Print all Substrings of length n possible from the given String️‍🔥

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Q: How to use Print all Substrings of length n possible from the given String in Data Structures and Algorithms?

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

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

Examples:

Input: str = “w3wiki”, 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:

C++

// CPP implementation of the approach
#include <iostream>
#include <string.h>
using namespace std;
 
// Drivers code
int main()
{
    string str = "w3wiki";
    int n = 3;
    for (int i = 0; i < str.length() - n + 1; i++)
        cout << str.substr(i, n) << " ";
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG {
    public static void main(String args[])
    {
        String str = "w3wiki";
        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

Python3

# Python implementation of the approach
str = "w3wiki"
n = 3
 
for i in range(len(str) - n + 1):
    print(str[i:i + n], end = " ")
 
# This code is contributed by Susobhan Akhuli

C#

// C# implementation of the approach
using System;
 
class MainClass {
    public static void Main(string[] args)
    {
        string str = "w3wiki";
        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

Javascript

<script>
// JavaScript implementation of the approach
let str = "w3wiki";
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>

Output

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:

C++

// CPP implementation of the approach
#include <iostream>
#include <string.h>
using namespace std;
 
int main()
{
    string str = "w3wiki";
    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

// Java implementation of the approach
import java.util.*;
 
public class GFG {
    public static void main(String[] args)
    {
        String str = "w3wiki";
        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

Python3

# Python implementation of the approach
str = "w3wiki"
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#

// C# implementation of the approach
using System;
 
class Program {
    static void Main(string[] args)
    {
        string str = "w3wiki";
        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

Javascript

<script>
// JavaScript implementation of the approach
var str = "w3wiki";
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>

Output

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:

C++

#include <iostream>
#include <vector>
using namespace std;
 
int main() {
    // C++ implementation of the approach
    string str = "w3wiki";
    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;
}

Java

/*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 = "w3wiki";
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);    }
}

Python3

# Python implementation of the approach
str = "w3wiki"
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

// JavaScript implementation of the approach
let str = "w3wiki";
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

C#

using System;
using System.Collections.Generic;
 
class GFG {
    static void Main(string[] args) {
        // C# implementation of the approach
        string str = "w3wiki";
        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));
    }
}

Output

['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.