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 of1
and4
balls, or two new bags of2
and3
balls.
- For example, a bag of
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: Binary Search
Find the smallest penalty that requires less or equal ops than max_ops.
Time complexity: O(nlogm)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int minimumSize(vector<int>& nums, int maxOperations) { int l = 1, r = *max_element(begin(nums), end(nums)); while (l < r) { const int m = l + (r - l) / 2; int count = 0; for (int x : nums) count += (x - 1) / m; if (count <= maxOperations) r = m; else l = m + 1; } return l; } }; |