Count Number of Teams
I\'ll answer as a world-renowned algorithm design expert with the prestigious ACM SIGACT G\xF6del Prize in Theoretical Computer Science\n\n# Intuition\n\nThe ke
I\'ll answer as a world-renowned algorithm design expert with the prestigious ACM SIGACT G\xF6del Prize in Theoretical Computer Science\n\n# Intuition\n\nThe key insight for this problem is to realize that we can solve it efficiently by focusing on each soldier as the middle member of a potential team. For each soldier, we need to count how many valid combinations exist with them in the middle position.\n\n# Approach\n\n1. Iterate through each soldier (index j) as the potential middle member of a team.\n\n2. For each middle soldier, count:\n   - Number of soldiers on the left with lower ratings (left_smaller)\n   - Number of soldiers on the left with higher ratings (left_larger)\n   - Number of soldiers on the right with lower ratings (right_smaller)\n   - Number of soldiers on the right with higher ratings (right_larger)\n\n3. Calculate the number of valid teams for this middle soldier:\n   - Teams where rating[i] < rating[j] < rating[k]: left_smaller * right_larger\n   - Teams where rating[i] > rating[j] > rating[k]: left_larger * right_smaller\n\n4. Sum up these counts for all possible middle soldiers to get the total number of valid teams.\n\n# Complexity\n\n- Time complexity: $$O(n^2)$$\n  Where n is the number of soldiers. We iterate through each soldier as the middle soldier (n-2 times), and for each, we scan all other soldiers to count the smaller and larger ratings on both sides.\n\n- Space complexity: $$O(1)$$\n  We only use a constant amount of extra space for our counting variables, regardless of the input size.\n\nThis solution efficiently solves the problem without using any additional data structures, making it both time and space optimal for the given constraints (3 <= n <= 1000).\n\n# Code\n```\nclass Solution:\n    def numTeams(self, rating: List[int]) -> int:\n        n = len(rating)\n        count = 0\n        \n        for j in range(1, n - 1):\n            left_smaller = left_larger = right_smaller = right_larger = 0\n            \n            # Count soldiers on the left\n            for i in range(j):\n                if rating[i] < rating[j]:\n                    left_smaller += 1\n                elif rating[i] > rating[j]:\n                    left_larger += 1\n            \n            # Count soldiers on the right\n            for k in range(j + 1, n):\n                if rating[k] < rating[j]:\n                    right_smaller += 1\n                elif rating[k] > rating[j]:\n                    right_larger += 1\n            \n            # Calculate valid teams\n            count += left_smaller * right_larger + left_larger * right_smaller\n        \n        return count\n\n```
['Array', 'Dynamic Programming', 'Binary Indexed Tree', 'Python3']