How do you solve Count Number of Teams on LeetCode?

Count Number of Teams (LeetCode #1395) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: The order of ratings matters significantly in forming valid teams.

What is the brute force solution for Count Number of Teams?

The brute force approach for Count Number of Teams is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of three soldiers to see if they satisfy the given conditions. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Count Number of Teams?

Count Number of Teams uses the following concepts: Array, Dynamic Programming, Binary Indexed Tree, Segment Tree. The recommended patterns to study are: Combination Counting, Dynamic Programming.

What are the interview tips for Count Number of Teams?

Always clarify the problem requirements before jumping to code. Think about edge cases and how your solution handles them. Discuss your thought process with the interviewer to showcase your understanding.

What are common mistakes when solving Count Number of Teams?

Not considering all combinations properly in brute force. Overlooking edge cases where no valid teams can be formed.

#1395

Count Number of Teams

Medium

ArrayDynamic ProgrammingBinary Indexed TreeSegment TreeCombination CountingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(1)

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Intuition

Time O(n²)Space O(1)

The optimal approach leverages counting how many soldiers have a lower or higher rating than the current soldier at index j. This way, we can efficiently calculate valid teams without checking every combination.

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Algorithm

5 steps

1Step 1: Initialize a counter to zero to keep track of valid teams.
2Step 2: For each soldier at index j, count how many soldiers before it have a lower rating and how many have a higher rating.
3Step 3: For each soldier at index j, count how many soldiers after it have a lower rating and how many have a higher rating.
4Step 4: Multiply the counts from Step 2 and Step 3 to get the number of valid teams for that soldier as the middle member.
5Step 5: Sum the results for all soldiers to get the total count.

solution.py15 lines

1# Full working Python code
2
3def countTeams(rating):
4    count = 0
5    n = len(rating)
6    for j in range(n):
7        lower_left = sum(1 for i in range(j) if rating[i] < rating[j])
8        higher_left = sum(1 for i in range(j) if rating[i] > rating[j])
9        lower_right = sum(1 for k in range(j + 1, n) if rating[k] < rating[j])
10        higher_right = sum(1 for k in range(j + 1, n) if rating[k] > rating[j])
11        count += lower_left * higher_right + higher_left * lower_right
12    return count
13
14# Example usage
15print(countTeams([2,5,3,4,1]))  # Output: 3

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Complexity note: The time complexity remains O(n²) due to the nested loops, but we avoid redundant checks by counting ratings, making it more efficient in practice. The space complexity is O(1) since we only use a few variables.

1The order of ratings matters significantly in forming valid teams.
2Counting lower and higher ratings efficiently reduces redundant checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.