Take any 2 input numbers and converts into binary format and find the number of digits flipped in between both numbers.

Question

Consider 2 numbers 7 and 10
Binary digits of 7 is 00000111
Binary digits of 10 is 00001010
Output is 3.from the above we can see that 3 digits are flipped.

Answer 1

x=int(input())
y=int(input())
a=[]
b=[]
a1=[]
b1=[]
k=0
while x>=1:
    m=x%2
    x=x//2
    a.append(m)

while y>=1:
    n=y%2
    y=y//2
    b.append(n)

if(len(a)<len(b)):
    while(len(a)!=len(b)):
        a.append(k)

if(len(b)<len(a)):
    while(len(b)!=len(a)):
        b.append(k)

a1=a[::-1]
b1=b[::-1]

for i in a1:
    print(i,end="")
print('\n')

for j in b1:
    print(j,end="")
print('\n')
    
count=0
for i in range(0,len(a1)):
    if(a1[i]!=b1[i]):
        count+=1
print("flips=",count)

Answer 2

#include <iostream>

unsigned int countSetBits(unsigned int n)

{

unsigned int count = 0;

while (n) {

count += n & 1;

n >>= 1;

}

return count;

}

int main()

{

int a, b;

std::cin >> a >>b;

int c = a^b;

std::cout << countSetBits(c);

return 0;

}

Answer 3

n,m=map(int,input().split())
b1=bin(n)
b2=bin(m)
c=0
b1=b1[2:]
b2=b2[2:]
n1=len(b1);n2=len(b2)

b1=str(0)*(8-n1)+b1
b2=str(0)*(8-n2)+b2

for i in range(len(b1)):
    if b1[i]!=b2[i]:
        c+=1;

print(c)