<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>

'''
Build 2 monotonically increasing/decreasing stacks from the end.

Use DP - Store even and odd jump results from each position

To find odd jump - arr[i] <= arr[j] and arr[j] is smallest possible value. We need monotonically decreasing stack.
example: 

arr[i] = 5, stack = [10, 6, 4, 2] -> We pop front until we reach 6 where 6 > 5.
What if next number in array is 2 though... we will have stack [10, 6, 5] but 2 is actually the lowest number higher.


'''
class TreeNode:
    def __init__(self, value, index, parent):
        self.value = value
        self.index = index
        self.left = None
        self.right = None
        self.parent = parent

class Solution:
    def insert(self, sortedVals, value, index, parent):
        if not sortedVals:
            result = TreeNode(value, index, parent)
            return result, result, -1
        else:
            current = sortedVals
            newNode = None
            displaced = -1
            if value == current.value:
                prev = current.index
                current.index = index
                return sortedVals, sortedVals, prev
            elif value < current.value:
                current.left, newNode, displaced = self.insert(current.left, value, index, current)
            elif value > current.value:
                current.right, newNode, displaced = self.insert(current.right, value, index, current)
            return current, newNode, displaced
    
    def getPredecessor(self, node):
        if node.left:
            current = node.left
            while current.right:
                current = current.right
            return current
        else:
            if node.parent and node.parent.right == node:
                return node.parent
            elif node.parent and node.parent.left == node:
                current = node.parent
                while current and current.value >= node.value:
                    current = current.parent
                if current and current.value <= node.value:
                    return current
                else:
                    return None
            else:
                return None
    
    def getSuccessor(self, node):
        if node.right:
            current = node.right
            while current.left:
                current = current.left
            return current
        else:
            if node.parent and node.parent.left == node:
                return node.parent
            elif node.parent and node.parent.right == node:
                current = node.parent
                while current and current.value <= node.value:
                    current = current.parent
                if current and current.value >= node.value:
                    return current
                else:
                    return None
            else:
                return None
    

    def oddEvenJumps(self, arr: List[int]) -> int:
        results = [[False,False] for i in range(len(arr))]
        results[-1] = [True, True]
        sortedVals = TreeNode(arr[-1], len(arr)-1, None)
        count = 1
        for i in range(len(arr) - 2, -1, -1):
            value = arr[i]
            sortedVals, newNode, displaced = self.insert(sortedVals, value, i, None)
            if displaced > 0:
                oddJumpIndex = TreeNode(0, displaced, None)
                evenJumpIndex = TreeNode(0, displaced, None)
            else:
                oddJumpIndex = self.getSuccessor(newNode)
                evenJumpIndex = self.getPredecessor(newNode)            
            if oddJumpIndex: 
                #print("odd", i, oddJumpIndex.index)
                newResult = results[oddJumpIndex.index][1]
                results[i][0] = newResult
                if newResult:
                    count += 1
            if evenJumpIndex:
                #print("even", i, evenJumpIndex.index)
                results[i][1] = results[evenJumpIndex.index][0]
        print(results)
        return count
            


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