Count Number of Teams

class Solution {
public:
    int countTeams(vector<int>& arr) {
        // Find the highest rating
        int highestRating = 0;
        for (int n : arr) {
            highestRating = max(highestRating, n);
        }

        // Initialize BITs for calculation
        vector<int> bitLeft(highestRating + 1, 0);
        vector<int> bitRight(highestRating + 1, 0);

        // Fill the right BIT with the initial values
        for (int n : arr) {
            bitUpdate(bitRight, n, 1);
        }

        int resultCount = 0;
        for (int r : arr) {
            // Remove the current element from the right BIT
            bitUpdate(bitRight, r, -1);

            // Calculate smaller and larger counts on both sides
            int leftLess = queryBIT(bitLeft, r - 1);
            int rightLess = queryBIT(bitRight, r - 1);
            int leftMore = queryBIT(bitLeft, highestRating) - queryBIT(bitLeft, r);
            int rightMore = queryBIT(bitRight, highestRating) - queryBIT(bitRight, r);

            // Compute results for valid sequences
            resultCount += (leftLess * rightMore) + (leftMore * rightLess);

            // Insert current into the BIT for left side
            bitUpdate(bitLeft, r, 1);
        }

        return resultCount;
    }

private:
    // Function to update a BIT
    void bitUpdate(vector<int>& bit, int index, int delta) {
        while (index < bit.size()) {
            bit[index] += delta;
            index += index & (-index); // Next index calculation
        }
    }

    // Function to compute prefix sum in a BIT
    int queryBIT(vector<int>& bit, int index) {
        int sum = 0;
        while (index > 0) {
            sum += bit[index];
            index -= index & (-index); // Move to the parent
        }
        return sum;
    }
};

The problem "Count Number of Teams" demands an efficient solution to count special sequences within a list of ratings. The given C++ solution implements this by utilizing Binary Indexed Trees (BITs) to handle some specific BIT manipulations like updates and queries for prefix sums.

Here’s a breakdown of how the solution works:

• Define the Highest Rating in the Rating Array: First, iterate over the ratings to determine the largest value, necessary for defining the size of two Binary Indexed Trees (BITs) used in the solution: bitLeft and bitRight.

• Initialize the BITs:

  • bitLeft is initialized to keep track of how many elements with a certain rating have been processed as we iterate from left to right.
  • bitRight is used to track how many elements of each rating exist to the right of the current element in the traversal.

  Populate bitRight with the initial counts of each rating.

• Iterate Over the Ratings to Compute Result: For each team member's rating in the array:

  1. Update bitRight to indicate that the current member's rating is now permanent and is moving from bitRight to bitLeft.
  2. Query both bitLeft and bitRight to ascertain the count of ratings both less than and greater than the current rating, to the left and right.
  3. Calculate the number of valid sequences using combinations of these counts.
  4. Update bitLeft to reflect the addition of the current rating now being considered as processed.

• Update and Query Functions: Define helper functions bitUpdate and queryBIT to respectively update the counts in the BITs and obtain prefix sum queries, both critical operations made efficient by the BIT structure.

This effectively allows maintaining and querying count information about ratings dynamically through the array traversal, crucial for calculating the result without resorting to less efficient brute force methods. The careful and efficient use of BIT operations provides the necessary mechanisms to compute the desired result of how many special sequences can be formed, optimizing the approach substantially over simpler iterative checks.

class Solution {

    public int calculateTeams(int[] ratings) {
        int highestRating = 0;
        for (int rate : ratings) {
            highestRating = Math.max(highestRating, rate);
        }

        int[] leftIndexTree = new int[highestRating + 1];
        int[] rightIndexTree = new int[highestRating + 1];

        for (int rate : ratings) {
            updateIndexTree(rightIndexTree, rate, 1);
        }

        int totalTeams = 0;
        for (int current : ratings) {
            updateIndexTree(rightIndexTree, current, -1);

            int leftLesser = sumPrefix(rightIndexTree, current - 1);
            int rightLesser = sumPrefix(leftIndexTree, current - 1);
            int leftGreater = sumPrefix(leftIndexTree, highestRating) - sumPrefix(leftIndexTree, current);
            int rightGreater = sumPrefix(rightIndexTree, highestRating) - sumPrefix(rightIndexTree, current);

            totalTeams += (leftLesser * rightGreater);
            totalTeams += (leftGreater * rightLesser);

            updateIndexTree(leftIndexTree, current, 1);
        }

        return totalTeams;
    }

    private void updateIndexTree(int[] tree, int idx, int val) {
        while (idx < tree.length) {
            tree[idx] += val;
            idx += idx & (-idx);
        }
    }

    private int sumPrefix(int[] tree, int idx) {
        int sum = 0;
        while (idx > 0) {
            sum += tree[idx];
            idx -= idx & (-idx);
        }
        return sum;
    }
}

In the provided Java code, the algorithm focuses on counting the number of valid teams based on a ranking system given in the array 'ratings'. The problem is approached by using two Fenwick Trees (Binary Indexed Trees), one to manage the left index updates and another for right index calculations.

The code functions based on the following sequence of operations:

• Initialize Variables: Two Fenwick Trees are initialized and the maximum value in the ratings array is determined to set the maximal bounds of the trees.

• Fill Right Index Tree: Initially, the right index tree is filled with the frequencies of respective ratings, enabling queries about how many soldiers have a rating less than or equal to the current during the main computation.

• Calculate Teams:

  1. Loop through each rating.

  2. Before processing each rating, decrement its count in the right index tree to exclude the current soldier from subsequent calculations.

  3. Calculate the number of valid combinations for forming teams by determining:

     • leftLesser: The number of soldiers to the left with a rating less than the current soldier.

     • rightLesser: Using the right tree's state at this point to determine how many soldiers can be right of the current with a lower rating.

     • leftGreater: Soldiers to the left with a larger rating.

     • rightGreater: Soldiers to the right with a larger rating.

  4. For each soldier, the number of teams that the soldier can form where he/she is the middle member is incremented by the product of the lesser counterparts on one side and the greater counterparts on the other side.

• Final Result: At the end of iterations, the totalTeams variable holds the total count of 3-member teams sorted by the rank that can be formed.

• Helper Functions: updateIndexTree facilitates updating values in the Fenwick Trees, while sumPrefix helps in querying the prefix sums to count how many ratings are below a certain threshold.

This method efficiently computes possible teams by leveraging the logarithmic update and query capabilities of the Fenwick Tree, thus ensuring that the solution is scalable for larger datasets.

class Solution:
    def countTeams(self, ratings: List[int]) -> int:
        highest_rating = 0
        for rate in ratings:
            highest_rating = max(highest_rating, rate)

        left_index_tree = [0] * (highest_rating + 1)
        right_index_tree = [0] * (highest_rating + 1)

        for rate in ratings:
            self._updateTree(right_index_tree, rate, 1)

        result = 0
        for rate in ratings:
            self._updateTree(right_index_tree, rate, -1)

            count_smaller_left = self._fetchSum(left_index_tree, rate - 1)
            count_smaller_right = self._fetchSum(right_index_tree, rate - 1)
            count_larger_left = self._fetchSum(left_index_tree, highest_rating) - self._fetchSum(left_index_tree, rate)
            count_larger_right = self._fetchSum(right_index_tree, highest_rating) - self._fetchSum(right_index_tree, rate)

            result += count_smaller_left * count_larger_right
            result += count_larger_left * count_smaller_right

            self._updateTree(left_index_tree, rate, 1)

        return result

    def _updateTree(self, index_tree: List[int], position: int, increment: int) -> None:
        while position < len(index_tree):
            index_tree[position] += increment
            position += position & (-position)

    def _fetchSum(self, index_tree: List[int], position: int) -> int:
        total = 0
        while position > 0:
            total += index_tree[position]
            position -= position & (-position)
        return total

This solution tackles the problem of counting the number of valid teams based on a rating system where a team can be formed by any three participants such that their ratings strictly increase or decrease.

The provided Python code uses a class Solution with the method countTeams which accepts a list of integers ratings representing the ratings of individuals. The process involves:

• Finding the highest rating to limit the size of Binary Indexed Trees (BIT) used for efficient range sum queries and updates.
• Initializing two BITs (left_index_tree and right_index_tree) to handle frequencies of ratings to the left and right of the current element respectively.
• Iterating over each rating to update right_index_tree indicating the presence of each rating.
• For each rating:
  • Decrease count in right_index_tree (because moving past the current rating).
  • Counting the numbers smaller and larger on both left and right of each rating using the helper method _fetchSum which returns the sum of the elements up to a given position in BIT.
  • Calculate potential valid teams and aggregate to previous results.
  • Update left_index_tree to include the current rating.
• The method _updateTree helps in modifying the BIT at a specified position, and _fetchSum is used to compute prefix sums from BIT for range queries.

This approach efficiently counts potential teams with the BIT managing frequency and partial sums computation, allowing quick updates and range sum calculations which are fundamental in this problem, accommodating dynamic changes in data within the rating sequence.