# Solution by RikeVoltz

• #### Approach #1 XOR Solution [Accepted]

Find the Hamming distance between two integers.

Idea of solution
Cast input integers to bitsets and use XOR operation. By definition of XOR operation it returns `false` if operands are the same and `true` otherwise. After XOR operation we'll get another bitset with amount of ones which is the same as Hamming distance. After that we'll use `count()` method of bitset to get the final result.

C++ algorithm

``````class Solution {
public:
int hammingDistance(int x, int y) {
bitset<32>xbit(x);
bitset<32>ybit(y);
bitset<32>zbit(xbit^ybit);
return zbit.count();
}
};
``````

In another point of view we can use 'one-line solution':

``````class Solution {
public:
int hammingDistance(int x, int y) {
return (bitset<32>(x)^bitset<32>(y)).count();
}
};
``````

Complexity Analysis

• Time complexity : `O(N_b)`, where `N_b` is the amount of bits in input numbers.
• Space complexity : `O(N_b)`, because we just use 3 bitsets.

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