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++

// 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 complement = k - rev;

    // If rev is odd parity
    if (rev % 2)
        reverse(s.begin(), s.end());

    // If complement is odd parity
    if (complement % 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;
}

Java

// 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 complement = k - rev;

    // If rev is odd parity
    if (rev % 2 == 1)
        s = reverse(s);

    // If complement is odd parity
    if (complement % 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

Python3

# 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
    complement = k - rev

    # If rev is odd parity
    if (rev % 2):
        s = s[::-1]

    # If complement is odd parity
    if (complement % 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

C#

// 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 complement = k - rev;

    // If rev is odd parity
    if (rev % 2 == 1)
        s = reverse(s);

    // If complement is odd parity
    if (complement % 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

Javascript

<script>

// Javascript 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
function ReverseComplement( s, n, k)
{

    // Number of reverse operations
    var rev = parseInt((k + 1) / 2);

    // Number of complement operations
    var complement = k - rev;

    // If rev is odd parity
    if (rev % 2)
    {
       s = s.split('').reverse().join('');
    }

    // If complement is odd parity
    if (complement % 2) {
        for (var 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
var str = "10011";
var k = 5;
var n = str.length;

// Function call
document.write( ReverseComplement(str, n, k));

// This code is contributed by famously.
</script>

Output:

11001

Time Complexity: O(N)
Auxiliary Space: O(1)