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Print an N x M matrix such that each row and column has all the vowels in it

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Given two integers N and M, the task is to print an N x M matrix such that each row and column contain all the vowels in it. If it is impossible to do so, then print -1

Examples: 

Input: N = 5, M = 5 
Output: 
a e i o u 
e i o u a 
i o u a e 
o u a e i 
u a e i o

Input: N = 6, M = 2 
Output: -1 

Approach: Since the number of vowels is 5, hence we need a minimum of 5 rows and 5 columns in order to generate a valid matrix. A pattern can be followed by filling “aeiouaeiou..” in the first row, “eiouaeio..” in the second row, and so on and the generated matrix will contain all the vowels in every row and column.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the required matrix
void printMatrix(int n, int m)
{
 
    // Impossible to generate
    // the required matrix
    if (n < 5 || m < 5) {
        cout << -1;
        return;
    }
 
    // Store all the vowels
    string s = "aeiou";
 
    // Print the matrix
    for (int i = 0; i < n; i++) {
 
        // Print vowels for every index
        for (int j = 0; j < m; j++) {
            cout << s[j % 5] << " ";
        }
        cout << endl;
        char c = s[0];
 
        // Shift the vowels by one
        for (int i = 0; i < 4; i++) {
            s[i] = s[i + 1];
        }
 
        s[4] = c;
    }
}
 
// Driver code
int main()
{
    int n = 5, m = 5;
 
    printMatrix(n, m);
 
    return 0;
}


Java




// Java implementation of the approach
import java.io.*;
 
public class GFG
{
 
// Function to print the required matrix
static void printMatrix(int n, int m)
{
 
    // Impossible to generate
    // the required matrix
    if (n < 5 || m < 5)
    {
        System.out.print(-1);
        return;
    }
 
    // Store all the vowels
    char[] s = "aeiou".toCharArray();
 
    // Print the matrix
    for (int i = 0; i < n; i++)
    {
 
        // Print vowels for every index
        for (int j = 0; j < m; j++)
        {
            System.out.print(s[j % 5] + " ");
        }
        System.out.println();
        char c = s[0];
 
        // Shift the vowels by one
        for (int k = 0; k < 4; k++)
        {
            s[k] = s[k + 1];
        }
 
        s[4] = c;
    }
}
 
// Driver code
public static void main(String[] args)
{
    int n = 5, m = 5;
 
    printMatrix(n, m);
}
}
 
// This code has been contributed by 29AjayKumar


Python3




# Python3 implementation of the approach
 
# Function to print the required matrix
def printMatrix(n, m) :
 
    # Impossible to generate
    # the required matrix
    if (n < 5 or m < 5) :
        print(-1,end = " ");
        return;
     
 
    # Store all the vowels
    s = "aeiou";
    s = list(s);
 
    # Print the matrix
    for i in range(n) :
         
        # Print vowels for every index
        for j in range(m) :
            print(s[j % 5],end= " ");
     
        print()
        c = s[0];
 
        # Shift the vowels by one
        for i in range(4) :
            s[i] = s[i + 1];
         
        s[4] = c;
 
 
# Driver code
if __name__ == "__main__" :
 
    n = 5; m = 5;
 
    printMatrix(n, m);
 
# This code is contributed by AnkitRai01


C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to print the required matrix
static void printMatrix(int n, int m)
{
 
    // Impossible to generate
    // the required matrix
    if (n < 5 || m < 5)
    {
        Console.Write(-1);
        return;
    }
 
    // Store all the vowels
    char[] s = "aeiou".ToCharArray();
 
    // Print the matrix
    for (int i = 0; i < n; i++)
    {
 
        // Print vowels for every index
        for (int j = 0; j < m; j++)
        {
            Console.Write(s[j % 5] + " ");
        }
        Console.WriteLine();
        char c = s[0];
 
        // Shift the vowels by one
        for (int k = 0; k < 4; k++)
        {
            s[k] = s[k + 1];
        }
 
        s[4] = c;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 5, m = 5;
 
    printMatrix(n, m);
}
}
 
/* This code contributed by PrinciRaj1992 */


PHP




<?php
// PHP implementation of the approach
 
// Function to print the required matrix
function printMatrix($n, $m)
{
 
    // Impossible to generate
    // the required matrix
    if ($n < 5 || $m < 5)
    {
        echo -1;
        return;
    }
 
    // Store all the vowels
    $s = "aeiou";
 
    // Print the matrix
    for ($i = 0; $i < $n; $i++)
    {
 
        // Print vowels for every index
        for ($j = 0; $j < $m; $j++)
        {
            echo $s[$j % 5] . " ";
        }
        echo "\n";
        $c = $s[0];
 
        // Shift the vowels by one
        for ($k = 0; $k < 4; $k++)
        {
            $s[$k] = $s[$k + 1];
        }
 
        $s[4] = $c;
    }
}
 
// Driver code
 
    $n = 5;
    $m = 5;
 
    printMatrix($n, $m);
 
    return 0;
     
// This code is contributed by ChitraNayal   
?>


Javascript




<script>
 
// JavaScript implementation of the approach   
 
// Function to print the required matrix
    function printMatrix(n , m) {
 
        // Impossible to generate
        // the required matrix
        if (n < 5 || m < 5) {
            document.write(-1);
            return;
        }
 
        // Store all the vowels
        var s = "aeiou";
 
        // Print the matrix
        for (var i = 0; i < n; i++) {
 
            // Print vowels for every index
            for (j = 0; j < m; j++) {
                document.write(s[j % 5] + " ");
            }
            document.write("<br/>");
            var c = s[0];
 
                s = s.substring(1,s.length)+s.substring(0,1);
         
 
            s[4] = c;
        }
    }
 
    // Driver code
     
        var n = 5, m = 5;
 
        printMatrix(n, m);
         
// This code contributed by Rajput-Ji
 
</script>


Output: 

a e i o u 
e i o u a 
i o u a e 
o u a e i 
u a e i o

 

Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
Auxiliary Space: O(1) as we are not using any extra space.



Last Updated : 08 Dec, 2022
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