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# 477. Total Hamming Distance

## LeetCode [477. Total Hamming Distance](https://leetcode-cn.com/problems/total-hamming-distance/)

### Description

The [Hamming distance](https://en.wikipedia.org/wiki/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`.

### Tags

Bit Manipulation

### Solution

We observe each place in binary form of all numbers, then count the number of 0s and 1s, and multiply both and add it onto the result.

### Complexity

* Time complexity: $$O(Cn)$$, `C` is the number of bits of integer type;
* Space complexity: $$O(1)$$

### Code

```go
func totalHammingDistance(nums []int) int {
	var ans int
	for i := 0; i < 30; i++ {
		var count int
		for _, num := range nums {
			count += num >> i & 1
		}
		ans += count * (len(nums) - count)
	}
	return ans
}
```