String obtained by reversing and complementing a Binary string K times

Given a binary string of size N and an integer K, the task is to perform K operations upon the string and print the final string:

  • If the operation number is odd, then reverse the string,
  • If the operation number even, then complement the string.

Examples:

Input: str = “1011”, K = 2
Output: 0010
After the first step, string will be reversed and becomes “1101”.
After the second step, the string will be complemented and becomes “0010”.

Input: str = “1001”, K = 4
Output: 1001
After all operation the string will remain same.

Naive Approach:
Traverse for all K steps and if the current step is odd then perform the reverse operation, otherwise complement the string.



Efficient Approach: Upon observing the given operation pattern:

  • If a string is reversed even number of times, the original string is obtained.
  • Similarly, if a string is complemented even number of times, the original string is obtained.
  • Therefore, these operations depends only upon the parity of K.
  • So we will count the number of reverse operations to be performed. If parity is odd, then we will reverse it. Else the string will remain unchanged.
  • Similarly we will count the number of complement operations to be performed. If parity is odd, then we will complement it. Else the string will remain unchanged.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to perform K operations upon
// the string and find the modified string
#include <bits/stdc++.h>
using namespace std;
  
// Function to perform K operations upon
// the string and find modified string
string ReverseComplement(
    string s, int n, int k)
{
  
    // Number of reverse operations
    int rev = (k + 1) / 2;
  
    // Number of complement operations
    int complment = k - rev;
  
    // If rev is odd parity
    if (rev % 2)
        reverse(s.begin(), s.end());
  
    // If complment is odd parity
    if (complment % 2) {
        for (int i = 0; i < n; i++) {
            // Complementing each position
            if (s[i] == '0')
                s[i] = '1';
            else
                s[i] = '0';
        }
    }
  
    // Return the modified string
    return s;
}
  
// Driver Code
int main()
{
    string str = "10011";
    int k = 5;
    int n = str.size();
  
    // Function call
    cout << ReverseComplement(str, n, k);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to perform K operations upon
// the String and find the modified String
class GFG{
   
// Function to perform K operations upon
// the String and find modified String
static String ReverseComplement(
    char []s, int n, int k)
{
   
    // Number of reverse operations
    int rev = (k + 1) / 2;
   
    // Number of complement operations
    int complment = k - rev;
   
    // If rev is odd parity
    if (rev % 2 == 1)
        s = reverse(s);
   
    // If complment is odd parity
    if (complment % 2 == 1) {
        for (int i = 0; i < n; i++) {
            // Complementing each position
            if (s[i] == '0')
                s[i] = '1';
            else
                s[i] = '0';
        }
    }
   
    // Return the modified String
    return String.valueOf(s);
}
  
static char[] reverse(char a[]) {
    int i, n = a.length;
    char t;
    for (i = 0; i < n / 2; i++) {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
  
// Driver Code
public static void main(String[] args)
{
    String str = "10011";
    int k = 5;
    int n = str.length();
   
    // Function call
    System.out.print(ReverseComplement(str.toCharArray(), n, k));
   
}
}
  
// This code is contributed by Rajput-Ji

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to perform K operations upon
# the string and find the modified string
  
# Function to perform K operations upon
# the string and find modified string
def ReverseComplement(s,n,k):
    # Number of reverse operations
    rev = (k + 1) // 2
  
    # Number of complement operations
    complment = k - rev
  
    # If rev is odd parity
    if (rev % 2):
        s = s[::-1]
  
    # If complment is odd parity
    if (complment % 2):
        for i in range(n):
            # Complementing each position
            if (s[i] == '0'):
                s[i] = '1'
            else:
                s[i] = '0'
  
    # Return the modified string
    return s
  
# Driver Code
if __name__ == '__main__':
    str1 = "10011"
    k = 5
    n = len(str1)
  
    # Function call
    print(ReverseComplement(str1, n, k))
  
# This code is contributed by Surendra_Gangwar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to perform K operations upon 
// the String and find the modified String 
using System;
class GFG{ 
      
// Function to perform K operations upon 
// the String and find modified String 
static string ReverseComplement(char []s, 
                                int n, int k) 
      
    // Number of reverse operations 
    int rev = (k + 1) / 2; 
      
    // Number of complement operations 
    int complment = k - rev; 
      
    // If rev is odd parity 
    if (rev % 2 == 1) 
        s = reverse(s); 
      
    // If complment is odd parity 
    if (complment % 2 == 1)
    
        for (int i = 0; i < n; i++) 
        
            // Complementing each position 
            if (s[i] == '0'
                s[i] = '1'
            else
                s[i] = '0'
        
    
      
    // Return the modified String 
    return (new string(s)); 
  
static char[] reverse(char[] a) 
    int i, n = a.Length; 
    char t; 
    for (i = 0; i < n / 2; i++) 
    
        t = a[i]; 
        a[i] = a[n - i - 1]; 
        a[n - i - 1] = t; 
    
    return a; 
  
// Driver Code 
public static void Main() 
    string str = "10011"
    int k = 5; 
    int n = str.Length; 
      
    // Function call 
    Console.Write(ReverseComplement(str.ToCharArray(), n, k)); 
  
// This code is contributed by rutvik_56

chevron_right


Output:

11001

Time Complexity: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.