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/*
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So in the question we want to find out the
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number of difference of bits between each pair
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so in the brute force we will iterate over the vector
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and for every pair we will calculate the Hamming  distance
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the Hamming distance will be calculated by taking XOR  between the two elements
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and then finding out the number of ones in the XOR of those two elements
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the intuition behind this method is that XOR will contain 1's at those places
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where the corresponding bits of  elements x & y are different
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therefore we will add this count
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to our answer
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*/
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class Solution {
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 public:
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  int hammingDistance(int x, int y) {
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    int XOR = x ^ y;
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    int count = 0;
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    while (XOR) {
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      if (XOR & 1) count++;
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      XOR = XOR >> 1;
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    }
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    return count;
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  }
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  int totalHammingDistance(vector<int> &nums) {
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    int ans = 0;
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    for (int i = 0; i < nums.size() - 1; i++) {
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      for (int j = i + 1; j < nums.size(); j++) {
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        ans += hammingDistance(nums[i], nums[j]);
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      }
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    }
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    return ans;
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  }
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};