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String with k distinct characters and no same characters adjacent

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Given n and k, print a string that has n characters. The string should have exactly k distinct characters and no adjacent positions.

Examples: 

Input  : n = 5, k = 3 
Output :  abcab
Explanation: 3 distinct character a, b, c
and n length string. 

Input: 3 2
Output: aba 
Explanation: 2 distinct character 'a' 
and 'b' and n length string.

Consider the first k Latin letters. We will add them to the answer in the order, firstly, we add a then b and so on. If letters are finished but the length of the answer is still less than the required ones, then we start again adding letters from the beginning of the alphabet. We repeat this process until the length of the answer becomes n and print it once done.

Below is the implementation of the above approach:

C++




// CPP program to construct a n length string
// with k distinct characters such that no two
// same characters are adjacent.
#include <iostream>
using namespace std;
  
// Function to find a string of length
// n with k distinct characters.
string findString(int n, int k)
{
    // Initialize result with first k
    // Latin letters
    string res = "";
    for (int i = 0; i < k; i++)
        res = res + (char)('a' + i);
  
    // Fill remaining n-k letters by
    // repeating k letters again and again.
    int count = 0;
    for (int i = 0; i < n - k; i++) {
        res = res + (char)('a' + count);
        count++;
        if (count == k)
            count = 0;
    }
    return res;
}
  
// Driver code
int main()
{
    int n = 5, k = 2;
    cout << findString(n, k);
    return 0;
}


Java




// Java program to construct a n length
// string with k distinct characters
// such that no two same characters
// are adjacent.
import java.io.*;
  
public class GFG {
      
    // Function to find a string of
    // length n with k distinct characters.
    static String findString(int n, int k)
    {
          
        // Initialize result with first k
        // Latin letters
        String res = "";
          
        for (int i = 0; i < k; i++)
            res = res + (char)('a' + i);
      
        // Fill remaining n-k letters by
        // repeating k letters again and 
        // again.
        int count = 0;
          
        for (int i = 0; i < n - k; i++)
        {
            res = res + (char)('a' + count);
            count++;
              
            if (count == k)
                count = 0;
        }
          
        return res;
    }
      
    // Driver code
    static public void main (String[] args)
    {
      
        int n = 5, k = 2;
          
        System.out.println(findString(n, k));
    }
}
  
// 


Python 3




# Python 3 program to construct a n 
# length string with k distinct characters 
# such that no two same characters are adjacent.
  
# Function to find a string of length
# n with k distinct characters.
def findString(n, k):
  
    # Initialize result with first k
    # Latin letters
    res = ""
    for i in range(k):
        res = res + chr(ord('a') + i)
  
    # Fill remaining n-k letters by
    # repeating k letters again and again.
    count = 0
    for i in range(n - k) :
        res = res + chr(ord('a') + count)
        count += 1
        if (count == k):
            count = 0;
      
    return res
  
# Driver code
if __name__ == "__main__":
      
    n = 5
    k = 2
    print(findString(n, k))
  
# This code is contributed by ita_c


C#




// C# program to construct a n length
// string with k distinct characters
// such that no two same characters
// are adjacent.
using System;
  
public class GFG {
      
    // Function to find a string
    // of length n with k distinct
    // characters.
    static string findString(int n, int k)
    {
          
        // Initialize result with
        // first k Latin letters
        string res = "";
          
        for (int i = 0; i < k; i++)
            res = res + (char)('a' + i);
      
        // Fill remaining n-k letters by
        // repeating k letters again and
        // again.
        int count = 0;
          
        for (int i = 0; i < n - k; i++)
        {
            res = res + (char)('a' + count);
            count++;
              
            if (count == k)
                count = 0;
        }
          
        return res;
    }
      
    // Driver code
    static public void Main ()
    {
          
        int n = 5, k = 2;
          
        Console.WriteLine(findString(n, k));
    }
}
  
// This code is contributed by vt_m.


PHP




<?php
// php program to construct a n length
// string with k distinct characters 
// such that no two same characters 
// are adjacent.
  
// Function to find a string of length
// n with k distinct characters.
function findString($n, $k)
{
    // Initialize result with first k
    // Latin letters
    $res = "";
    for ($i = 0; $i < $k; $i++)
        $res = $res . chr(ord('a') + $i);
  
    // Fill remaining n-k letters by
    // repeating k letters again and
    // again.
    $count = 0;
    for ($i = 0; $i < $n - $k; $i++) 
    {
        $res = $res . chr(ord('a'
                          + $count);
        $count++;
        if ($count == $k)
            $count = 0;
    }
    return $res;
}
  
// Driver code
    $n = 5;
    $k = 2;
      
    echo findString($n, $k);
          
// This code is contributed by Sam007
?>


Javascript




<script>
  
// Javascript program to construct a n length
// string with k distinct characters
// such that no two same characters
// are adjacent.
      
    // Function to find a string of
    // length n with k distinct characters.
    function findString(n, k)
    {
      
        // Initialize result with first k
        // Latin letters
        let res = "";
            
        for (let i = 0; i < k; i++)
            res = res + String.fromCharCode('a'.charCodeAt(0) + i);
        
        // Fill remaining n-k letters by
        // repeating k letters again and 
        // again.
        let count = 0;
            
        for (let i = 0; i < n - k; i++)
        {
            res = res + String.fromCharCode('a'.charCodeAt(0) + count);
            count++;
                
            if (count == k)
                count = 0;
        }
            
        return res;
    }
      
     // Driver code
     let n = 5, k = 2;
     document.write(findString(n, k));
      
    // This code is contributed by rag2127
</script>


Output

ababa

Time complexity : O(n), where n is the given integer.
Auxiliary Space: O(n), where n is the given integer.

This article is contributed by Raja Vikramaditya.  



Last Updated : 18 Sep, 2023
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