Skip to content
Related Articles
String with k distinct characters and no same characters adjacent
• Difficulty Level : Easy
• Last Updated : 04 May, 2021

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 one, 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 ``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));``    ``}``}` `// This article is contributed by vt_m.`

## 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

 ``

## Javascript

 ``

Output:

`ababa`

Time complexity : O(n)
This article is contributed by Raja Vikramaditya. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes arrow_drop_up