# Contiguous subsegments of a string having distinct subsequent characters

• Last Updated : 20 Dec, 2022

Given string str of length L and an integer N, the task is to form a total of (L / N) contiguous sub-segments of the string which contain distinct subsequent characters.
Note: that the integer N will be a factor of the length of the string i.e L.
Examples:

Input: str = “geeksforgeeksgfg”, N = 4
Output:
gek
sfor
gek
sgf
The length of “geeksforgeeksgfg” is 16, therefore there will be 4 subsegments.
The first subsegment will contain the characters g, e, e and k but the alphabet ‘e’ is repeated,
So we discard one ‘e’. Therefore, the final subsegment will be “gek”.
Similarly, the other three subsegments will be “sfor”, “gek” and “sgf”.
Each subsegment should have subsequent distinct characters.

Input: str = “aabdekfgf”, N = 3
Output:
ab
dek
fg

Approach: An array is created every time the iteration is started over the new sub segment. Those elements are stored in a sequence in that array. If any element already exists in the array, then it is not pushed into the array.
Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the approach``#include``using` `namespace` `std;` `// Function that prints the segments``void` `sub_segments(string str, ``int` `n)``{``    ``int` `l = str.length();``    ``for` `(``int` `x = 0; x < l; x += n)``    ``{``        ``string newlist = str.substr(x, n);``        ` `        ``// New array for every iteration``        ``list<``char``> arr;` `        ``// Iterator for new array``        ``list<``char``>::iterator it;` `        ``for``(``auto` `y:newlist)``        ``{``            ``it = find(arr.begin(), arr.end(), y);``            ` `            ``// Check if iterator points to end or not``            ``if``(it == arr.end())``                ``arr.push_back(y);``        ``}` `        ``for``(``auto` `y:arr)``            ``cout << y;``        ``cout << endl;``    ``}``}` `// Driver code``int` `main()``{``    ``string str = ``"geeksforgeeksgfg"``;``    ``int` `n = 4;``    ``sub_segments(str, n);``}` `// This code is contributed by Sanjit_Prasad`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function that prints the segments``static` `void` `sub_segments(String str, ``int` `n)``{``    ``int` `l = str.length();``    ``for` `(``int` `x = ``0``; x < l; x += n)``    ``{``        ``String newlist = str.substring(x, x + n);` `        ``// New array for every iteration``        ``List arr = ``new` `ArrayList();``        ``for` `(``char` `y : newlist.toCharArray())``        ``{` `            ``// Check if the character is in the array``            ``if` `(!arr.contains(y))``                ``arr.add(y);``        ``}``        ``for` `(``char` `y : arr)``            ``System.out.print(y);``        ``System.out.println();``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String str = ``"geeksforgeeksgfg"``;``    ``int` `n = ``4``;``    ``sub_segments(str, n);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach` `# Function that prints the segments``def` `sub_segments (string, n):``    ``l ``=` `len` `(string)``    ``for` `x ``in` `range` `(``0``, l, n):``        ``newlist ``=` `string[x : x ``+` `n]` `        ``# New array for every iteration``        ``arr ``=` `[]``        ``for` `y ``in` `newlist:` `           ``# Check if the character is in the array``            ``if` `y ``not` `in` `arr:``                ``arr.append (y)``       ` `        ``print` `(''.join (arr))` `# Driver code``string ``=` `"geeksforgeeksgfg"``n ``=` `4``sub_segments (string, n)`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function that prints the segments``static` `void` `sub_segments(String str, ``int` `n)``{``    ``int` `l = str.Length;``    ``for` `(``int` `x = 0; x < l; x += n)``    ``{``        ``String newlist = str.Substring(x, n);` `        ``// New array for every iteration``        ``List<``char``> arr = ``new` `List<``char``>();``        ``foreach` `(``char` `y ``in` `newlist.ToCharArray())``        ``{` `            ``// Check if the character is in the array``            ``if` `(!arr.Contains(y))``                ``arr.Add(y);``        ``}``        ``foreach` `(``char` `y ``in` `arr)``            ``Console.Write(y);``        ``Console.WriteLine();``    ``}``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``String str = ``"geeksforgeeksgfg"``;``    ``int` `n = 4;``    ``sub_segments(str, n);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

```gek
sfor
gek
sgf
```

Time Complexity: O(L), Here L is the length of the string.
Auxiliary Space: O(n), The extra space is used to store the substrings.

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