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 stringstring 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 Codeint 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 Stringclass GFG{ // Function to perform K operations upon// the String and find modified Stringstatic 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 Codepublic 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 stringdef 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 Codeif __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 stringfunction 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 Codevar str = "10011";var k = 5;var n = str.length;// Function calldocument.write( ReverseComplement(str, n, k));// This code is contributed by famously.</script> |
Output:
11001
Time Complexity: O(N)
Auxiliary Space: O(1)
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