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 an you should return 2.
Example 3:
Input: nums = [7,17], maxOperations = 2
Output: 7
Constraints:
1 <= nums.length <= 105
1 <= maxOperations, nums[i] <= 109
Solution:
we can calculate the operations needed to reach target by (num - 1) / target target is defined by largest penalty (max number) in the array
For example, the mid = 3, A[i] = 2, we split it into [2], and operations = 0 A[i] = 3, we split it into [3], and operations = 0 A[i] = 4, we split it into [3,1], and operations = 1 A[i] = 5, we split it into [3,2], and operations = 1 A[i] = 6, we split it into [3,3], and operations = 1 A[i] = 7, we split it into [3,3,1], and operations = 2
class Solution {
public int minimumSize(int[] nums, int maxOperations) {
int left = 1, right = Integer.MAX_VALUE - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
int ops = 0;
for (int n : nums) {
ops += minOpsToReach(n, mid);
}
// System.out.println(mid + ", " + ops);
// mid can be smaller
if (ops <= maxOperations) {
right = mid - 1;
} else {
left = mid + 1;
}
}
// System.out.println(minOpsToReach(9 ,1));
return left;
}
public int minOpsToReach(int num, int target) {
return (num - 1) / target;
}
}