Number of rows and columns in a Matrix that contain repeated values
Last Updated :
10 Jan, 2024
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;
}
// 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};
}
}
def repeated_val(N, matrix):
# Initialize counters for rows and columns with repeated values
column = 0
row = 0
# Iterate through each row in the matrix
for i in range(N):
# For every row, create a set to track unique elements.
row_set = set(matrix[i])
# The length of the set should be equal to N for all unique elements.
# If not, increment the row counter.
if len(row_set) != N:
row += 1
# For every column, create a set to track unique elements.
col_set = set()
# Iterate through each element in the current column
for j in range(N):
col_set.add(matrix[j][i])
# The length of the set should be equal to N for all unique elements.
# If not, increment the column counter.
if len(col_set) != N:
column += 1
# Returning the count of rows and columns with repeated values
return row, column
if __name__ == '__main__':
# Sample input
N = 3
matrix = [[2, 1, 3], [1, 3, 2], [1, 2, 3]]
# Call the function and get the result
result = repeated_val(N, matrix)
# Print the result
print(result[0], result[1])
using System;
using System.Collections.Generic;
class Program {
// Function to count the number of rows and columns
// that contain repeated values in a square matrix.
static Tuple<int, int>
RepeatedValues(int N, List<List<int> > matrix)
{
int column = 0;
int row = 0;
for (int i = 0; i < N; i++) {
// For every row, create a HashSet to track
// unique elements.
HashSet<int> rowSet
= new HashSet<int>(matrix[i]);
// The size of the HashSet should be equal to N
// for all unique elements.
if (rowSet.Count != N) {
row++;
}
// For every column, create a HashSet to track
// unique elements.
HashSet<int> colSet = new HashSet<int>();
for (int j = 0; j < N; j++) {
colSet.Add(matrix[j][i]);
}
// The size of the HashSet should be equal to N
// for all unique elements.
if (colSet.Count != N) {
column++;
}
}
// Returning the count of rows and columns
return new Tuple<int, int>(row, column);
}
static void Main()
{
int N = 3;
List<List<int> > matrix = new List<List<int> >{
new List<int>{ 2, 1, 3 },
new List<int>{ 1, 3, 2 },
new List<int>{ 1, 2, 3 }
};
Tuple<int, int> result = RepeatedValues(N, matrix);
Console.WriteLine(result.Item1 + " "
+ result.Item2);
}
}
// 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();
Time Complexity: O(N2)
Auxiliary Space: O(N), where N is the size of the square matrix
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