## Description

https://leetcode.com/problems/odd-even-jump/

You are given an integer array `A`

. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, …) jumps in the series are called **odd-numbered jumps**, and the (2nd, 4th, 6th, …) jumps in the series are called **even-numbered jumps**. Note that the **jumps** are numbered, not the indices.

You may jump forward from index `i`

to index `j`

(with `i < j`

) in the following way:

- During
**odd-numbered jumps**(i.e., jumps 1, 3, 5, …), you jump to the index`j`

such that`A[i] <= A[j]`

and`A[j]`

is the smallest possible value. If there are multiple such indices`j`

, you can only jump to the**smallest**such index`j`

. - During
**even-numbered jumps**(i.e., jumps 2, 4, 6, …), you jump to the index`j`

such that`A[i] >= A[j]`

and`A[j]`

is the largest possible value. If there are multiple such indices`j`

, you can only jump to the**smallest**such index`j`

. - It may be the case that for some index
`i`

, there are no legal jumps.

A starting index is **good** if, starting from that index, you can reach the end of the array (index `A.length - 1`

) by jumping some number of times (possibly 0 or more than once).

Return *the number of good starting indices*.

**Example 1:**

Input:A = [10,13,12,14,15]Output:2Explanation:From starting index i = 0, we can make our 1st jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[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.

**Example 2:**

Input:A = [2,3,1,1,4]Output:3Explanation: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 A[1] is the smallest value in [A[1], A[2], A[3], A[4]] that is greater than or equal to A[0]. During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in [A[2], A[3], A[4]] that is less than or equal to A[1]. A[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 A[3] is the smallest value in [A[3], A[4]] that is greater than or equal to A[2]. We can'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'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.

**Example 3:**

Input:A = [5,1,3,4,2]Output:3Explanation:We can reach the end from starting indices 1, 2, and 4.

**Constraints:**

`1 <= A.length <= 2 * 10`

^{4}`0 <= A[i] < 10`

^{5}

## Explanation

## Python Solution

```
class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
n = len(A)
next_higher, next_lower = [0] * n, [0] * n
stack = []
for a, i in sorted([a, i] for i, a in enumerate(A)):
while stack and stack[-1] < i:
next_higher[stack.pop()] = i
stack.append(i)
for a, i in sorted([-a, i] for i, a in enumerate(A)):
while stack and stack[-1] < i:
next_lower[stack.pop()] = i
stack.append(i)
higher, lower = [0] * n, [0] * n
higher[-1] = lower[-1] = 1
for i in range(n - 1)[::-1]:
higher[i] = lower[next_higher[i]]
lower[i] = higher[next_lower[i]]
return sum(higher)
```

- Time Complexity: ~N
- Space Complexity: ~N