Count of cyclic permutations having XOR with other binary string as 0

Given two binary strings $A$ and $B$ . Let $S$ be set of all the cyclic permutations of string $B$ . The task is to find how many strings in set $S$ when XORed with $A$ give $0$ as result.

Examples:

Input: A = “101”, B = “101”
Output: 1
S = {“101”, “011”, “110”}
Only “101” XOR “101” = 0

Input: A = “111”, B = “111”
Output: 3

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

Approach: Concatenate $B$ with $B$ so that it contains all of it’s cyclic permutations as sub-strings. Now the problem is reduced to find the number of occurrences of pattern $A$ in string $B$ which can be solved using Z algorithm (for linear time pattern searching).

Below is the implementation of the above approach:

C++

// C++ program to find bitwise XOR between binary 
// string A and all the cyclic permutations 
// of binary string B 
#include <bits/stdc++.h> 
using namespace std; 
  
// Implementation of Z-algorithm 
// for linear time pattern searching 
void compute_z(string s, int z[]) 
{ 
    int l = 0, r = 0; 
    int n = s.length(); 
    for (int i = 1; i <= n - 1; i++) { 
        if (i > r) { 
            l = i, r = i; 
            while (r < n && s[r - l] == s[r]) 
                r++; 
            z[i] = r - l; 
            r--; 
        } 
        else { 
            int k = i - l; 
            if (z[k] < r - i + 1) { 
                z[i] = z[k]; 
            } 
            else { 
                l = i; 
                while (r < n && s[r - l] == s[r]) 
                    r++; 
                z[i] = r - l; 
                r--; 
            } 
        } 
    } 
} 
  
// Function to get the count of the cyclic 
// permutations of b that given 0 when XORed with a 
int countPermutation(string a, string b) 
{ 
    // concatenate b with b 
    b = b + b; 
  
    // new b now contains all the cyclic 
    // permutations of old b as it's sub-strings 
    b = b.substr(0, b.size() - 1); 
  
    // concatenate pattern with text 
    int ans = 0; 
    string s = a + "$" + b; 
    int n = s.length(); 
  
    // Fill z array used in Z algorithm    
    int z[n]; 
    compute_z(s, z); 
  
    for (int i = 1; i <= n - 1; i++) { 
  
        // pattern occurs at index i since 
        // z value of i equals pattern length 
        if (z[i] == a.length()) 
            ans++; 
    } 
    return ans; 
} 
  
// Driver code 
int main() 
{ 
    string a = "101"; 
    string b = "101"; 
  
    cout << countPermutation(a, b) << endl; 
  
    return 0; 
} 

Java

// Java program to find bitwise XOR between binary 
// string A and all the cyclic permutations 
// of binary string B 
  
public class GFG{ 
      
    // Implementation of Z-algorithm 
    // for linear time pattern searching 
    static void compute_z(String s, int z[]) 
    { 
        int l = 0, r = 0; 
        int n = s.length(); 
        for (int i = 1; i <= n - 1; i++) { 
            if (i > r) { 
                l = i; 
                r = i; 
                while (r < n && s.charAt(r - l) == s.charAt(r)) 
                    r++; 
                z[i] = r - l; 
                r--; 
            } 
            else { 
                int k = i - l; 
                if (z[k] < r - i + 1) { 
                    z[i] = z[k]; 
                } 
                else { 
                    l = i; 
                    while (r < n && s.charAt(r - l) == s.charAt(r)) 
                        r++; 
                    z[i] = r - l; 
                    r--; 
                } 
            } 
        } 
    } 
      
    // Function to get the count of the cyclic 
    // permutations of b that given 0 when XORed with a 
    static int countPermutation(String a, String b) 
    { 
        // concatenate b with b 
        b = b + b; 
        // new b now contains all the cyclic 
        // permutations of old b as it's sub-strings 
        b = b.substring(0, b.length() - 1); 
      
        // concatenate pattern with text 
        int ans = 0; 
        String s = a + "$" + b; 
        int n = s.length(); 
      
        // Fill z array used in Z algorithm    
        int z[] = new int[n]; 
        compute_z(s, z); 
      
        for (int i = 1; i <= n - 1; i++) { 
      
            // pattern occurs at index i since 
            // z value of i equals pattern length 
            if (z[i] == a.length()) 
                ans++; 
        } 
        return ans; 
    } 
      
      
    // Driver Code 
     public static void main(String []args){ 
           
               String a = "101"; 
               String b = "101"; 
               System.out.println(countPermutation(a, b)) ; 
     } 
     // This code is contributed by ANKITRAI1 
} 

Python3

# Python 3 program to find bitwise XOR  
# between binary string A and all the  
# cyclic permutations of binary string B 
  
# Implementation of Z-algorithm 
# for linear time pattern searching 
def compute_z(s, z): 
    l = 0
    r = 0
    n = len(s) 
    for i in range(1, n, 1): 
        if (i > r): 
            l = i 
            r = i 
            while (r < n and s[r - l] == s[r]): 
                r += 1
            z[i] = r - l 
            r -= 1
      
        else: 
            k = i - l 
            if (z[k] < r - i + 1): 
                z[i] = z[k] 
              
            else: 
                l = i 
                while (r < n and s[r - l] == s[r]): 
                    r += 1
                z[i] = r - l 
                r -= 1
      
# Function to get the count of the cyclic 
# permutations of b that given 0 when XORed with a 
def countPermutation(a, b): 
      
    # concatenate b with b 
    b = b + b 
  
    # new b now contains all the cyclic 
    # permutations of old b as it's sub-strings 
    b = b[0:len(b) - 1] 
  
    # concatenate pattern with text 
    ans = 0
    s = a + "$" + b 
    n = len(s) 
  
    # Fill z array used in Z algorithm  
    z = [0 for i in range(n)] 
    compute_z(s, z) 
  
    for i in range(1, n, 1): 
          
        # pattern occurs at index i since 
        # z value of i equals pattern length 
        if (z[i] == len(a)): 
            ans += 1
      
    return ans 
  
# Driver code 
if __name__ == '__main__': 
    a = "101"
    b = "101"
  
    print(countPermutation(a, b)) 
      
# This code is contributed by 
# Surendra_Gangwar 

C#

// C# program to find bitwise XOR between  
// binary string A and all the cyclic  
// permutations of binary string B  
using System; 
  
class GFG 
{ 
  
// Implementation of Z-algorithm  
// for linear time pattern searching  
public static void compute_z(string s,  
                             int[] z) 
{ 
    int l = 0, r = 0; 
    int n = s.Length; 
    for (int i = 1; i <= n - 1; i++) 
    { 
        if (i > r) 
        { 
            l = i; 
            r = i; 
            while (r < n && s[r - l] == s[r]) 
            { 
                r++; 
            } 
            z[i] = r - l; 
            r--; 
        } 
        else
        { 
            int k = i - l; 
            if (z[k] < r - i + 1) 
            { 
                z[i] = z[k]; 
            } 
            else
            { 
                l = i; 
                while (r < n && s[r - l] == s[r]) 
                { 
                    r++; 
                } 
                z[i] = r - l; 
                r--; 
            } 
        } 
    } 
} 
  
// Function to get the count of the  
// cyclic permutations of b that 
// given 0 when XORed with a  
public static int countPermutation(string a,  
                                   string b) 
{ 
    // concatenate b with b  
    b = b + b; 
      
    // new b now contains all the cyclic  
    // permutations of old b as it's sub-strings  
    b = b.Substring(0, b.Length - 1); 
  
    // concatenate pattern with text  
    int ans = 0; 
    string s = a + "$" + b; 
    int n = s.Length; 
  
    // Fill z array used in Z algorithm  
    int[] z = new int[n]; 
    compute_z(s, z); 
  
    for (int i = 1; i <= n - 1; i++) 
    { 
  
        // pattern occurs at index i since  
        // z value of i equals pattern length  
        if (z[i] == a.Length) 
        { 
            ans++; 
        } 
    } 
    return ans; 
} 
  
// Driver Code  
public static void Main(string[] args) 
{ 
    string a = "101"; 
    string b = "101"; 
    Console.WriteLine(countPermutation(a, b)); 
} 
} 
  
// This code is contributed by Shrikant13 

Output:

My Personal Notes

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

C++

Java

Python3

C#

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