Shrink the Reachable Value Range

### Problem For positive integer array `nums`, you may repeatedly halve an even entry or double an odd entry. The deviation is the largest value minus the smallest. Return the least deviation reachable with any number of these operations. ```cpp class Solution { public: int minimumDeviation(vector<int>& nums); }; ``` ### Examples **Example 1:** ```text nums = [3,6,12] Output: 0 Explanation: Halve 6 once and 12 twice to obtain [3,3,3]. ``` **Example 2:** ```text nums = [5,8] Output: 1 Explanation: Halve 8 to 4, giving a difference of 1. The reachable values are {5,10} and {1,2,4,8}, so zero is impossible. ``` ### Constraints - `n == nums.length`, with `2 <= n <= 5*10^4`. - `1 <= nums[i] <= 10^9`. ### Clarifying Questions An odd value can only double to an even value, whose next halving returns to that odd value. Operations do not generate arbitrarily large powers of two. ```hint Inspect the structure Which operation is legal at the current value, and when has a claimed target configuration actually been reached? ``` ### Reasoning checkpoints - Characterize which values are reachable under the parity-dependent operations. - Show that the claimed deviation is attainable and no other reachable configuration is narrower. ### Optional extensions What dual strategy starts from each entry’s minimum reachable value?