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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)
1import java.util.*;
2
3public class MinimumPenalty {
4 private static boolean canAchievePenalty(int[] nums, int maxOperations
The Java solution uses a binary search approach to find the minimum penalty. It checks if we can split the balls into bags each with a maximum count of mid using the canAchievePenalty function. Based on the result, the search range is adjusted until convergence.
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#include <queue>
#include <iostream>
using namespace std;
int minimumPenalty(vector<int>& nums, int maxOperations) {
priority_queue<int> pq(nums.begin(), nums.end());
while (maxOperations > 0) {
int largest = pq.top();
pq.pop();
int split = (largest + 1) / 2;
pq.push(split);
pq.push(largest - split);
--maxOperations;
}
return pq.top();
}
int main() {
vector<int> nums = {9};
int maxOperations = 2;
cout << minimumPenalty(nums, maxOperations) << endl; // Output: 3
return 0;
}This C++ solution uses a priority queue (max-heap) to always divide the largest bag available. The goal is to maintain the smallest possible maximum size of a bag after all operations have been completed or all available splits are exhausted.