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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 togrid[i][j + 1]
. - Element at
grid[i][n - 1]
moves togrid[i + 1][0]
. - Element at
grid[m - 1][n - 1]
moves togrid[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; } }
-
// 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