477. Total Hamming Distance

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

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.

4 (0100) 14 (1110) 2(0010) 可以觀察最右邊第一個bit，所有的數都為0，所以距離是0。 右邊數來第二個bit，1個0，2個1，這個bit造成的Hamming Distance為2，1 * 2得來的。 右邊數來第三個bit，1個0，2個1，這個bit形成的hamming distance為2，1 * 2得來的。 最左邊的bit，2個0，1個1，形成的hamming distance為2 * 1 = 2。

int totalHammingDistance(vector<int>& nums) { // time: O(n); space: O(1)
    int n = nums.size(), res = 0;
    for (int i = 0; i < 32; ++i) {
        int cnt = 0;
        for (int num : nums) {
            if (num & (1 << i)) ++cnt;
        }
        res += cnt * (n - cnt);
    }
    return res;
}