Minimize the Longest Jump on a Round Trip through Ordered Stones

### Problem The frog starts at stones[0]=0, must reach the last stone, then return to the first. It may jump to each stone at most once; the initial placement is not a landing jump, so the required return to the first stone is allowed. Each jump costs the absolute difference in positions. Return the smallest possible maximum jump over the round trip. ```cpp class Solution { public: int maxJump(vector<int>& stones); }; ``` ### Examples **Example 1:** ```text Input: stones = [0,1,4,6] Output: 5 Explanation: Route 0->4->6->1->0 has jumps4,2,5,1 and visits each interior stone once. ``` **Example 2:** ```text Input: stones = [0,8] Output: 8 Explanation: Both the outward and return jumps span 8. ``` ### Constraints - `2 <= stones.length <= 100000`. - `0 <= stones[i] <= 10^9`, with stones[0]=0. - Positions are strictly increasing. ### Clarifying Questions - Is the return to the first stone required? Yes. The initial placement is not a landing, so returning there is permitted. - Must all other stones be used? No. - Can an intermediate stone be landed on twice? No. - Is cost the sum of jump lengths? No; it is the longest jump in the round trip. ```hint Check the boundary Check the complete outward-and-return journey, rather than just one direction. ```