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 was k units, then its next jump must be either k - 1, k, or k + 1 units. Note that the frog can only jump in the forward direction.

Note:

• The number of stones is in [2, 1100).
• Each stone's position will be a non-negative integer < $2^{31}$.
• 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. The frog can jump to the last stone by jumping 1 unit to 1, then 2 units to 3, then 2 unit to 5, then 3 units to 8, 4 units to 12, and 5 units to 17.

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.

Solution: DP

class Solution(object):
    def canCross(self, stones):
        """
        :type stones: List[int]
        :rtype: bool
        """
        jumps = {stone: set() for stone in stones}
        jumps[0].add(1)
        for stone in stones:
            for jump in jumps[stone]:
                if jump > 0:
                    next_stone = stone + jump
                    if next_stone in jumps:
                        jumps[next_stone].add(jump - 1)
                        jumps[next_stone].add(jump)
                        jumps[next_stone].add(jump + 1)
        return bool(jumps[stones[-1]])