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.

Solution:

Any number can be represented as a 32-bit binary code.

If we have n numbers, for each binary position, we count the 1s' of all n numbers as bitCount.

The hamming distance between all these n numbers at this position is bitCount * (n - bitCount).

Therefore, after going through all 32 positions, we add up the hamming distances in each position and output the result.

The time complexity is O(n) and the space complexity is O(1).



Code:

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