Given string str, the task is to print the pattern given in the examples below:
Examples:
Input: str = “neveropen”
Output:
neveropen
*kee*
**e**
The reverse of “neveropen” 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 patternvoid 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 codeint main(){ char s[] = "neveropen"; 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 Codepublic static void main(String []args){ char[] s = "neveropen".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 = "neveropen" 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 Codepublic static void Main(String []args){ char[] s = "neveropen".ToCharArray(); int n = s.Length; printPattern(s, n);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to print the required pattern// Function to print the required patternfunction 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 codevar s = "neveropen".split('');var n = s.length;printPattern(s, n);</script> |
Output
neveropen *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
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
