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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 <vector>
2#include <iostream>
3using namespace std;
4
5bool canAchievePenalty(const vector<int>& nums, int maxOperations, int penalty) {
6 int operations = 0;
7 for (int num : nums) {
8 if (num > penalty) {
9 operations += (num - 1) / penalty;
10 }
11 if (operations > maxOperations) {
12 return false;
13 }
14 }
return true;
}
int minimumPenalty(vector<int>& nums, int maxOperations) {
int left = 1, right = 1e9;
while (left < right) {
int mid = left + (right - left) / 2;
if (canAchievePenalty(nums, maxOperations, mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
int main() {
vector<int> nums = {9};
int maxOperations = 2;
cout << minimumPenalty(nums, maxOperations) << endl; // Output: 3
return 0;
}The C++ solution is similar to the C version but uses a vector for ease of use with dynamic array sizes. Binary search is applied over the potential maximum number of balls in the bags (penalty), adjusting the penalty value based on feasibility checks using the canAchievePenalty function.
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.