2012. Sum of Beauty in the Array

You are given a 0-indexed integer array nums. For each index i (1 <= i <= nums.length - 2) the beauty of nums[i] equals:

• 2, if nums[j] < nums[i] < nums[k], for all 0 <= j < i and for all i < k <= nums.length - 1.
• 1, if nums[i - 1] < nums[i] < nums[i + 1], and the previous condition is not satisfied.
• 0, if none of the previous conditions holds.

Return the sum of beauty of all nums[i] where 1 <= i <= nums.length - 2.

class Solution {
public:
    int sumOfBeauties(vector<int>& A) {
        int res = 0, n = A.size();
        vector<int> l(n), r(n);
        for(int i = 1, ma = A[0]; i < n; i++) {
            l[i] = ma;
            ma = max(ma, A[i]);
        }
        for(int i = n - 2, mi = A[n - 1]; i >= 0; i--) {
            r[i] = mi;
            mi = min(mi, A[i]);
        }
        for(int i = 1; i <= n - 2; i++) {
            if(l[i] < A[i] and A[i] < r[i]) res += 2;
            else if(A[i-1] < A[i] and A[i] < A[i+1]) res += 1;
        }
        return res;
    }
};