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# Lexicographically smallest string of maximum length made up of first K alphabets that does not contain any repeating substring

• Last Updated : 27 May, 2021

Given a positive integer K, the task is to find lexicographically the smallest string that can be generated by using the first K lowercase alphabets such that no substring of length at least 2 is repeated in the generated string.

Examples:

Input: K = 3
Output: aabacbbcca
Explanation:
In the string “aabacbbcca”, all possible substrings of length at least 2 is repeated more than once.

Input: K = 4

Approach: The given problem can be solved based on the following observations:

• If all substrings of length 2 are unique, then all substrings of length greater than 2 will also be unique.
• Hence, the maximum length string should contain all unique substrings of length 2 arrange in lexicographic order such that no 3 consecutive characters in the string are the same.

Follow the steps below to solve the problem:

• Initialize a string, say S as the empty string that stores the resultant string.
• Iterate over all the first K characters of the lowercase alphabet using the variable, say i and perform the following steps:
• Append the current character i to the string S.
• Iterate from the (i + 1)th character to the Kth character, and append the character i followed by character j to the string S.
• Add the character ‘a’ to the string S so that substring consisting of the last and the first alphabet is also present in the resultant string.
• After completing the above steps, print the string S as the resultant string.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the lexicographically``// smallest string of the first K lower``// case alphabets having unique substrings``void` `generateString(``int` `K)``{``    ``// Stores the resultant string``    ``string s = ``""``;` `    ``// Iterate through all the characters``    ``for` `(``int` `i = 97; i < 97 + K; i++) {` `        ``s = s + ``char``(i);` `        ``// Inner Loop for making pairs``        ``// and adding them into string``        ``for` `(``int` `j = i + 1;``             ``j < 97 + K; j++) {``            ``s += ``char``(i);``            ``s += ``char``(j);``        ``}``    ``}` `    ``// Adding first character so that``    ``// substring consisting of the last``    ``// the first alphabet is present``    ``s += ``char``(97);` `    ``// Print the resultant string``    ``cout << s;``}` `// Driver Code``int` `main()``{``    ``int` `K = 4;``    ``generateString(K);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{``    ` `// Function to find the lexicographically``// smallest string of the first K lower``// case alphabets having unique substrings``static` `void` `generateString(``int` `K)``{``    ` `    ``// Stores the resultant string``    ``String s = ``""``;``  ` `    ``// Iterate through all the characters``    ``for``(``int` `i = ``97``; i < ``97` `+ K; i++)``    ``{``        ``s = s + (``char``)(i);``  ` `        ``// Inner Loop for making pairs``        ``// and adding them into string``        ``for``(``int` `j = i + ``1``; j < ``97` `+ K; j++)``        ``{``            ``s += (``char``)(i);``            ``s += (``char``)(j);``        ``}``    ``}``  ` `    ``// Adding first character so that``    ``// substring consisting of the last``    ``// the first alphabet is present``    ``s += (``char``)(``97``);``  ` `    ``// Print the resultant string``    ``System.out.println(s);``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `K = ``4``;``    ` `    ``generateString(K);``}``}` `// This code is contributed by sanjoy_62`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to find the lexicographically``// smallest string of the first K lower``// case alphabets having unique substrings``static` `void` `generateString(``int` `K)``{``    ` `    ``// Stores the resultant string``    ``string` `s = ``""``;` `    ``// Iterate through all the characters``    ``for``(``int` `i = 97; i < 97 + K; i++)``    ``{``        ``s = s + (``char``)(i);` `        ``// Inner Loop for making pairs``        ``// and adding them into string``        ``for``(``int` `j = i + 1; j < 97 + K; j++)``        ``{``            ``s += (``char``)(i);``            ``s += (``char``)(j);``        ``}``    ``}` `    ``// Adding first character so that``    ``// substring consisting of the last``    ``// the first alphabet is present``    ``s += (``char``)(97);` `    ``// Print the resultant string``    ``Console.Write(s);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `K = 4;``    ``generateString(K);``}``}` `// This code is contributed by ukasp`

## Python3

 `# python 3 program for the above approach` `# Function to find the lexicographically``# smallest string of the first K lower``# case alphabets having unique substrings``def` `generateString(K):``  ` `    ``# Stores the resultant string``    ``s ``=` `""` `    ``# Iterate through all the chracters``    ``for` `i ``in` `range``(``97``,``97` `+` `K,``1``):``        ``s ``=` `s ``+` `chr``(i);` `        ``# Inner Loop for making pairs``        ``# and adding them into string``        ``for` `j ``in` `range``(i ``+` `1``,``97` `+` `K,``1``):``            ``s ``+``=` `chr``(i)``            ``s ``+``=` `chr``(j)` `    ``# Adding first chracter so that``    ``# substring consisting of the last``    ``# the first alphabet is present``    ``s ``+``=` `chr``(``97``)` `    ``# Print the resultant string``    ``print``(s)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``K ``=` `4``    ``generateString(K)` `    ``# This code is contributed by SURENDRA_GANGWAR.`

## Javascript

 ``
Output:
`aabacadbbcbdccdda`

Time Complexity: O(K2)
Auxiliary Space: O(K2)

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