Given two binary strings A and B of equal length N, the task is to find the number of distinct XORs possible by arbitrarily reordering the two binary strings. Since the number can be large enough, find the number modulo 109 + 7
Examples:
Input: A = “00”, B = “01”
Output: 2
Explanation:
There are two possible results by rearranging the digits of the string B.
They are: “10” and “01”Input: A = “010”, B = “100”
Output: 4
Explanation:
There are four possible results possible by rearranging the digits of both the strings.
They are: “000”, “110”, “011”, “101”
Approach:
- Since we know that
0 XOR 0 = 0 0 XOR 1 = 1 1 XOR 0 = 1 1 XOR 1 = 0
Therefore, to get an XOR value as ‘1’ at any index of the result string, the input strings must have odd number of 1s at that index.
- Now, we will try to rearrange the Binary strings in a way, that maximum number of indices have an odd number of 1s at them. This can be visualized by the following example:
- Therefore, from the above observation, the idea is to find the minimum and the maximum number of 1’s possible by reordering the strings.
- To find the Maximum ‘1’s: The maximum ‘1’s in the result would occur when maximum {0, 1} and {1, 0} pairs are formed. Therefore,
Maximum number of {0, 1} pairs = minimum(count of ‘0’ in A, count of ‘1’ in B)
Maximum number of {1, 0} pairs = minimum(count of ‘1’ in A, count of ‘0’ in B)
Therefore, Maximum number of ‘1’s in the XOR = Maximum number of {0, 1} pairs + Maximum number of {1, 0} pairs
- To find the Minimum ‘1’s: This case can be seen as the converse of the maximum number of ‘0’s in the result. Similarly, the maximum ‘0’s in the result would occur when maximum {0, 0} and {1, 1} pairs are formed. Therefore,
Maximum number of {0, 0} pairs = minimum(count of ‘0’ in A, count of ‘0’ in B)
Maximum number of {1, 1} pairs = minimum(count of ‘1’ in A, count of ‘1’ in B)
Maximum number of ‘0’s in the XOR = Maximum number of {0, 0} pairs + Maximum number of {1, 1} pairs
Therefore, Minimum number of ‘1’s in the XOR = N – Maximum number of ‘0’s in the XOR
- All the combinations of 1’s can be formed in between these two numbers (minimum and maximum) with the difference of 2.
- Finally, the total number of possible ways to get the result can be calculated by the number of combinations from the minimum number of 1’s and the maximum number of 1’s with a step of 2.
Below is the implementation of the above approach:
C++14
// C++ program to find the number of // distinct XORs formed by rearranging // two binary strings #include <bits/stdc++.h> using namespace std; // function to compute modulo power long long power( long long a, long long b, long long mod) { long long aa = 1; while (b) { if (b&1) { aa = aa * a; aa %= mod; } a = a * a; a %= mod; b /= 2; } return aa; } // Function to calculate nCr % p // over a range long long nCrRangeSum( long long n, long long r1, long long r2, long long p) { // Initialize the numerator // and denominator long long num = 1, den = 1; // Initialize the sum long long sum = 0; // nC0 is 1 if (r1 == 0) sum += 1; // Traversing till the range for ( int i = 0; i < r2; i++) { // Computing the numerator num = (num * (n - i)) % p; // Computing the denominator den = (den * (i + 1)) % p; // If 'i' lies between the given range // and is at an even long long interval from // the starting range because // the combinations at a step of 2 // is required if (i - r1 >= -1 and (i - r1 + 1) % 2 == 0) { // Computing nCr and adding the value // sum sum += (num * power(den, p - 2, p)) % p; sum %= p; } } return sum; } // Function to find the number of // distinct XORs formed by // rearranging two binary strings int compute(string A, string B, int N) { // Initializing the count variable // to 0 int c0A = 0, c1A = 0, c0B = 0, c1B = 0; // Iterating through A for ( char c:A) { // Increment the c1A variable // if the current element is 1 if (c == '1' ) c1A += 1; // Increment the c0A variable // if the current element is 0 else if (c == '0' ) c0A += 1; } // Iterating through B for ( char c:B){ // Increment the c1B variable // if the current element is 1 if (c == '1' ) c1B += 1; // Increment the c0A variable // if the current element is 0 else if (c == '0' ) c0B += 1; } // Finding the minimum number of '1's in the XOR // and the maximum number of '1's in the XOR int max1xor = min(c0A, c1B) + min(c1A, c0B); int min1xor = N - min(c0B, c0A) - min(c1A, c1B); // Compute the number of combinations between // the minimum number of 1's and the maximum // number of 1's and perform % with 10^9 + 7 int ans = nCrRangeSum(N, min1xor, max1xor, 1000000000 + 7); // Return the answer return ans; } // Driver code int main() { long long N = 3; string A = "010" ; string B = "100" ; cout << compute(A, B,N); return 0; } // This code is contributed by mohit kumar 29 |
Java
// JAVA program to find the number of // distinct Bitwise XORs formed by rearranging // two binary strings class GFG { // function to compute modular exponentiation // i.e. to find (a^b) % mod static long mod_power( long a, long b, long mod) { long result = 1l; while (b > 0 ) { if ((b& 1 ) == 0 ) // b is even { result = a * a; a %= mod; b /= 2 ; } else // b is odd { result = result * a; result %= mod; } } return result; } // method to evaluate nCr modulo p // over an interval static long nCr_RangeSum( long n, long r1, long r2, long p) { // initializing numerator // and denominator long num = 1 , den = 1 ; // initialize the sum long sum = 0l; // nC0 is 1 if (r1 == 0 ) sum += 1l; // Iterating through the range for ( int i = 0 ; i < r2; i++) { // computing the numerator num = (num * (n - i)) % p; // computing the denominator den = (den * (i + 1 )) % p; // If 'i' lies between the given range // and is at an even interval from // the starting range because // the combinations at a step of 2 // is required if (i - r1 >= - 1 && (i - r1 + 1 ) % 2 == 0 ) { // Computing nCr and adding the value // to the sum sum += (num * mod_power(den, p - 2 , p)) % p; sum %= p; } } return sum; } // method to find the number of // distinct XORs formed by // rearrangement of two binary strings static long compute(String A, String B, int N) { // Initializing the counter variables // to 0 int c0A = 0 , c1A = 0 , c0B = 0 , c1B = 0 ; // Iterating through A's characters for ( char c : A.toCharArray()) { // Increment the c1A variable // if the current element is 1 if (c == '1' ) c1A += 1 ; // Increment the c0A variable // if the current element is 0 else if (c == '0' ) c0A += 1 ; } // Iterating through B's characters for ( char c : B.toCharArray()) { // Increment the c1B variable // if the current element is 1 if (c == '1' ) c1B += 1 ; // Increment the c0A variable // if the current element is 0 else if (c == '0' ) c0B += 1 ; } // Finding the minimum number of '1's in the XOR // and the maximum number of '1's in the XOR int max1xor = Math.min(c0A, c1B) + Math.min(c1A, c0B); int min1xor = N - Math.min(c0B, c0A) - Math.min(c1A, c1B); // Compute the number of combinations between // the minimum number of 1's and the maximum // number of 1's and perform modulo with 10^9 + 7 long ans = nCr_RangeSum(N, min1xor, max1xor, 1000000000 + 7 ); // Return the answer return ans; } // Driver code public static void main(String[] args) { int N = 3 ; // length of each string String A = "010" ; String B = "100" ; System.out.print(compute(A, B, N)); } } // This Code is contributed by Soumitri Chattopadhyay. |
Python3
# Python3 program to find the number of # distinct XORs formed by rearranging # two binary strings # Function to calculate nCr % p # over a range def nCrRangeSum(n, r1, r2, p): # Initialize the numerator # and denominator num = den = 1 # Initialize the sum sum = 0 # nC0 is 1 if r1 = = 0 : sum + = 1 # Traversing till the range for i in range (r2): # Computing the numerator num = (num * (n - i)) % p # Computing the denominator den = (den * (i + 1 )) % p # If 'i' lies between the given range # and is at an even interval from # the starting range because # the combinations at a step of 2 # is required if (i - r1 > = - 1 and (i - r1 + 1 ) % 2 = = 0 ): # Computing nCr and adding the value # sum sum + = (num * pow (den, p - 2 , p)) % p sum % = p return sum # Function to find the number of # distinct XORs formed by # rearranging two binary strings def compute(A, B): # Initializing the count variable # to 0 c0A = c1A = c0B = c1B = 0 # Iterating through A for c in A: # Increment the c1A variable # if the current element is 1 if c = = '1' : c1A + = 1 # Increment the c0A variable # if the current element is 0 elif c = = '0' : c0A + = 1 # Iterating through B for c in B: # Increment the c1B variable # if the current element is 1 if c = = '1' : c1B + = 1 # Increment the c0A variable # if the current element is 0 elif c = = '0' : c0B + = 1 # Finding the minimum number of '1's in the XOR # and the maximum number of '1's in the XOR max1xor = min (c0A, c1B) + min (c1A, c0B) min1xor = N - min (c0B, c0A) - min (c1A, c1B) # Compute the number of combinations between # the minimum number of 1's and the maximum # number of 1's and perform % with 10^9 + 7 ans = nCrRangeSum(N, min1xor, max1xor, 10 * * 9 + 7 ) # Return the answer return ans # Driver code if __name__ = = "__main__" : N = 3 A = "010" B = "100" print (compute(A, B)) |
C#
// C# program to find the number of // distinct Bitwise XORs formed by // rearranging two binary strings using System; class GFG{ // Function to compute modular exponentiation // i.e. to find (a^b) % mod static long mod_power( long a, long b, long mod) { long result = 1; while (b > 0) { if ((b & 1) == 0) // b is even { result = a * a; a %= mod; b /= 2; } else // b is odd { result = result * a; result %= mod; } } return result; } // Function to evaluate nCr modulo p // over an interval static long nCr_RangeSum( long n, long r1, long r2, long p) { // Initializing numerator // and denominator long num = 1, den = 1; // Initialize the sum long sum = 0; // nC0 is 1 if (r1 == 0) sum += 1; // Iterating through the range for ( int i = 0; i < r2; i++) { // Computing the numerator num = (num * (n - i)) % p; // Computing the denominator den = (den * (i + 1)) % p; // If 'i' lies between the given range // and is at an even interval from // the starting range because // the combinations at a step of 2 // is required if (i - r1 >= -1 && (i - r1 + 1) % 2 == 0) { // Computing nCr and adding the value // to the sum sum += (num * mod_power( den, p - 2, p)) % p; sum %= p; } } return sum; } // Function to find the number of distinct // XORs formed by rearrangement of two // binary strings static long compute( string A, string B, int N) { // Initializing the counter variables // to 0 int c0A = 0, c1A = 0, c0B = 0, c1B = 0; // Iterating through A's characters foreach ( char c in A) { // Increment the c1A variable // if the current element is 1 if (c == '1' ) c1A += 1; // Increment the c0A variable // if the current element is 0 else if (c == '0' ) c0A += 1; } // Iterating through B's characters foreach ( char c in B) { // Increment the c1B variable // if the current element is 1 if (c == '1' ) c1B += 1; // Increment the c0A variable // if the current element is 0 else if (c == '0' ) c0B += 1; } // Finding the minimum number of // '1's in the XOR and the maximum // number of '1's in the XOR int max1xor = Math.Min(c0A, c1B) + Math.Min(c1A, c0B); int min1xor = N - Math.Min(c0B, c0A) - Math.Min(c1A, c1B); // Compute the number of combinations // between the minimum number of 1's // and the maximum number of 1's and // perform modulo with 10^9 + 7 long ans = nCr_RangeSum(N, min1xor, max1xor, 1000000000 + 7); // Return the answer return ans; } // Driver code static public void Main() { // Length of each string int N = 3; string A = "010" ; string B = "100" ; Console.WriteLine(compute(A, B, N)); } } // This code is contributed by offbeat |
Javascript
// JavaScript program to find the number of // distinct XORs formed by rearranging // two binary strings // function to compute modulo power function power(a, b, mod) { var aa = 1n; while (b) { if (BigInt(b) & 1n) { aa = aa * a; aa %= mod; } a = a * a; a %= mod; b = Math.floor(Number(BigInt(b) / 2n)); } return aa; } // Function to calculate nCr % p // over a range function nCrRangeSum(n, r1, r2, p) { // Initialize the numerator // and denominator var num = 1n; var den = 1n; // Initialize the sum var sum = 0n; // nC0 is 1 if (r1 == 0) sum += 1n; // Traversing till the range for ( var i = 0; i < r2; i++) { // Computing the numerator num = (num * (BigInt(n) - BigInt(i))) % p; // Computing the denominator den = BigInt(den * (BigInt(i) + 1n)) % p; // If 'i' lies between the given range // and is at an even long long interval from // the starting range because // the combinations at a step of 2 // is required if (i - r1 >= -1 && (i - r1 + 1) % 2 == 0) { // Computing nCr and adding the value // sum sum += BigInt(num * power(den, p - 2n, p)) % p; sum %= p; } } return Number(sum); } // Function to find the number of // distinct XORs formed by // rearranging two binary strings function compute(A, B, N) { // Initializing the count variable // to 0 var c0A = 0; var c1A = 0; var c0B = 0; var c1B = 0; // Iterating through A for ( var c of A) { // Increment the c1A variable // if the current element is 1 if (c == '1' ) c1A += 1; // Increment the c0A variable // if the current element is 0 else if (c == '0' ) c0A += 1; } // Iterating through B for ( var c of B) { // Increment the c1B variable // if the current element is 1 if (c == '1' ) c1B += 1; // Increment the c0A variable // if the current element is 0 else if (c == '0' ) c0B += 1; } // Finding the minimum number of '1's in the XOR // and the maximum number of '1's in the XOR var max1xor = Math.min(c0A, c1B) + Math.min(c1A, c0B); var min1xor = N - Math.min(c0B, c0A) - Math.min(c1A, c1B); // Compute the number of combinations between // the minimum number of 1's and the maximum // number of 1's and perform % with 10^9 + 7 var ans = nCrRangeSum(N, min1xor, max1xor, BigInt(1000000000 + 7)); // Return the answer return ans; } // Driver code var N = 3; var A = "010" ; var B = "100" ; // Function call console.log(compute(A, B, N)); // This code is contributed by phasing17 |
4
Time Complexity : O( N )
Space Complexity : O( 1 )
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