Comment on page

# 403 Frog Jump

A frog is crossing a river. The river is divided into x 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 is able to cross the river by landing on the last stone. Initially, the frog is on the first stone and assume the first jump must be 1 unit.
If the frog's last jump waskunits, then its next jump must be eitherk- 1,k, ork+ 1 units. Note that the frog can only jump in the forward direction.
Note:
• The number of stones is ≥ 2 and is < 1,100.
• Each stone's position will be a non-negative integer < 231
• The first stone's position is always 0.
Example 1:
[0,1,3,5,6,8,12,17]
There are a total of 8 stones.
The first stone at the 0th unit, second stone at the 1st unit,
third stone at the 3rd unit, and so on...
The last stone at the 17th unit.
Return true
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:
[0,1,2,3,4,8,9,11]
Return false
. There is no way to jump to the last stone as
the gap between the 5th and 6th stone is too large.
The Idea: Perform BFS. (note this this algorithm exceeds the OJ time requirement)
From unit 1 (if it exists), we can jump k-1, k or k+1 units. Each of the possibilities branch out to what potentially could be another stone. We introduce a few constraints into the system. 1) We must always move forward (so avoid when k-1 moves us backwards), and 2) If there exists a stone jump combination that exists in another path, avoid remaking that calculation. If any of these paths end up at the index of the final stone, then return true. Otherwise return false. Complexity: O(3^n) time: 3 branches per branch
import queue
class Solution:
def canCross(self, stones):
"""
:type stones: List[int]
:rtype: bool
"""
# first jump has to be to unit 1
if stones != 1:
return False
final_unit = stones[-1]
s = set(stones)
q = queue.Queue()
q.put((1, 1)) # (index, jump)
options = [-1, 0, 1]
memory = set()
while not q.empty():
front = q.get()
par_index = front
par_jump = front
if par_index == final_unit:
return True
# otherwise can jump k-1, k, k+1 units
for opt in options:
next_jump_i = par_index + par_jump + opt
i_jump = (next_jump_i, par_jump + opt)
if next_jump_i in s and next_jump_i > par_index and i_jump not in memory:
q.put((next_jump_i, par_jump + opt))