Print all Substrings of length n possible from the given String

Given a string str and an integer N, the task is to print all possible sub-strings of length N.

Examples:

Input: str = “geeksforgeeks”, N = 3
Output: gee eek eks ksf sfo for org rge gee eek eks
Explanations: All possible sub-strings of length 3 are “gee”, “eek”, “eks”, “ksf”, “sfo”, “for”, “org”, “rge”, "gee", "eek" and “eks”.

Input: str = “GFG”, N = 2 
Output: GF FG
Explanations: All possible sub-strings of length 2 are “GF”, “FG”

Method 1: Using slicing

Approach: To solve the problem follow the below steps:

• Initialize a variable 'n' to the desired length of the substrings.
• Use a for loop to iterate through the characters of the original string, starting from the first character.
• In each iteration of the loop, use slicing to extract a substring of length 'n' from the original string, starting from the current character. The slicing can be done by using the syntax 'string[start:end]' where the start and end are the indices of the first and last character of the substring, respectively.
• In each iteration, the variable 'i' will be the starting index of the substring, and 'i + n' will be the ending index of the substring, so the substring can be extracted by using the slicing syntax 'string[i: i + n]'.
• Print or store the extracted substring for further processing.
• Repeat the process for all characters in the original string.

Below is the implementation of the above approach:

// CPP implementation of the approach
#include <iostream>
#include <string.h>
using namespace std;

// Drivers code
int main()
{
    string str = "geeksforgeeks";
    int n = 3;
    for (int i = 0; i < str.length() - n + 1; i++)
        cout << str.substr(i, n) << " ";
    return 0;
}

// Java implementation of the approach
import java.util.*;

class GFG {
    public static void main(String args[])
    {
        String str = "geeksforgeeks";
        int n = 3;
        for (int i = 0; i < str.length() - n + 1; i++)
            System.out.print(str.substring(i, i + n) + " ");
    }
}
// This code is contributed by Susobhan Akhuli

# Python implementation of the approach
str = "geeksforgeeks"
n = 3

for i in range(len(str) - n + 1):
    print(str[i:i + n], end = " ")

# This code is contributed by Susobhan Akhuli

// C# implementation of the approach
using System;

class MainClass {
    public static void Main(string[] args)
    {
        string str = "geeksforgeeks";
        int n = 3;
        for (int i = 0; i < str.Length - n + 1; i++)
            Console.Write(str.Substring(i, n) + " ");
    }
}

// This code is contributed by Susobhan Akhuli

<script>
// JavaScript implementation of the approach
let str = "geeksforgeeks";
let n = 3;
for (let i = 0; i < str.length - n + 1; i++) {
    let substring = str.substring(i, i+n);
    document.write(substring+" ");
}

//  This code is contributed by Susobhan Akhuli
</script>

gee eek eks ksf sfo for org rge gee eek eks 

Time Complexity: O(|str|*n), where |str| is the length of the string
Auxiliary Space: O(n)

Method 2: Using a for loop

Approach: To solve the problem follow the below steps:

• Initialize a variable 'n' to the desired length of the substrings.
• Use a for loop to iterate through the characters of the original string, starting from the first character.
• In each iteration of the outer loop, initialize an empty variable 'substring' to store the extracted substring.
• Use a nested for loop to iterate through the next 'n' characters of the original string, starting from the current character.
• In each iteration of the inner loop, add the current character to the 'substring' variable.
• After the inner loop finishes, print or store the extracted substring for further processing.
• Repeat the process for all characters in the original string.

Below is the implementation of the above approach:

// CPP implementation of the approach
#include <iostream>
#include <string.h>
using namespace std;

int main()
{
    string str = "geeksforgeeks";
    int n = 3;
    for (int i = 0; i < str.length() - n + 1; i++) {
        string substring = "";
        for (int j = i; j < i + n; j++)
            substring += str[j];
        cout << substring << " ";
    }
    return 0;
}
// This code is contributed by Susobhan Akhuli

// Java implementation of the approach
import java.util.*;

public class GFG {
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        int n = 3;
        for (int i = 0; i < str.length() - n + 1; i++) {
            String substring = "";
            for (int j = i; j < i + n; j++)
                substring += str.charAt(j);
            System.out.print(substring + " ");
        }
    }
}

// This code is contributed by Susobhan Akhuli

# Python implementation of the approach
str = "geeksforgeeks"
n = 3

for i in range(len(str) - n + 1):
    substring = ""
    for j in range(i, i + n):
        substring += str[j]
    print(substring, end =' ')

# This code is contributed by Susobhan Akhuli

// C# implementation of the approach
using System;

class Program {
    static void Main(string[] args)
    {
        string str = "geeksforgeeks";
        int n = 3;
        for (int i = 0; i < str.Length - n + 1; i++) {
            string substring = "";
            for (int j = i; j < i + n; j++)
                substring += str[j];
            Console.Write(substring + " ");
        }
    }
}

// This code is contributed by Susobhan Akhuli

<script>
// JavaScript implementation of the approach
var str = "geeksforgeeks";
var n = 3;
for (let i = 0; i < str.length - n + 1; i++) {
    var substring = str[i];
    for (let j = i+1; j < i + n; j++)
        substring += str[j];
    document.write(substring + " ");
}

//  This code is contributed by Susobhan Akhuli
</script>

gee eek eks ksf sfo for org rge gee eek eks 

Time Complexity: O(|str|*n), where |str| is the length of the string
Auxiliary Space: O(n)

Method 3: Using list comprehension (Only for Python)

Approach: To solve the problem follow the below steps:

• Initialize a variable 'n' to the desired length of the substrings.
• Use a list comprehension to iterate through the characters of the original string and extract substrings of length 'n' in one line.
• The list comprehension uses the slicing syntax 'string[start:end]' where the start and end are the indices of the first and last character of the substring, respectively.
• The variable 'i' will be the starting index of the substring and 'i+n' will be the ending index of the substring, so the substring can be extracted by using the slicing syntax 'string[i:i+n]' inside the list comprehension.
• Assign the output of the list comprehension to a variable e.g. substrings = [string[i:i+n] for i in range(len(string) - n + 1)]
• Then you can print or access individual substrings from the list by indexing.

Below is the implementation of the above approach:

#include <iostream>
#include <vector>
using namespace std;

int main() {
    // C++ implementation of the approach
    string str = "geeksforgeeks";
    int n = 3;
    vector<string> substrings;

    for (int i = 0; i <= str.length() - n; i++) {
        substrings.push_back(str.substr(i, n));
    }

    for (auto s : substrings) {
        cout << s << " ";
    }
    cout << endl;

    return 0;
}

/*package whatever //do not write package name here */

import java.util.*;

class GFG {
    public static void main (String[] args) {
//Java implementation of the approach
String str = "geeksforgeeks";
int n = 3;
List<String> substrings = new ArrayList<>();

for (int i = 0; i <= str.length() - n; i++) {
    substrings.add(str.substring(i, i + n));
}

System.out.println(substrings);    }
}

# Python implementation of the approach
str = "geeksforgeeks"
n = 3
substrings = [str[i:i + n] for i in range(len(str) - n + 1)]
print(substrings)

# This code is contributed by Susobhan Akhuli

// JavaScript implementation of the approach
let str = "geeksforgeeks";
let n = 3;
let substrings = [];

for (let i = 0; i <= str.length - n; i++) {
substrings.push(str.slice(i, i + n));
}

console.log(substrings);

// This code is contributed by codebraxnzt

using System;
using System.Collections.Generic;

class GFG {
    static void Main(string[] args) {
        // C# implementation of the approach
        string str = "geeksforgeeks";
        int n = 3;
        List<string> substrings = new List<string>();

        for (int i = 0; i <= str.Length - n; i++) {
            substrings.Add(str.Substring(i, n));
        }

        Console.WriteLine(string.Join(", ", substrings));
    }
}

['gee', 'eek', 'eks', 'ksf', 'sfo', 'for', 'org', 'rge', 'gee', 'eek', 'eks']

Time Complexity: O(|str|*n), where |str| is the length of the string
Auxiliary Space: O(|str|*n)

N.B.: It is important to note that the above approach will only work if the length of the string is greater than n, otherwise it will throw index out of range error.