Substring Reverse Pattern (original) (raw)

Last Updated : 12 Sep, 2022

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 <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

// 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

`

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