Introduction

In this blog post, we will explore how to solve the LeetCode problem "1760" 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 an integer array nums where the ith bag contains nums[i] balls. You are also given an integer maxOperations. You can perform the following operation at most maxOperations times: Take any bag of balls and divide it into two new bags with a positive number of balls. For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls. Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations. Return the minimum possible penalty after performing the operations. Example 1: Input: nums = [9], maxOperations = 2 Output: 3 Explanation: - Divide the bag with 9 balls into two bags of sizes 6 and 3. [9] -> [6,3]. - Divide the bag with 6 balls into two bags of sizes 3 and 3. [6,3] -> [3,3,3]. The bag with the most number of balls has 3 balls, so your penalty is 3 and you should return 3. Example 2: Input: nums = [2,4,8,2], maxOperations = 4 Output: 2 Explanation: - Divide the bag with 8 balls into two bags of sizes 4 and 4. [2,4,8,2] -> [2,4,4,4,2]. - Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,4,4,4,2] -> [2,2,2,4,4,2]. - Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,4,4,2] -> [2,2,2,2,2,4,2]. - Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,2,2,4,2] -> [2,2,2,2,2,2,2,2]. The bag with the most number of balls has 2 balls, so your penalty is 2, and you should return 2. Constraints: 1 <= nums.length <= 105 1 <= maxOperations, nums[i] <= 109

Explanation

Here's the breakdown of the solution:

• Binary Search: The problem can be solved using binary search. We search for the minimum possible penalty within the range of 1 to the maximum value in the nums array.

• Check Feasibility: For each potential penalty value (mid value in binary search), we determine if it's possible to achieve that penalty within the given maxOperations. We do this by calculating how many operations are needed to reduce each bag's size to be less than or equal to the potential penalty.

• Adjust Search Range: Based on whether the potential penalty is feasible or not, we adjust the binary search range (either search in the lower half or the upper half).

• Time & Space Complexity: O(n log(max(nums))), O(1)

Code

    def minimumSize(nums, maxOperations):
    """
    Finds the minimum possible penalty after performing the operations.

    Args:
        nums: A list of integers representing the number of balls in each bag.
        maxOperations: The maximum number of operations allowed.

    Returns:
        The minimum possible penalty after performing the operations.
    """

    def check(penalty):
        """
        Checks if it's possible to achieve the given penalty within the allowed operations.
        """
        operations = 0
        for num in nums:
            operations += (num - 1) // penalty  # Integer division to get the number of operations needed
        return operations <= maxOperations

    left, right = 1, max(nums)
    ans = right  # Initialize the answer to the maximum possible penalty

    while left <= right:
        mid = (left + right) // 2

        if check(mid):
            ans = mid
            right = mid - 1  # Try to find an even smaller penalty
        else:
            left = mid + 1  # The current penalty is too small, need to increase it

    return ans