Print all Substrings of length n possible from the given String
Last Updated :
31 Mar, 2023
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:
C++
// 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
// 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
Python3
# 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#
// 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
JavaScript
<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>
Outputgee 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 = "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
// 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
Python3
# 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#
// 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
JavaScript
<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>
Outputgee 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 = "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;
}
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 = "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); }
}
Python3
# 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
// 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
C#
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));
}
}
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.
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