Remove the Shortest Block to Restore Divisibility

### Problem Remove a contiguous block from the positive integer array `nums` so that the sum left behind is divisible by `p`. Removing nothing is permitted; removing the entire array is forbidden. Return the minimum removed length, or `-1` if no allowed removal works. ```cpp class Solution { public: int minSubarray(vector<int>& nums, int p); }; ``` ### Examples **Example 1:** ```text nums = [2,5,3], p = 4 Output: 1 Explanation: Remove 2, leaving a sum of 8. ``` **Example 2:** ```text nums = [1,2], p = 5 Output: -1 Explanation: The only block with the required remainder is the entire array, which cannot be removed. ``` ### Constraints - `1 <= nums.length <= 10^5`. - `1 <= nums[i] <= 10^9`. - `1 <= p <= 10^9`. ### Clarifying Questions A block is contiguous. If the original sum is divisible by p, return zero even when positive-length removals also work. ```hint Inspect the structure Which removals are permitted, and how would you distinguish a valid shortest removal from the forbidden entire array? ``` ### Reasoning checkpoints - Establish divisibility after each claimed removal and justify the minimum removed length. - Distinguish removing nothing, a proper block, and the forbidden whole-array removal. ### Optional extensions Why does the shortest-removal objective favor the latest rather than earliest matching prefix?