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 <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));` ` ` `}` `}` ` ` `// 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

`<?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` `?>` |

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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