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Solving Leetcode Interviews in Seconds with AI: Minimum Number of Groups to Create a Valid Assignment

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3 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "2910" 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 collection of numbered balls and instructed to sort them into boxes for a nearly balanced distribution. There are two rules you must follow: Balls with the same box must have the same value. But, if you have more than one ball with the same number, you can put them in different boxes. The biggest box can only have one more ball than the smallest box. ​Return the fewest number of boxes to sort these balls following these rules. Example 1: Input: balls = [3,2,3,2,3] Output: 2 Explanation: We can sort balls into boxes as follows: [3,3,3] [2,2] The size difference between the two boxes doesn't exceed one. Example 2: Input: balls = [10,10,10,3,1,1] Output: 4 Explanation: We can sort balls into boxes as follows: [10] [10,10] [3] [1,1] You can't use fewer than four boxes while still following the rules. For example, putting all three balls numbered 10 in one box would break the rule about the maximum size difference between boxes. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109

Explanation

Here's the breakdown of the solution:

  • Count Ball Frequencies: Determine the count of each distinct ball number in the input array.
  • Iterate Possible Box Counts: Iterate through possible numbers of boxes, starting from 1, and check if the given ball counts can be distributed among that many boxes, adhering to the size constraints.
  • Check Feasibility: For each possible number of boxes, determine the ideal box sizes (either x or x+1) and check if we can allocate each ball type into some boxes of these sizes.

  • Runtime Complexity: O(N * sqrt(N)), where N is the length of the input array.

  • Storage Complexity: O(M), where M is the number of distinct elements in the input array.

Code

    from collections import Counter
import math

def min_boxes(balls):
    """
    Calculates the minimum number of boxes needed to sort the balls according to the given rules.

    Args:
      balls: A list of integers representing the ball numbers.

    Returns:
      The minimum number of boxes required.
    """

    counts = Counter(balls)
    n = len(balls)

    for num_boxes in range(1, n + 1):
        possible = True
        for count in counts.values():
            if not can_distribute(count, num_boxes):
                possible = False
                break
        if possible:
            return num_boxes
    return n  # Worst case: each ball in its own box


def can_distribute(count, num_boxes):
    """
    Checks if 'count' balls can be distributed into 'num_boxes' boxes such that
    the size difference between the largest and smallest box is at most 1.

    Args:
      count: The number of balls of a particular type.
      num_boxes: The number of boxes to distribute the balls into.

    Returns:
      True if the balls can be distributed, False otherwise.
    """

    min_size = count // num_boxes
    max_size = min_size + 1
    num_max_size = count % num_boxes
    num_min_size = num_boxes - num_max_size

    if num_max_size * max_size + num_min_size * min_size == count:
        return True
    else:
        return False

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