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# Substring Reverse Pattern

• Difficulty Level : Easy
• Last Updated : 21 May, 2021

Given string str, the task is to print the pattern given in the examples below:

Examples:

Input: str = “geeks”
Output:
geeks
*kee*
**e**
The reverse of “geeks” is “skeeg”
Replace the first and last characters with ‘*’ i.e. *kee*
Replace the second and second last character in the modified string i.e. **e**
And so on.

Input: str = “first”
Output:
first
*sri*
**r**

Approach:

• Print the unmodified string.
• Reverse the string and initialize i = 0 and j = n – 1.
• Replace s[i] = ‘*’ and s[j] = ‘*’ and update i = i + 1 and j = j – 1 then print the modified string.
• Repeat the above steps while j – i > 1.

Below is the implementation of the above approach:

## C++

 `// C++ program to print the required pattern``#include ``using` `namespace` `std;` `// Function to print the required pattern``void` `printPattern(``char` `s[], ``int` `n)``{` `    ``// Print the unmodified string``    ``cout << s << ``"\n"``;` `    ``// Reverse the string``    ``int` `i = 0, j = n - 2;``    ``while` `(i < j) {``        ``char` `c = s[i];``        ``s[i] = s[j];``        ``s[j] = c;``        ``i++;``        ``j--;``    ``}` `    ``// Replace the first and last character by '*' then``    ``// second and second last character and so on``    ``// until the string has characters remaining``    ``i = 0;``    ``j = n - 2;``    ``while` `(j - i > 1) {``        ``s[i] = s[j] = ``'*'``;``        ``cout << s << ``"\n"``;``        ``i++;``        ``j--;``    ``}``}` `// Driver code``int` `main()``{``    ``char` `s[] = ``"geeks"``;``    ``int` `n = ``sizeof``(s) / ``sizeof``(s);` `    ``printPattern(s, n);``    ``return` `0;``}`

## Java

 `// Java program to print the required pattern``class` `GFG``{``    ` `// Function to print the required pattern``static` `void` `printPattern(``char``[] s, ``int` `n)``{``    ``// Print the unmodified string``    ``System.out.println(s);` `    ``// Reverse the string``    ``int` `i = ``0``, j = n - ``1``;``    ``while` `(i < j)``    ``{``        ``char` `c = s[i];``        ``s[i] = s[j];``        ``s[j] = c;``        ``i++;``        ``j--;``    ``}` `    ``// Replace the first and last character``    ``// by '*' then second and second last``    ``// character and so on until the string``    ``// has characters remaining``    ``i = ``0``;``    ``j = n - ``1``;``    ``while` `(j - i > ``1``)``    ``{``        ``s[i] = s[j] = ``'*'``;``        ``System.out.println(s);``        ``i++;``        ``j--;``    ``}``}` `// Driver Code``public` `static` `void` `main(String []args)``{``    ``char``[] s = ``"geeks"``.toCharArray();``    ``int` `n = s.length;` `    ``printPattern(s, n);``}``}` `// This code is contributed by Rituraj Jain`

## Python3

 `# Python3 program to print the required pattern` `# Function to print the required pattern``def` `printPattern(s, n):` `    ``# Print the unmodified string``    ``print``(''.join(s))` `    ``# Reverse the string``    ``i, j ``=` `0``, n ``-` `1``    ` `    ``while` `i < j:``        ``s[i], s[j] ``=` `s[j], s[i]``        ``i ``+``=` `1``        ``j ``-``=` `1``    ` `    ``# Replace the first and last character``    ``# by '*' then second and second last``    ``# character and so on until the string``    ``# has characters remaining``    ``i, j ``=` `0``, n ``-` `1``    ` `    ``while` `j ``-` `i > ``1``:``        ``s[i], s[j] ``=` `'*'``, ``'*'``        ``print``(''.join(s))``        ``i ``+``=` `1``        ``j ``-``=` `1` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``s ``=` `"geeks"``    ``n ``=` `len``(s)` `    ``printPattern(``list``(s), n)``    ` `# This code is contributed``# by Rituraj Jain`

## C#

 `// C# program to print the required pattern``using` `System;` `class` `GFG``{``    ` `// Function to print the required pattern``static` `void` `printPattern(``char``[] s, ``int` `n)``{``    ``// Print the unmodified string``    ``Console.WriteLine(s);` `    ``// Reverse the string``    ``int` `i = 0, j = n - 1;``    ``while` `(i < j)``    ``{``        ``char` `c = s[i];``        ``s[i] = s[j];``        ``s[j] = c;``        ``i++;``        ``j--;``    ``}` `    ``// Replace the first and last character``    ``// by '*' then second and second last``    ``// character and so on until the string``    ``// has characters remaining``    ``i = 0;``    ``j = n - 1;``    ``while` `(j - i > 1)``    ``{``        ``s[i] = s[j] = ``'*'``;``        ``Console.WriteLine(s);``        ``i++;``        ``j--;``    ``}``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``char``[] s = ``"geeks"``.ToCharArray();``    ``int` `n = s.Length;` `    ``printPattern(s, n);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
```geeks
*kee*
**e**```

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