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Python Program for Mirror of matrix across diagonal

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Given a 2-D array of order N x N, print a matrix that is the mirror of the given tree across the diagonal. We need to print the result in a way: swap the values of the triangle above the diagonal with the values of the triangle below it like a mirror image swap. Print the 2-D array obtained in a matrix layout.

Examples:  

Input : int mat[][] = {{1 2 4 }
                       {5 9 0}
                       { 3 1 7}}
Output :  1 5 3 
          2 9 1
          4 0 7
Input : mat[][] = {{1  2  3  4 }
                   {5  6  7  8 }
                   {9  10 11 12}
                   {13 14 15 16} }
Output : 1 5 9 13 
         2 6 10 14  
         3 7 11 15 
         4 8 12 16 

A simple solution to this problem involves extra space. We traverse all right diagonal (right-to-left) one by one. During the traversal of the diagonal, first, we push all the elements into the stack and then we traverse it again and replace every element of the diagonal with the stack element. 

Below is the implementation of the above idea. 

Python3




# Simple Python3 program to find mirror of
# matrix across diagonal.
MAX = 100
  
  
def imageSwap(mat, n):
  
    # for diagonal which start from at
    # first row of matrix
    row = 0
  
    # traverse all top right diagonal
    for j in range(n):
  
        # here we use stack for reversing
        # the element of diagonal
        s = []
        i = row
        k = j
        while (i < n and k >= 0):
            s.append(mat[i][k])
            i += 1
            k -= 1
  
        # push all element back to matrix
        # in reverse order
        i = row
        k = j
        while (i < n and k >= 0):
            mat[i][k] = s[-1]
            k -= 1
            i += 1
            s.pop()
  
    # do the same process for all the
    # diagonal which start from last
    # column
    column = n - 1
    for j in range(1, n):
  
        # here we use stack for reversing
        # the elements of diagonal
        s = []
        i = j
        k = column
        while (i < n and k >= 0):
            s.append(mat[i][k])
            i += 1
            k -= 1
  
        # push all element back to matrix
        # in reverse order
        i = j
        k = column
        while (i < n and k >= 0):
            mat[i][k] = s[-1]
            i += 1
            k -= 1
            s.pop()
  
# Utility function to pra matrix
  
  
def printMatrix(mat, n):
    for i in range(n):
        for j in range(n):
            print(mat[i][j], end =" ")
        print()
  
  
# Driver code
mat = [[1, 2, 3, 4], [5, 6, 7, 8],
       [9, 10, 11, 12], [13, 14, 15, 16]]
n = 4
imageSwap(mat, n)
printMatrix(mat, n)
  
# This code is contributed by shubhamsingh10


Output

1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16 

Time complexity: O(n*n)

Method 2:

 we just have to swap (mat[i][j] to mat[j][i]). 

Below is the implementation of the above idea. 

Python3




# Efficient Python3 program to find mirror of
# matrix across diagonal.
from builtins import range
MAX = 100
  
  
def imageSwap(mat, n):
  
    # traverse a matrix and swap
    # mat[i][j] with mat[j][i]
    for i in range(n):
        for j in range(i + 1):
            t = mat[i][j]
            mat[i][j] = mat[j][i]
            mat[j][i] = t
  
# Utility function to pra matrix
  
  
def printMatrix(mat, n):
    for i in range(n):
        for j in range(n):
            print(mat[i][j], end =" ")
        print()
  
  
# Driver code
if __name__ == '__main__':
    mat = [1, 2, 3, 4], \
        [5, 6, 7, 8], \
        [9, 10, 11, 12], \
        [13, 14, 15, 16]
    n = 4
    imageSwap(mat, n)
    printMatrix(mat, n)


Output: 

1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16 

Time complexity: O(n*n)

Auxiliary space: O(1) as it is using constant space.

Method 3: Using List Comprehension

Python3




# Utility function to print matrix
def printMatrix(mat, n):
    for i in range(n):
        for j in range(n):
            print(mat[i][j], end =" ")
        print()
  
  
# Driver code
mat = [[1, 2, 3, 4], [5, 6, 7, 8],
       [9, 10, 11, 12], [13, 14, 15, 16]]
n = 4
mat = [[(mat[j][i]) for j in range(n)] for i in range(n)]
  
printMatrix(mat, n)
  
# This code is contributed by vikkycirus


Output

1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16 

Time Complexity: O(n*n)

Space Complexity: O(n)

Please refer complete article on Mirror of matrix across diagonal for more details!



Last Updated : 24 Mar, 2023
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