class Solution {
public:
int hammingDistance(int x, int y) {
int dist = 0, n = x ^ y;
while (n) {
++dist;
n &= n  1;
}
return dist;
}
};
My C++ solution using bit manipulation

@jonvasquez1
So the while works by turning off the right most bit in n in each iteration, the details are as follows.the "while(n)" par is equivalent to while there is a bit set go into the loop
when were in the loop the "++dist" just counts how may bits we have turned off(set to 0) so far
the "n &= n1" turns off(set to 0) the right most 1 bit, you can see this by just trying a few examples.
so where we exit the loop we know that n must be zero and hence dist will contain the number of bits set to one in x^y.
That's how the loop works.



@jonvasquez1 Considering this example. Let's say
n = 10101
, anddist = 0
asccording to above code of @pengr7. Before we go into the
while
loop.n = 10101, dist = 0
 Loop 1.
dist = 1, n =10101 & 10100 = 10100
 Loop 2.
dist = 2, n =10100 & 10011 = 10000
 Loop 3.
dist = 3, n =10000 & (0)1111 = 0
 Loop ends.
dist = 3
The change of
n
:10101
10100
10000
00000
Conclusion:
n & (n1)
converts the right most1
to0
.
This is the key idea solving the problem. By counting how many times we can perform this operation, we know how many1
exists in the binary representation ofn
 Before we go into the


@pengr7 Your solution is very fast and performant, I like it a lot!!! In my original solution I used bitwise shift operator and counted 1s when they became LSB. Which turns out to be the slowest solution of all because number of operations is quite large (when your numbers are big).