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 = 4imageSwap(mat, n) printMatrix(mat, n) # This code is contributed by shubhamsingh10 |
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 rangeMAX = 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 = 4mat = [[(mat[j][i]) for j in range(n)] for i in range(n)] printMatrix(mat, n) # This code is contributed by vikkycirus |
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!
