<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:
        if len(stones) < 2:
            return False

        # Convert the stones array to a set for O(1) lookups
        stoneset = set(stones)

        # Recursive function to make jumps, memoized using @cache
        @cache
        def makeJump(i, k):
            # If the frog reaches the last stone, return True
            if i == len(stones) - 1:
                return True

            # Try to jump to the next stones with jump sizes k-1, k, k+1
            for j in range(k - 1, k + 2):
                if j <= 0:
                    continue  # Can't make non-positive jumps

                next_pos = stones[i] + j
                if next_pos in stoneset:
                    next_index = bisect_left(stones, next_pos)
                    if next_index < len(stones) and stones[next_index] == next_pos:
                        if makeJump(next_index, j):
                            return True

            return False

        # Start from the first stone with an initial jump of size 1
        return makeJump(0, 0)



"""
INPUT: stones: int[], asc order
OUTPUT: bool: if frog can reach last stone

Qs/Constraints
- stones[i] is unit of stone i
- cannot jump into water
- must land on last stone
- start on first stone
- first jump MUST be 1 unit only
- if last jump was k, next jump is k-1, k, k+1
- NO BACKWARD
- CAN skip stones

- if units to next stone > k+1 OR < k-1 then invalid and stop

0 1 2 10 12
0 1 2 3 4 5 6
0 1

APPROACH
- k starts at 0?
- must take positive +1 jump
- check next stone against min/max k values
- if equal to any then take the jump
- if distance is greater than k+1 then return False
- if not greater than keep checking next values in stones
- backtracking, recursive
        - O(3^n)


"""

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