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:

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

Solution

Loop Through 32-bit Solution: O(n) time

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

Analysis

Since int has 32 bits, we loop through each bit of every number in given nums
Then we count # of bit ends with 1 and 0
Since each bit ends with 1 and 0 can construct a different hamming distance
We count the res as hasOne * notHasOne (n - hasOne)