## Monday, February 20, 2017

### LeetCode OJ - Total Hamming Distance

Problem:

Analysis:

The input could have 10,000 elements, we need to figure out a way to compute the total hamming distance without going through all pairs.

Solution:

Imagine for a bit they are all 1 bit number (i.e. either 0 or 1), then it is easy. Just count the number of zeros and number of ones and multiply them together.

For a full solution, just do this for all bits.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/total-hamming-distance

#include "LEET_TOTAL_HAMMING_DISTANCE.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_TOTAL_HAMMING_DISTANCE
{
class Solution
{
public:
int totalHammingDistance(vector<int>& nums)
{
int dist = 0;
for (int i = 0; i < 32; i++)
{
int zero = 0;
int ones = 0;
for (size_t j = 0; j < nums.size(); j++)
{
bool is_zero = (nums[j] & mask) == 0;
if (is_zero) { zero++; } else { ones++; }
}
dist += zero * ones;
}
return dist;
}
};
};

using namespace _LEET_TOTAL_HAMMING_DISTANCE;

int LEET_TOTAL_HAMMING_DISTANCE()
{
Solution solution;
int input_array[] = { 4, 14, 2 };
vector<int> input(input_array, input_array + _countof(input_array));
cout << solution.totalHammingDistance(input) << endl;
return 0;
}