# 975 Odd Even Jump

You are given an integer `array` arr. 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 `arr[i] <= arr[j]` and `arr[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 arr`[i] >= arr[j]` and `arr[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 `arr.length - 1`) by jumping some number of times (possibly 0 or more than once).

Return the number of good starting indices.

Example 1:

``````Input: arr = [10,13,12,14,15]
Output: 2
Explanation:
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.
``````

Example 2:

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

Constraints:

• `1 <= arr.length <= 2 * 104`
• `0 <= arr[i] < 105`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 `````` ``````class Solution: def oddEvenJumps(self, arr: List[int]) -> int: n = len(arr) next_higher, next_lower = [0] * n, [0] * n stack = [] for a, i in sorted([a, i] for i, a in enumerate(arr)): while stack and stack[-1] < i: next_higher[stack.pop()] = i stack.append(i) stack = [] for a, i in sorted([-a, i] for i, a in enumerate(arr)): 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)``````