Split Bags to Minimize the Largest Bag

### Problem `nums[i]` is the number of balls in one bag. One operation replaces a bag by two bags containing positive numbers of balls. Perform at most `maxOperations` splits and return the smallest possible largest bag size. ```cpp class Solution { public: int minimumSize(vector<int>& nums, int maxOperations); }; ``` ### Examples **Example 1:** ```text nums = [10], maxOperations = 3 Output: 3 Explanation: Four bags of sizes 3, 3, 2, and 2 are possible. A maximum of 2 would need five bags and four splits. ``` **Example 2:** ```text nums = [3,7], maxOperations = 2 Output: 3 Explanation: Split 7 into 3, 2, and 2, leaving the original 3. ``` ### Constraints - `1 <= nums.length <= 10^5`. - `1 <= maxOperations <= 10^9`. - `1 <= nums[i] <= 10^9`. ### Clarifying Questions Each split increases the bag count by one. Empty bags are forbidden, and bags cannot be merged. ```hint Inspect the structure Which operations change the number of bags, and what must remain true about every bag's contents? ``` ### Reasoning checkpoints - Show that the proposed final bag sizes are positive and reachable within the split budget. - Prove that no smaller largest bag size is feasible. ### Optional extensions What construction realizes the minimum required splits for each original bag?