Number of rows and columns in a Matrix that contain repeated values

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:

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# 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))  

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Output:

(0, 2)

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