1395. Count Number of Teams

Problem Description

The problem presented deals with forming teams of 3 soldiers from a line-up where each soldier has a unique rating value. The goal is to form a team following certain rules. A team is considered valid if one of these two conditions are satisfied:

• the ratings of the three soldiers (let's call them rating[i], rating[j], and rating[k]) are in increasing order (rating[i] < rating[j] < rating[k])
• or the ratings are in decreasing order (rating[i] > rating[j] > rating[k]).

The key constraint is that we must choose three soldiers with indices i, j, k (0 ≤ i < j < k < n) to form a valid team. The question asks for the total number of valid teams that we can create, given that soldiers can be part of multiple teams.

Intuition

The solution uses a Binary Indexed Tree (BIT), also known as a Fenwick Tree, to efficiently calculate the cumulative frequencies (or, in this context, the number of soldiers with certain ratings) and modifies them over time.

To solve this problem, it becomes necessary to consider each valid triplet (i, j, k) and check if they form an increasing or decreasing sequence. However, doing so with a brute force approach would take O(n^3) time, which is too slow for large inputs.

Instead of checking every possible triplet, we leverage the BIT to store and quickly compute the number of ratings that are less than or greater than a given rating (while maintaining the ordering constraint). This is helpful because, for a given soldier j, we are interested in how many soldiers to the left have a lower rating (for an increasing sequence) or a higher rating (for a decreasing sequence), and similarly, how many soldiers to the right have a higher rating (increasing sequence) or a lower rating (decreasing sequence).

Here’s the intuition behind our solution steps:

1. We store all unique ratings in a sorted array to handle ratings[i] in a relative manner.
2. We create two BITs, tree1 and tree2, for cumulative frequency calculations.
3. Before we start iterating over each soldier, we update tree2 with their ratings to represent the total number of soldiers having a certain rating.
4. For each soldier i in rating, we do the following:
   • Using binary search, find the index x of the current rating in the sorted array of ratings.
   • Update BIT tree1 by adding 1 to index x because we now have one more soldier with a rating up to x.
   • Query BIT tree1 to get the count of soldiers to the left of soldier i with a smaller rating (l), and similarly, query tree2 to get the count of soldiers to the right of i with a greater rating (r).
   • Calculate the number of valid increasing or decreasing triplets that soldier i can form as the center by multiplying l and r for the increasing sequence and (i-l) and (n-i-1-r) for the decreasing sequence, adding both results to the total count ans.
   • Update tree2 by subtracting 1 from index x because we have now counted soldier i and it should not be included in the calculation for future soldiers.

Finally, ans will contain the total number of valid teams that meet the problem's conditions.

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Solution Approach

The implementation of the solution involves the usage of Binary Indexed Trees (BITs) to perform efficient range queries and updates. The Binary Indexed Tree is a data structure that allows getting the prefix sum up to a certain index and updating the frequencies, both operations in O(log n) time. This feature is particularly useful in this problem when we need to calculate the count of soldiers with ratings less than or greater than the current rating quickly.

Here's a step-by-step breakdown of the solution's implementation:

1. Initialization:

   • We first create a set from the rating array and then convert it into a sorted list called nums. This step effectively maps the actual ratings to indexes, where nums[0] would contain the smallest rating, and so on.
   • Two Binary Indexed Trees, tree1 and tree2, are initialized. tree1 is used to track the counts of soldiers when traversing from left to right, and tree2 would be used in the opposite direction.

2. Populating tree2:

   • Before iterating over the soldiers, we populate tree2 with the frequency of each soldier's rating. This preparation step will help us later when we want to count the number of soldiers on the right of a particular soldier with a certain condition (rating higher or lower).

3. Iterating Over Soldiers:

   • The main loop of the implementation goes over each soldier's rating. The variable n holds the total number of soldiers, and ans is initialized to 0, to accumulate the number of valid teams.
   • For each soldier with rating v, we perform the following actions:
     • Find the index x representing the soldier's rating in the nums array using binary search.
     • Use tree1 to find the number of soldiers to the left of the current soldier with a lower rating (l) by querying it up to index x-1.
     • Find the number of soldiers to the right of the current soldier with a higher rating (r by querying tree2 from index x+1 to the end and subtracting it from the total remaining soldiers to the right.
     • Add the products l * r and (i - l) * (n - i - 1 - r) to ans. The first product corresponds to the number of increasing sequences that can be formed, and the second is for the decreasing sequences.
     • Update tree1 by adding 1 to index x to include the current soldier's rating in the left count for subsequent soldiers.
     • Decrement the frequency of the current soldier's rating in tree2 by updating index x with -1 to exclude it from the right count for subsequent soldiers.

This approach efficiently computes the valid teams by focusing on each soldier and considering them as the middle soldier of a team, checking how many teams they can form with soldiers to their left and right.

After iterating through all the soldiers, the variable ans holds the final count of all valid teams, according to the problem's conditions.

Example Walkthrough

Let us take an example rating array: [2, 5, 3, 4, 1].

Following the solution approach:

1. Initialization:

   • We first create a set from the rating array: {1, 2, 3, 4, 5}.
   • Convert this set into a sorted list nums: [1, 2, 3, 4, 5].
   • Initialize two Binary Indexed Trees, tree1 and tree2.

2. Populating tree2:

   • Before the iteration, we populate tree2 using the indices of nums. The frequencies of the ratings in tree2 will look like this: {1: 1, 2: 1, 3: 1, 4: 1, 5: 1} since each rating appears exactly once.

3. Iterating Over Soldiers:

   • Start iterating over each soldier's rating at index i and ans is set to 0.
   • For i = 0 (rating 2):
     • Find the index x: 1 (nums[1] = 2).
     • Find l (left count) using tree1: query(0) = 0 (no soldiers to the left with lower ratings).
     • Find r (right count) using tree2: query(5) - query(1) = 4 - 1 = 3 (three soldiers to the right with higher ratings).
     • Add l * r + (0 - l) * (4 - 0 - 1 - r) = 0 * 3 + 0 * 0 = 0 to ans.
     • Update tree1 at index x: update(1, 1).
     • Update tree2 at index x: update(1, -1).
   • Continue for the rest of the indices:
     • i = 1 (rating 5): x = 4 ...
     • i = 2 (rating 3): x = 2 ...
     • i = 3 (rating 4): x = 3 ...
     • i = 4 (rating 1): x = 0 ...

   For the purpose of the example walkthrough, let's detail the step for i = 2 (rating 3):

   • Find the index x: 2 (nums[2] = 3).
   • Find l using tree1: query(1) = 1 (one soldier to the left with a lower rating 2).
   • Find r using tree2: query(5) - query(2) = 2 - 1 = 1 (one soldier to the right with a higher rating 4).
   • Add l * r + (2 - 1) * (4 - 2 - 1 - 1) = 1 * 1 + 1 * 0 = 1 to ans.
   • Update tree1 at index x with 1 and tree2 at index x with -1.

At the end of iteration, we add up all the contributions to ans from each i, which gives us the total count of valid teams.

In our example, the final ans would be the total number of valid sequences that can be formed using the ratings provided, following the rules described in the problem.

Solution Implementation

Time and Space Complexity

The code provided uses a Binary Indexed Tree (BIT) structure to solve the problem of counting the number of teams that can be formed according to the given rating constraints. Let's analyze the time complexity and space complexity of the code:

Time Complexity

• The initialization of the BinaryIndexedTree takes O(n) time, both for tree1 and tree2, where n is the length of the provided ratings array.
• Sorting the distinct values in rating takes O(m log m) time, where m is the number of distinct ratings.
• Updating and querying operations on a BIT take O(log m), where m is the number of nodes (or the length of the BIT array).
• The for loop to update tree2 runs n times, making the complexity O(n log m).
• The main for loop also runs n times, with each iteration performing a constant number of updates and queries on the BIT, each taking O(log m). This results into another O(n log m).

The overall time complexity can be calculated by adding all the individual complexities: O(n) + O(m log m) + O(n log m) + O(n log m) = O((2n + m) log m). With the constraint that m <= n, the expression simplifies to O(n log n).

Space Complexity

• The space complexity of the BIT tree1 and tree2 is O(m) each since each tree holds an array of size m + 1.
• The space used for sorting to create the nums array is O(m).
• Auxiliary space used for variables is O(1).

Therefore, the total space complexity is primarily determined by the BIT arrays and the sorted list of unique elements. Hence, the final space complexity is O(m) + O(m) + O(m) = O(3m), which simplifies to O(m). Since m <= n, this can also be simplified to O(n).

Both complexities are dependent on the size of the rating input array and the number of distinct ratings within it.

Learn more about how to find time and space complexity quickly using problem constraints.