode 1260 Solution Explanation
Explanation
To shift the 2D grid k times, we can first flatten the grid into a 1D list, then perform the shifts in this 1D list, and finally reshape the list back into the original 2D grid shape. Shifting k times is equivalent to shifting k % (m * n) times.
- Flatten the grid into a 1D list.
- Perform k shifts on the 1D list.
- Reshape the modified 1D list back into the 2D grid shape.
Time complexity: O(m * n) where m is the number of rows and n is the number of columns in the grid. Space complexity: O(m * n) for storing the flattened grid.
LeetCode 1260 Solutions in Java, C++, Python
class Solution {
public List<List<Integer>> shiftGrid(int[][] grid, int k) {
int m = grid.length;
int n = grid[0].length;
List<List<Integer>> result = new ArrayList<>();
// Flatten the grid into a 1D list
List<Integer> flatGrid = new ArrayList<>();
for (int[] row : grid) {
for (int num : row) {
flatGrid.add(num);
}
}
// Perform k shifts on the 1D list
k = k % (m * n);
Collections.rotate(flatGrid, k);
// Reshape the modified 1D list back into the 2D grid shape
for (int i = 0; i < m; i++) {
List<Integer> row = new ArrayList<>();
for (int j = 0; j < n; j++) {
row.add(flatGrid.get(i * n + j));
}
result.add(row);
}
return result;
}
}Interactive Code Editor for LeetCode 1260
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