Count Number of Bad Pairs LeetCode solution in C++

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class Solution {
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 public:
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  long long countBadPairs(vector<int> &nums) {
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    // j - i != nums[j] - nums[i] means  nums[i]-i != nums[j]-j
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    map<long long, long long> mp;
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    for (int i = 0; i < nums.size(); i++) {
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      nums[i] = nums[i] - i;
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      mp[nums[i]]++;
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    }
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    long long n = nums.size();
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    long long totalPair = n * (n - 1) / 2;
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    for (auto &it : mp) {
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      if (it.second > 1) {
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        totalPair -= (it.second) * (it.second - 1) / 2;
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      }
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    }
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    return totalPair;
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  }
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};