Introduction

In this blog post, we will explore how to solve the LeetCode problem "2790" 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

You are given a 0-indexed array usageLimits of length n. Your task is to create groups using numbers from 0 to n - 1, ensuring that each number, i, is used no more than usageLimits[i] times in total across all groups. You must also satisfy the following conditions: Each group must consist of distinct numbers, meaning that no duplicate numbers are allowed within a single group. Each group (except the first one) must have a length strictly greater than the previous group. Return an integer denoting the maximum number of groups you can create while satisfying these conditions. Example 1: Input: usageLimits = [1,2,5] Output: 3 Explanation: In this example, we can use 0 at most once, 1 at most twice, and 2 at most five times. One way of creating the maximum number of groups while satisfying the conditions is: Group 1 contains the number [2]. Group 2 contains the numbers [1,2]. Group 3 contains the numbers [0,1,2]. It can be shown that the maximum number of groups is 3. So, the output is 3. Example 2: Input: usageLimits = [2,1,2] Output: 2 Explanation: In this example, we can use 0 at most twice, 1 at most once, and 2 at most twice. One way of creating the maximum number of groups while satisfying the conditions is: Group 1 contains the number [0]. Group 2 contains the numbers [1,2]. It can be shown that the maximum number of groups is 2. So, the output is 2. Example 3: Input: usageLimits = [1,1] Output: 1 Explanation: In this example, we can use both 0 and 1 at most once. One way of creating the maximum number of groups while satisfying the conditions is: Group 1 contains the number [0]. It can be shown that the maximum number of groups is 1. So, the output is 1. Constraints: 1 <= usageLimits.length <= 105 1 <= usageLimits[i] <= 109

Explanation

Here's the solution, incorporating efficiency and adhering to the prompt's requirements:

• Key Idea: Sort the usageLimits array to prioritize using elements with smaller limits first. Then, iteratively build groups, ensuring each new group is larger than the previous. A greedy approach is employed here.
• Binary Search Optimization: Use binary search to efficiently determine the maximum possible number of groups. This avoids a linear iteration and significantly improves performance, especially for large input arrays.

• Early Exit: Exit early when creating groups becomes impossible (insufficient capacity).

• Runtime & Storage Complexity: O(n log n) time for sorting and binary search operations, O(1) storage (excluding the input array).

Code

    def maximumGroups(usageLimits):
    """
    Calculates the maximum number of groups that can be created given usage limits.

    Args:
        usageLimits: A list of integers representing the usage limits for each number.

    Returns:
        The maximum number of groups that can be created.
    """

    usageLimits.sort()  # Sort usage limits in ascending order

    n = len(usageLimits)
    left, right = 0, n

    while left <= right:
        mid = (left + right) // 2
        total_needed = mid * (mid + 1) // 2
        total_available = 0

        possible = True
        for i in range(n):
            total_available += usageLimits[i]

        if total_available < total_needed:
            right = mid -1
            continue    

        current_sum = 0
        for i in range(mid):
          current_sum += usageLimits[i]

        needed_capacity = mid * (mid+1) // 2

        sufficient_capacity = True

        remaining_capacity = current_sum
        group_size = 1

        for i in range(1, mid + 1):
          remaining_capacity -= i
          if remaining_capacity < 0:
            sufficient_capacity = False
            break


        if sufficient_capacity:
          left = mid + 1
        else:
          right = mid - 1

    return right