There are n soldiers standing in a line. Each soldier is assigned a unique rating value.

You have to form a team of 3 soldiers amongst them under the following rules:

• Choose 3 soldiers with index (i, j, k) with rating (rating[i], rating[j], rating[k]).
• A team is valid if:  (rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k]) where (0 <= i < j < k < n).

Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).

 

Example 1:

Input: rating = [2,5,3,4,1]
Output: 3
Explanation: We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1). 

Example 2:

Input: rating = [2,1,3]
Output: 0
Explanation: We can't form any team given the conditions.

Example 3:

Input: rating = [1,2,3,4]
Output: 4

 

Constraints:

• n == rating.length
• 1 <= n <= 200
• 1 <= rating[i] <= 10^5

巧妙算法。对于每个soldier j, 计算他的左边有l个人rating比它小，右边有r个人rating比他大。那对于rating[i]<rating[j]<rating[k]的情况，就从左边l人里任取一个pair with右边r个人里任取一个，总共l*r种。已知l, r，那左边有j-l个人rating比她大，右边有n-1-j-r个人rating比它小，所以另一种情况是总共 (j-l)*(n-1-j-r)种。   实现： O(n^2)

class Solution {
public:
    int numTeams(vector<int>& rating) {
        
        int n = rating.size();
        int res = 0;
        
        for (int j = 0; j<n; j++){
            int l = 0;
            int r = 0;
            for (int i=0; i<j; i++){
                if (rating[i] < rating[j])
                    l++;
            }
            for (int k=j+1; k<n; k++){
                if (rating[k] > rating[j])
                    r++;
            }
            res += l*r + (j-l)*(n-1-j-r);
        }
        
        return res;
    }
};