Minimum Limit Of Balls In A Bag - Leetcode Solution

Problem Statement: You are given an integer array nums where each integer represents the number of balls in the ith bag. You are also given an integer maxOperations. You can perform the following operation at most maxOperations times: Choose any bag of balls and remove maxLimit number of balls from it. If a bag contains fewer than maxLimit balls, you can remove all the balls from it. Return the minimum number of balls you can have after applying the above operations any number of times.

Example: Input: nums = [9], maxOperations = 2 Output: 3 Explanation:

• Remove 3 balls from the first bag. Now nums = [6].
• Remove 3 balls from the first bag. Now nums = [3]. The bag with the smallest number of balls has 3 balls, so the answer is 3.

Solution: First, we need to find the range of the minimum and maximum number of balls that can be left in a bag after applying the operations. The minimum possible value of balls in a bag is 1 because we cannot remove any more balls from an empty bag. The maximum possible value of balls in a bag is the maximum number of balls in the given array because we cannot add more balls to any bag. Therefore, the range of possible values for the minimum number of balls is [1, max(nums)].

We can use binary search to find the minimum number of balls that can be left in a bag. For each guess of the minimum number of balls, we simulate the removal of balls from each bag until we reach the desired number of operations or the number of balls in all bags is less than the guessed minimum. If the number of operations is less than or equal to the given value, then we know that we can achieve the guessed minimum number of balls. Otherwise, we need to increase our guess of the minimum number of balls.

Time Complexity: O(n * log_2(max(nums))*log_2(n)), where n is the length of the input array. The outer binary search loop runs log_2(max(nums)) times because this is the maximum possible range of the binary search. The inner loop simulates the removal of balls from each bag and runs log_2(n) times because this is the maximum possible number of times we can simulate the removal of balls from a bag.

Space Complexity: O(1) because we only need constant extra space to store the variables used in the binary search and simulation.

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