Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:

Input: 4, 14, 2

Output: 6

Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case). So the answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

Note:

1. Elements of the given array are in the range of 0 to 10^9
2. Length of the array will not exceed 10^4.

class Solution1(object):
    def totalHammingDistance(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res = 0
        bins = []
        for num in nums:
            bins
 
.append(‘{0:b}‘.format(num).zfill(32))
        for i in range(32):
            bitCount = 0
            # zeroCount
            for num in bins:
                if num[i] is ‘1‘:
                    bitCount += 1
            # zeroCount = (len(nums) - bitCount)
            res += bitCount * (len(nums) - bitCount
 
)
        return res
class Solution2(object):
    def totalHammingDistance(self, nums):
        res = 0
        for i in range(32):
            bitCount = 0
            for num in nums:
                bitCount += num >> i & 1
            res += bitCount * (len(nums) - bitCount)
        return res
nums = [4, 14, 2]
s = Solution1()
res = s.totalHammingDistance(nums)
print(res)

477. Total Hamming Distance 總的漢明距離