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This approach leverages binary search to efficiently minimize the penalty. The idea is to use binary search to determine the minimum possible 'penalty', where penalty is defined as the maximum number of balls allowed in any bag.
We set our binary search range from 1 to the maximum number of balls in any bag initially. For each mid value (potential penalty), we check if it's possible to achieve this penalty with the given number of operations. We do this by calculating how many operations we'd need to make every bag have at most 'mid' number of balls. If we can do that within the allowed number of operations, we adjust our search range to potentially lower penalties; otherwise, we increase our search range.
Time Complexity: O(n log(max(nums)))
Space Complexity: O(1)
1#include <stdio.h>
2
3int canAchievePenalty(int* nums, int numsSize, int maxOperations, int penalty) {
4 int operations = 0
The C solution uses a helper function canAchievePenalty which computes whether it's possible to split the balls such that the maximum number of balls in any bag does not exceed the proposed penalty, within the given number of operations. The main function minimumPenalty then uses binary search over this helper to find the minimum possible penalty.
Another approach is to continuously divide the largest bags using a priority queue to keep track of the largest bag sizes. By always splitting the largest bag first, we aim to reduce the penalty faster. This works similarly to the binary search approach in terms of dividing the problem space but uses a different method to handle priorities.
Time Complexity: O(n log n + n log(max(nums)))
Space Complexity: O(1)
1
In Java, we utilize a PriorityQueue to manage and split the largest bag present until no more operations can be performed. The priority queue helps in dynamically finding the largest element efficiently.