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:

C++

// Nikunj Sonigara

#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;

// Function to count the number of rows and columns
// that contain repeated values in a square matrix.
pair<int, int> repeated_val(int N, vector<vector<int>>& matrix) {
    int column = 0;
    int row = 0;

    for (int i = 0; i < N; i++) {
        // For every row, create an unordered_set to track unique elements.
        unordered_set<int> rowSet(matrix[i].begin(), matrix[i].end());

        // The size of the unordered_set should be equal to N for all unique elements.
        if (rowSet.size() != N) {
            row++;
        }

        // For every column, create an unordered_set to track unique elements.
        unordered_set<int> colSet;

        for (int j = 0; j < N; j++) {
            colSet.insert(matrix[j][i]);
        }

        // The size of the unordered_set should be equal to N for all unique elements.
        if (colSet.size() != N) {
            column++;
        }
    }

    // Returning the count of rows and columns
    return make_pair(row, column);
}

int main() {
    int N = 3;
    vector<vector<int>> matrix = {{2, 1, 3}, {1, 3, 2}, {1, 2, 3}};

    pair<int, int> result = repeated_val(N, matrix);

    cout << result.first << " " << result.second << endl;
    return 0;
}

Java

// Nikunj Sonigara

import java.util.HashSet;
import java.util.Set;

public class Main {
    public static void main(String[] args) {
        int N = 3;
        int[][] matrix = {{2, 1, 3}, {1, 3, 2}, {1, 2, 3}};

        int[] result = repeatedVal(N, matrix);

        System.out.println(result[0] + " " + result[1]);
    }

    public static int[] repeatedVal(int N, int[][] matrix) {
        int column = 0;
        int row = 0;

        for (int i = 0; i < N; i++) {
            // For every row, create a set to track unique elements.
            Set<Integer> rowSet = new HashSet<>();
            for (int j = 0; j < N; j++) {
                rowSet.add(matrix[i][j]);
            }

            // The size of the set should be equal to N for all unique elements.
            if (rowSet.size() != N) {
                row++;
            }

            // For every column, create a set to track unique elements.
            Set<Integer> colSet = new HashSet<>();
            for (int j = 0; j < N; j++) {
                colSet.add(matrix[j][i]);
            }

            // The size of the set should be equal to N for all unique elements.
            if (colSet.size() != N) {
                column++;
            }
        }

        return new int[] {row, column};
    }
}

Python3

# 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))

Javascript

// Nikunj Sonigara

function repeatedVal(N, matrix) {
    let column = 0;
    let row = 0;

    for (let i = 0; i < N; i++) {
        // For every row, create a Set to track unique elements.
        let rowSet = new Set(matrix[i]);

        // The size of the Set should be equal to N for all unique elements.
        if (rowSet.size !== N) {
            row++;
        }

        // For every column, create a Set to track unique elements.
        let colSet = new Set();
        for (let j = 0; j < N; j++) {
            colSet.add(matrix[j][i]);
        }

        // The size of the Set should be equal to N for all unique elements.
        if (colSet.size !== N) {
            column++;
        }
    }

    return [row, column];
}

function main() {
    const N = 3;
    const matrix = [[2, 1, 3], [1, 3, 2], [1, 2, 3]];

    const result = repeatedVal(N, matrix);

    console.log(result[0] + " " + result[1]);
}

main();

Output

(0, 2)