Given a 2D array, print it in reverse spiral form. We have already discussed Print a given matrix in spiral form. This article discusses how to do the reverse printing. See the following examples.
Input:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output:
10 11 7 6 5 9 13 14 15 16 12 8 4 3 2 1
Input:
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
Output:
11 10 9 8 7 13 14 15 16 17 18 12 6 5 4 3 2 1
Python3
# Python3 Code to Print a given # matrix in reverse spiral form# This is a modified code of# https:#www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/R, C = 3, 6def ReversespiralPrint(m, n, a): # Large array to initialize it # with elements of matrix b = [0 for i in range(100)] #/* k - starting row index #l - starting column index*/ i, k, l = 0, 0, 0 # Counter for single dimension array # in which elements will be stored z = 0 # Total elements in matrix size = m * n while (k < m and l < n): # Variable to store value of matrix. val = 0 # Print the first row # from the remaining rows for i in range(l, n): # printf("%d ", a[k][i]) val = a[k][i] b[z] = val z += 1 k += 1 # Print the last column # from the remaining columns for i in range(k, m): # printf("%d ", a[i][n-1]) val = a[i][n - 1] b[z] = val z += 1 n -= 1 # Print the last row # from the remaining rows if (k < m): for i in range(n - 1, l - 1, -1): # printf("%d ", a[m-1][i]) val = a[m - 1][i] b[z] = val z += 1 m -= 1 # Print the first column # from the remaining columns if (l < n): for i in range(m - 1, k - 1, -1): # printf("%d ", a[i][l]) val = a[i][l] b[z] = val z += 1 l += 1 for i in range(size - 1, -1, -1): print(b[i], end = " ")# Driver Codea = [[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18]]ReversespiralPrint(R, C, a)# This code is contributed by mohit kumar |
Output:
11 10 9 8 7 13 14 15 16 17 18 12 6 5 4 3 2 1
Time complexity: O(m*n) where m is number of rows and n is number of columns of a given matrix
Auxiliary Space: O(1) because constant space has been used
Please refer complete article on Print a given matrix in reverse spiral form for more details!
Using Recursion:
Approach:
This approach uses recursion to traverse the matrix in reverse spiral form. We start by printing the last row from right to left, then the last column from bottom to top, then the first row from left to right, and finally the first column from top to bottom. We then recursively call the function on the remaining submatrix.
The function print_reverse_spiral takes a matrix as input and initializes variables m and n to hold the number of rows and columns in the matrix, respectively.
The function defines a helper function called print_spiral_helper which takes four parameters: row_start, row_end, col_start, and col_end. These parameters define the boundaries of the current spiral.
The helper function first checks if the current spiral is valid (i.e., if row_start is less than or equal to row_end and col_start is less than or equal to col_end). If the spiral is not valid, the function simply returns.
The helper function then prints the elements of the last row of the spiral in reverse order, followed by the elements of the first column of the spiral (excluding the last element), followed by the elements of the first row of the spiral (excluding the first element), and finally the elements of the last column of the spiral (excluding the last element).
The helper function then calls itself recursively with updated parameters to traverse the next inner spiral.
Finally, the print_reverse_spiral function calls the print_spiral_helper function with the initial boundaries of the matrix to print its elements in reverse spiral order.
Python3
def print_reverse_spiral(matrix): m = len(matrix) n = len(matrix[0]) def print_spiral_helper(row_start, row_end, col_start, col_end): if row_start > row_end or col_start > col_end: return for j in range(col_end, col_start - 1, -1): print(matrix[row_end][j], end=' ') for i in range(row_end - 1, row_start - 1, -1): print(matrix[i][col_start], end=' ') for j in range(col_start + 1, col_end + 1): print(matrix[row_start][j], end=' ') for i in range(row_start + 1, row_end): print(matrix[i][col_end], end=' ') print_spiral_helper(row_start + 1, row_end - 1, col_start + 1, col_end - 1) print_spiral_helper(0, m - 1, 0, n - 1)matrix = [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]print_reverse_spiral(matrix) |
16 15 14 13 9 5 1 2 3 4 8 12 11 10 6 7
Time Complexity: O(mn) – where m is the number of rows and n is the number of columns in the matrix.
Auxiliary Space: O(mn) – due to the recursive call stack.
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