477. Total Hamming Distance

Problem:

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

Analysis:

4:     0 1 0 0

14:   1 1 1 0

2:     0 0 1 0

1:     0 0 0 1

Count bit by bit. Take look at the first bit. There is one 1. The hamming distance for first bit is 3. one 1 to three 0. Second bit has two 1, hamming distance is 4. one 1 to two 0 twice. We can conclude that the hamming distance of specific bit is number of 1 times number of 0.

Solution:

class Solution {
    public int totalHammingDistance(int[] nums) {
        int res = 0, len = nums.length;
        for (int i = 0; i < 32; i++) {
            int count = 0;
            for (int n: nums) {
                count += ((n >> i) & 1);
            }
            res += count * (len - count);
        }
        return res;
    }
}