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 of0to10^9
2. Length of the array will not exceed10^4

Solution:

The idea is to count the number of ones and zeros for each of 32 bits. At each bit the total distance is NZeros*NOnes. Because if and only if there is a 0,1 or 1,0 pair, the total distance increments by 1.

    public int totalHammingDistance(int[] nums) {
        int res = 0;
        int len = nums.length;
        for (int i = 0; i <= 31; i++) {
            int zeros = 0;
            for (int j = 0; j < nums.length; j++) {
                if ((nums[j]&(1<<i)) == 0) {
                    zeros++;
                }
            }
            res += zeros*(len - zeros);
        }
        return res;
    }