Contiguous subsegments of a string having distinct subsequent characters

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<bits/stdc++.h>
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<Character> arr = new ArrayList<Character>();
        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

<script>
// Javascript implementation of the approach


// Function that prints the segments
function sub_segments(str, n) {
    let l = str.length;
    for (let x = 0; x < l; x += n) {
        let newlist = str.substr(x, n);

        // New array for every iteration
        let arr = [];

        for (let y of newlist) {

            // Check if the character is in the array
            if (!arr.includes(y))
                arr.push(y);
        }
        for (let y of arr)
            document.write(y);
        document.write("<br>");
    }
}

// Driver code

let str = "geeksforgeeksgfg";
let n = 4;
sub_segments(str, n);

// This code is contributed by gfgking

</script>

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.