Given a N x N square matrix arr[][] which contains only integers between 1 and N, the task is to compute the number of rows and the number of columns in the matrix that contain repeated values.
Examples:
Input: N = 4, arr[][] = {{1, 2, 3, 4}, {2, 1, 4, 3}, {3, 4, 1, 2}, {4, 3, 2, 1}}
Output: 0 0
Explanation:
None of the rows or columns contain repeated values.Input: N = 4, arr[][]= {{2, 2, 2, 2}, {2, 3, 2, 3}, {2, 2, 2, 3}, {2, 2, 2, 2}}
Output: 4 4
Explanation:
In every column and every row of the square matrix, the values are repeated.
Therefore, the total count is 4 for both rows and columns.
Approach: The idea is to use the NumPy library.
 Make a NumPy array of every row and every column in the square matrix.
 Find the length of the unique elements.
 If the length is equal to N then, there are no repeated values present in that particular row or column.
Below is the implementation of the above approach:
# Python program to count the number of # rows and columns in a square matrix # that contain repeated values import numpy as np
# Function to count the number of rows # and number of columns that contain # repeated values in a square matrix. def repeated_val(N, matrix):
column = 0
row = 0
for i in range (N):
# For every row, an array is formed.
# The length of the unique elements
# is calculated, which if not equal
# to 'N' then the row has repeated values.
if ( len (np.unique(np.array(matrix[i])))! = N):
row + = 1 # For every column, an array is formed.
# The length of the unique elements
# is calculated, which if not equal
# to N then the column has repeated values.
for j in range (N):
if ( len (np.unique(np.array([m[j] for m in matrix])))! = N):
column + = 1
# Returning the count of
# rows and columns
return row, column
# Driver code if __name__ = = '__main__' :
N = 3
matrix = [ [ 2 , 1 , 3 ], [ 1 , 3 , 2 ], [ 1 , 2 , 3 ] ]
print (repeated_val(N, matrix))

(0, 2)
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