<p>You are given an integer array <code>arr</code>. From some starting index, you can make a series of jumps. The (1<sup>st</sup>, 3<sup>rd</sup>, 5<sup>th</sup>, ...) jumps in the series are called <strong>odd-numbered jumps</strong>, and the (2<sup>nd</sup>, 4<sup>th</sup>, 6<sup>th</sup>, ...) jumps in the series are called <strong>even-numbered jumps</strong>. Note that the <strong>jumps</strong> are numbered, not the indices.</p>

<p>You may jump forward from index <code>i</code> to index <code>j</code> (with <code>i &lt; j</code>) in the following way:</p>

<ul>
	<li>During <strong>odd-numbered jumps</strong> (i.e., jumps 1, 3, 5, ...), you jump to the index <code>j</code> such that <code>arr[i] &lt;= arr[j]</code> and <code>arr[j]</code> is the smallest possible value. If there are multiple such indices <code>j</code>, you can only jump to the <strong>smallest</strong> such index <code>j</code>.</li>
	<li>During <strong>even-numbered jumps</strong> (i.e., jumps 2, 4, 6, ...), you jump to the index <code>j</code> such that <code>arr[i] &gt;= arr[j]</code> and <code>arr[j]</code> is the largest possible value. If there are multiple such indices <code>j</code>, you can only jump to the <strong>smallest</strong> such index <code>j</code>.</li>
	<li>It may be the case that for some index <code>i</code>, there are no legal jumps.</li>
</ul>

<p>A starting index is <strong>good</strong> if, starting from that index, you can reach the end of the array (index <code>arr.length - 1</code>) by jumping some number of times (possibly 0 or more than once).</p>

<p>Return <em>the number of <strong>good</strong> starting indices</em>.</p>

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

<pre>
<strong>Input:</strong> arr = [10,13,12,14,15]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
From starting index i = 0, we can make our 1st jump to i = 2 (since arr[2] is the smallest among arr[1], arr[2], arr[3], arr[4] that is greater or equal to arr[0]), then we cannot jump any more.
From starting index i = 1 and i = 2, we can make our 1st jump to i = 3, then we cannot jump any more.
From starting index i = 3, we can make our 1st jump to i = 4, so we have reached the end.
From starting index i = 4, we have reached the end already.
In total, there are 2 different starting indices i = 3 and i = 4, where we can reach the end with some number of
jumps.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,3,1,1,4]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd-numbered), we first jump to i = 1 because arr[1] is the smallest value in [arr[1], arr[2], arr[3], arr[4]] that is greater than or equal to arr[0].
During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because arr[2] is the largest value in [arr[2], arr[3], arr[4]] that is less than or equal to arr[1]. arr[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3
During our 3rd jump (odd-numbered), we jump from i = 2 to i = 3 because arr[3] is the smallest value in [arr[3], arr[4]] that is greater than or equal to arr[2].
We can&#39;t jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can&#39;t jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indices i = 1, i = 3, and i = 4, where we can reach the end with some
number of jumps.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [5,1,3,4,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can reach the end from starting indices 1, 2, and 4.
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt; 10<sup>5</sup></code></li>
</ul>

'''

[1,5,4,3,6,2]


[1,5,4,6,3,4,2]
[5,6,4,x,4,x,Z]

BST
Scan from R --> L.
When we see a new value, we need to search the BST for the right index.
'''

class MyBST():
    
    def __init__(self, val: int, index: int):
        self.val = val
        self.index = index
        self.left = None
        self.right = None
        
    def insert(self, val: int, index: int) -> None:
        if val == self.val:
            # just update the index
            self.index = index
            return
        if val < self.val:
            if not self.left:
                self.left = MyBST(val, index)
            else:
                self.left.insert(val, index)
        else:
            if not self.right:
                self.right = MyBST(val, index)
            else:
                self.right.insert(val, index)
    
    @staticmethod
    def find(root: Optional['MyBST'], val: int) -> Optional[int]:
        if not root:
            return None
        if root.val == val:
            return root.index
        if root.val < val:
            # root is too small
            # result has to be on the right
            return MyBST.find(root.right, val)
        
        # root is large
        # could be root or on the left
        r_index = MyBST.find(root.left, val)
        if r_index is not None:
            return r_index
        return root.index
        

# implemented for odd jumps; only non-decreasing
def compute_jump_table_odd(arr: list[int]) -> list[int]:
    n = len(arr)
    jump_pos = [-1] * n
    jump_pos[-1] = n - 1
    
    bst = MyBST(arr[n-1], n-1)
    for i in range(n - 2, -1, -1):
        jump_index = MyBST.find(bst, arr[i])
        print(arr[i], jump_index)
        if jump_index is not None:
            jump_pos[i] = jump_index
        bst.insert(arr[i], i)
    return jump_pos
    
def compute_jump_table_even(arr: list[int]) -> list[int]:
    return compute_jump_table_odd([-x for x in arr])

def can_reach_end(pos: int,
                  parity: int,
                  jump_tables: tuple[list[int], list[int]],
                  cache: dict[tuple[int, int], bool]) -> True:
    cache_key = (pos, parity)
    if cache_key in cache:
        return cache[cache_key]
    
    n = len(jump_tables[0])
    if pos == n - 1:
        cache[cache_key] = True
        return True
    
    next_pos = jump_tables[parity][pos]
    if next_pos < 0:
        cache[cache_key] = False
        return False
    
    res = can_reach_end(next_pos, (parity + 1) % 2, jump_tables, cache)
    cache[cache_key] = res
    return res
    
class Solution:
    def oddEvenJumps(self, arr: List[int]) -> int:
        odd_jumps = compute_jump_table_odd(arr)
        even_jumps = compute_jump_table_even(arr)
        print(odd_jumps)
        print(even_jumps)
        
        cache = {}
        jump_tables = (odd_jumps, even_jumps)
        res = 0
        for i in range(len(arr)):
            if can_reach_end(i, 0, jump_tables, cache):
                res += 1
        return res
        
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