<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:
#         n = len(stones)
#         #dp = collections.defaultdict(dict)
#         dp = {}
#         if stones[1] !=1 :
#             return False 
#         if n == 2 :
#             return True

#         def dfs(curval,k):
#             ans = False
#             #print(curval,k)
#             if curval in dp:
#                 if k in dp[curval]:
#                     print("found", curval, k, dp)
#                     ans = dp[curval][k]
#             else:
#                 for step in (k-1,k, k+1):
#                     val = curval+step
#                     print(curval,k,val,step)
#                     if val == stones[-1] :
#                         print("if")
#                         ans = ans or True
#                     elif val > stones[-1] :
#                         continue
#                     elif val in stones and val > curval :
#                         print("else")
#                         ans = ans or dfs(val, step) 
#                     dp[curval][k] = ans
#                 #print(ans,dp)
#             return ans       

#         return dfs(1,1)     
        

class Solution:
    def canCross(self, stones: List[int]) -> bool:
        n = len(stones)
        dp = {}
        if stones[1] != 1:
            return False
        if n == 2:
            return True

        def dfs(curval, k):
            if (curval, k) in dp:
                return dp[(curval, k)]
            ans = False
            if curval == stones[-1]:
                return True
            for step in (k-1, k, k+1):
                val = curval + step
                if val in stones and val > curval:
                    ans |= dfs(val, step)
            dp[(curval, k)] = ans
            return ans

        return dfs(1, 1)
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