Introduction

In this blog post, we will explore how to solve the LeetCode problem "1395" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 3 <= n <= 1000 1 <= rating[i] <= 105 All the integers in rating are unique.

Explanation

Here's the breakdown of the problem and the solution:

• High-Level Approach: The core idea is to iterate through each soldier as the potential middle soldier (rating[j]). For each j, we count how many soldiers to the left have a smaller rating (less_left) and how many have a larger rating (greater_left). Similarly, we count how many soldiers to the right have a smaller (less_right) and larger (greater_right) rating. Finally, we calculate the number of increasing teams (less_left * greater_right) and decreasing teams (greater_left * less_right) that can be formed with this soldier as the middle one.

• Complexity:

  • Runtime: O(n²) - due to nested loops.
  • Storage: O(1) - constant extra space.

Code

    def numTeams(rating):
    n = len(rating)
    count = 0
    for j in range(1, n - 1):
        less_left = 0
        greater_left = 0
        less_right = 0
        greater_right = 0
        for i in range(j):
            if rating[i] < rating[j]:
                less_left += 1
            elif rating[i] > rating[j]:
                greater_left += 1
        for k in range(j + 1, n):
            if rating[k] < rating[j]:
                less_right += 1
            elif rating[k] > rating[j]:
                greater_right += 1
        count += (less_left * greater_right) + (greater_left * less_right)
    return count