int Solution::candy(vector<int> &A) {
    int n = A.size();
    int ans = 1;
    if(!n) return 0;
    vector<int> inc(n); inc[0] = 0;
    vector<int> curr(n); curr[0] = 1;
    for(int i = 1; i < n; i++){
        if(A[i] >= A[i-1]) inc[i] = 0;
        else inc[i] = inc[i - 1] + 1;
    }
    for(int i = 1; i < n; i++){
        int last = curr[i - 1];
        if(A[i] > A[i - 1]){
            curr[i] = last + 1;
            ans += curr[i];
        } else{
            ans++;
            if(last == 1){
                // Adding 1 to each one behind me
                // Example :    4 3 2 1 1 <- me
                // Since the last one is 1 and me is also 1, I need to add one to it
                // becomes 2, but the last of the last could be 2 and so on
                // so inc[i] will store how many in incresing order from right to left
                // I have so then I will increment all of them in O(1)
                ans += inc[i];

                // This is a tricky if for a special case
                // Exmaple :  1 2 3 4 5 6 3 2 1 1 <- me
                // With inc[i] I would add 1 to all my last 4 numbers, since I had
                // 6 3 2 1 1 to become 7 4 3 2 1 <- me, but I didnt had to inc
                // 6 to become 7 because 3 + 1 < 6, so I check if my last element of that
                // sequence is greater than the value that it should be, to uncount that
                if(curr[i - inc[i]] > inc[i]) ans--;
            }
            curr[i] = 1;
        }
    }
    return ans;
}