In C++, translate the following into Chinese: Calculate the minimum number of bit flips such that the XOR result of A and B is equal to C

Given three binary sequences A, B and C of length N. Each sequence represents a Binary number. We have to find out no. The number of flips required of the bits in A and B such that the XOR of A and B gives C. A XOR B becomes C.

First let us understand the truth table of the XOR operation-

X	Y	X XOR Y
0	0	0
0	1	1
1	0	1
1	1	0

#From the above table we observe that for the same values ​​in X and Y, X XOR Y results is 0, otherwise Result 1. So this will help to find the bits that flip in A and B to reach C. The situation would be

• if A[i]==B[i] and C[i]==0, no flipping is needed,
• if A[i]== B[i] and C[i]==1, then flip A[i] or B[i] and increase the flip count by 1
• If A[i]!=B[i] and C[ i]==0, then flip A[i] or B[i] and increase the flip count by 1
• If A[i]!=B[i] and C[i]==1, then No flipping required.

Input

A[]= { 0,0,0,0 } B[]= { 1,0,1,0 } C= {1,1,1,1}

Output

Required flips : 2

Explanation

’s Chinese translation is:

Explanation

A[0] xor B[0] 0 xor 1 = 1 C[0]=1 no flip
A[1] xor B[1] 0 xor 0 = 0 C[0]=1 flip count=1
A[2] xor B[2] 0 xor 1 = 1 C[0]=1 no flip
A[3] xor B[3] 0 xor 0 = 0 C[0]=1flip count=2

Input

A[]= { 0,0,1,1 } B[]= { 0,0,1,1 } C= {0,0,1,1}

Output

Required flips : 2

Explanation

’s Chinese translation is:

Explanation

A[0] xor B[0] 0 xor 0 = 0 C[0]=0 no flip
A[1] xor B[1] 0 xor 0 = 0 C[0]=0 no flip
A[2] xor B[2] 1 xor 1 = 0 C[0]=1 flip count=1
A[3] xor B[3] 1 xor 1 = 0 C[0]=1 flip count=2

The methods used in the following program are as follows

• Arrays a[], b[] and c[] are used to store binary numbers. < /p>

• Function FlipCount(int A[], int B[], int C[], int n) takes the arrays a, b, c and their length n as Enter and return the number of flips required on the bits of A[] or B[] so that C[] equals A XOR B B

• The variable count represents the flip count and is initialized to 0.

• Use a for loop to iterate through each bit in the cell starting from i = 0 to i

• For each bit A[i ] and B[i]. If they are equal and C[i] is 1, increment the count.

• For each bit A[i] and B[i]. If they are not equal and C[i] is 0, increment the count.

• Returns the count of the desired results.

Example

Live demonstration

#include<bits/stdc++.h>
using namespace std;
int flipCount(int A[], int B[], int C[], int N){
   int count = 0;
   for (int i=0; i < N; ++i){
      // If both A[i] and B[i] are equal then XOR results 0, if C[i] is 1 flip
      if (A[i] == B[i] && C[i] == 1)
         ++count;
         // If Both A and B are unequal then XOR results 1 , if C[i] is 0 flip
      else if (A[i] != B[i] && C[i] == 0)
         ++count;
   }
   return count;
}
int main(){
   //N represent total count of Bits
   int N = 5;
   int a[] ={1,0,0,0,0};
   int b[] ={0,0,0,1,0};
   int c[] ={1,0,1,1,1};
   cout <<"Minimum bits to flip such that XOR of A and B equal to C :"<<flipCount(a, b, c,N);
   return 0;
}

Output

Minimum bits to flip such that XOR of A and B equal to C :2

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