<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:
    @cache
    def find_next_step(self, target, position):
        
        low, high = position + 1, self.n - 1
        
        while low <= high:
            
            mid = low + (high - low) // 2
            
            
            if self.stones[mid] == target: return mid
            
            if self.stones[mid] < target: low = mid + 1
                
            else:
                high = mid - 1
                
                
        return -1
                
    
    
    @cache
    def rec_find(self, position, prev_steps):
        n = self.n
        
        if position == n - 1: return True
        
        
        if self.stones[position] + prev_steps + 1 < self.stones[position + 1]: return False
            
        if self.stones[position] + prev_steps + 1 == self.stones[-1]: return True
            
            
        option_1 = self.find_next_step( self.stones[position] + prev_steps - 1, position)
        option_2 = self.find_next_step( self.stones[position] + prev_steps + 1, position)
        option_3 = self.find_next_step( self.stones[position] + prev_steps, position)
        
        print(f"Stone: {position+1, self.stones[position]}, Prev_steps: {prev_steps}, Options: {option_1 + 1}, {option_2 + 1}, {option_3 + 1}")
        
        ans1, ans2, ans3 = False, False, False
        
        if option_1 != -1:
            ans1 = self.rec_find(option_1, prev_steps - 1)
        
        if ans1: return True
        
        
        
        if option_2 != -1:
            ans2 = self.rec_find(option_2, prev_steps + 1)
        
        if ans2: return True
        
        
        
        if option_3 != -1:
            ans3 = self.rec_find(option_3, prev_steps)
        
        if ans3: return True
        
        return False
        
    
    
    def canCross(self, stones: List[int]) -> bool:
        
        self.stones = stones
        self.n = len(stones)
        
        if self.n < 2: return False
        if self.stones[1] != 1: return False
        
        
        return self.rec_find(1, 1)
    
        
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