Answer to Question #262020 in Python for prime

Question #262020

For this program, I want an exact output with all test cases passed and it should not have any brackets.


def rotate1(matrix, n):

 a = len(matrix)

 n = int(a)

 b = int(n/2)

 

 for row in range(0, b):

 for col in range(row, (n- row - 1)):

  temp = matrix[row][col]

  matrix[row][col] = matrix[n - 1 - col][row] 

  matrix[n - 1 - col][row] = matrix[n - 1 - row][n- 1 - col] 

  matrix[n - 1 - row][n - 1 - col] = matrix[col][n - 1 - row] 

  matrix[col][n - 1 - row] = temp 

 return matrix


matrix = [[1, 2, 3, 4],[5, 6, 7, 8],[9, 10, 11, 12],[13, 14, 15, 16]]

n = len(matrix)

array = rotate1(matrix, n)

for col in array:

 print(col, sep=" ")


Sample Input 1

4 4

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16

3

Sample Output 1

13 9 5 1

14 7 11 2

15 6 10 3

16 12 8 4

Sample Input 2

3 4

1 2 3 4

10 11 12 5

9 8 7 6

2

Sample Output 2

9 10 1 2

8 11 12 3

7 6 5 4

 



1
Expert's answer
2021-11-06T12:07:45-0400
def rotate1(matrix, n):


    a = len(matrix)
    
    n = int(a)
    
    b = int(n/2)


    for row in range(0, b):
    
        for col in range(row, (n- row - 1)):
    
            temp = matrix[row][col]
            
            matrix[row][col] = matrix[n - 1 - col][row] 
            
            matrix[n - 1 - col][row] = matrix[n - 1 - row][n- 1 - col] 
            
            matrix[n - 1 - row][n - 1 - col] = matrix[col][n - 1 - row] 
            
            matrix[col][n - 1 - row] = temp 


    return matrix






matrix = [[1, 2, 3, 4],[5, 6, 7, 8],[9, 10, 11, 12],[13, 14, 15, 16]]


n = len(matrix)


array = rotate1(matrix, n)


for i in range(4):
    for j in range(4):
        print(array[i][j],end=" ")
    print()




 

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