# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` `} `

## 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` `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 `

## 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 ` `    ``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 `

## 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` `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 `

Output:

```11001
```

Time Complexity: O(N)

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