<p>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.</p>

<p>Given a list of <code>stones</code>&nbsp;positions (in units) in sorted <strong>ascending order</strong>, 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 <code>1</code> unit.</p>

<p>If the frog&#39;s last jump was <code>k</code> units, its next jump must be either <code>k - 1</code>, <code>k</code>, or <code>k + 1</code> units. The frog can only jump in the forward direction.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> stones = [0,1,3,5,6,8,12,17]
<strong>Output:</strong> true
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stones = [0,1,2,3,4,8,9,11]
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= stones.length &lt;= 2000</code></li>
	<li><code>0 &lt;= stones[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>stones[0] == 0</code></li>
	<li><code>stones</code>&nbsp;is sorted in a strictly increasing order.</li>
</ul>

class Solution:
    def canCross(self, stones: List[int]) -> bool:
        # brute force
        # - generate all paths that are valid, and see if they will cover all stones
        # - 1 x {0, 1, 2} x {0, 1, 2, 3} etc

        # table stone -> index
        stone_to_index = {}
        for i, stone in enumerate(stones):
            stone_to_index[stone] = i # stone is unique

        # naive - just brute force try all paths - O(3^n) ??
        # let's try to memoize
        mem = {} # (prev_jump, current_index) -> result

        def backtrack(prev_jump, i):
            next_jumps = []
            if prev_jump == 1: # no point jumping backward
                next_jumps = [prev_jump, prev_jump+1]
            elif prev_jump > 1:
                next_jumps = [prev_jump-1, prev_jump, prev_jump+1]

            # exit condition
            if i == len(stones)-1:
                return True # we are done

            # check if any of the next jumps are valid (i.e. we'd land on a rock)
            curr_stone = stones[i]
            res = False
            for next_jump in next_jumps:
                next_stone = curr_stone+next_jump
                if next_stone in stone_to_index.keys(): # next stone does exist!
                    # check if we have computed it before
                    next_index = stone_to_index[next_stone]
                    if (next_jump, next_index) not in mem.keys():
                        mem[(next_jump, next_index)] = backtrack(next_jump, next_index)
                    
                    res = res | mem[(next_jump, next_index)]

            return res # if any of the returns are true, we are done

        #  we are assuming the frog's first jump must be 1, and we are starting from the second stone.
        # confirm this.
        if stones[1]-stones[0] > 1:
            return False

        return backtrack(1, 1)            
        

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