Problem Description
A frog starts on the first stone (at position stones[0]) and must travel to the last stone and return to the first stone. The frog cannot land on any stone more than once, and each jump’s length is the absolute difference between the positions of the two stones. The cost of a path is defined as the maximum jump length in that path. The goal is to find the minimum possible cost over all valid round-trip paths.
Key Insights
- Since the stones array is sorted in strictly increasing order, the frog’s optimal path can be derived by carefully choosing an order that minimizes long jumps.
- An optimal strategy (often called a “zig-zag” approach) is to visit stones in an order that alternates between ends, thereby ensuring that no jump covers too large a gap.
- The maximum jump in the optimal “zig-zag” path is determined by differences between stones that are two indices apart.
- Special care is needed for boundary cases (for example, when there are only two stones).
Space and Time Complexity
Time Complexity: O(n) – we make a single pass through the stones array. Space Complexity: O(1) – only constant extra space is used.
Solution
We can solve the problem by reordering the stones in a specific “zig-zag” order. Notice that if we traverse the stones such that we alternatively pick stones from the left and right, the maximum jump occurs between stones that are two positions apart in the original sorted order. Thus, we can compute the answer by taking the maximum of:
- The jump from the first stone to the second stone.
- For each stone from index 2 onward: the gap between stones[i] and stones[i-2].
This approach avoids explicitly constructing the complex path and instead leverages the sorted order of the stones along with a greedy observation.