LeetCode Practice 55.
Question
A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.
Given a list of stones' positions (in units) in sorted ascending order, determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit.
If the frog’s last jump was k units, its next jump must be either k - 1, k, or k + 1 units. The frog can only jump in the forward direction.
Example 1
Input: stones = [0,1,3,5,6,8,12,17]
Output: true
Explanation: The frog can jump to the last stone by jumping 1 unit to the 2nd stone, then 2 units to the 3rd stone, then 2 units to the 4th stone, then 3 units to the 6th stone, 4 units to the 7th stone, and 5 units to the 8th stone.Example 2
Input: stones = [0,1,2,3,4,8,9,11]
Output: false
Explanation: There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.Thought 1 — Backtracking
Of course, everyone knows it we need the backtracking method to find out the answer. Move forward to the starting index to check whether it can jump to the final place. At first, we can check if the destination is achievable by calculating the gap than the previous one. If the gap value is larger than the index value, it’s impossible to reach. It also can be used to check each place on the stones.
Result — Time Limit Exceeded
Thought 2 — Backtracking + memory table
It seems the first implementation isn’t good enough. We need some tricks. Bring a memorized table to reduce redundant calculation. And much better in the result.
Runtime: 104 ms, faster than 94.98% of Python3 online submissions for Frog Jump.
Memory Usage: 16.3 MB, less than 65.63% of Python3 online submissions for Frog Jump.
Reference
If you like this post, you can buy me a coffee! :)
