LintCode "Permutation Index II" !

Simply a variation to "Permutation Index". When calculating current digit index, we consider duplicated case.

Again, similar as "Digit Counts", it is another counting problem and stil digit by digit.

And please note: we can use Fenwick Tree to calculate rank by O(nlgn)

class Solution 
{
    long long dupPerm(unordered_map<int, int> &hash) 
    {
        if (hash.empty()) return 1;

        long long dup = 1;
        for (auto it = hash.begin(); it != hash.end(); ++it) 
        {
        dup *= w(it->second);
        }

        return dup;
    }
    long long w(int n) // n*(n-1)*..*2*1
    {
        long long ret = 1;
        while(n > 1)
        {
            ret *= n;
            n --;
        }
        return ret;
    }
public:
    /**
     * @param A an integer array
     * @return a long integer
     */
    long long permutationIndexII(vector<int>& A) 
    {
        int n = A.size();
        if(!n) return 0;
        
        long long index = 1;
        long long factor= w(n - 1);
        
        //  MSB -> LSB
        for(int i = 0; i < n; i ++)    
        {
            // Calc rank
          int rank = 0;
          for(int j = i + 1; j < n; j ++)
          {            
            if(A[i] > A[j])
              rank ++;
          }
          
          // Calc dup factor
            unordered_map<int, int> hm;          
            for(int j = i; j < n; j ++)
                ++hm[A[j]];
      
            index += rank * factor / dupPerm(hm);
            if(i < (n - 1))factor /= n - i - 1;
        }
        
        return index;
    }
};