Substring Reverse Pattern

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++ program to print the required pattern
#include <bits/stdc++.h>
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[0]);

    printPattern(s, n);
    return 0;
}

// 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 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# 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

<script>

// Javascript program to print the required pattern

// Function to print the required pattern
function printPattern(s, n)
{

    // Print the unmodified string
    document.write( s.join('') + "<br>");

    // Reverse the string
    var i = 0, j = n - 1;
    while (i < j) {
        var 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] = '*';
        document.write( s.join('') + "<br>");
        i++;
        j--;
    }
}

// Driver code
var s = "geeks".split('');
var n = s.length;
printPattern(s, n);


</script> 

geeks
*kee*
**e**

Complexity Analysis:

• Time Complexity: O(N) since one traversal of the string is required to complete all operations hence the overall time required by the algorithm is linear
• Auxiliary Space: O(1) since no extra array is used so the space taken by the algorithm is constant