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

 

Ideas:

题目意思是，第一次(奇数跳的时候)跳要找大的higher，第二次(偶数跳)跳要找小的lower，然后交替进行；

1. 对于奇数跳， 我们要找这个index之后的最小值可以大于等于这个index的数

2. 对于偶数跳，我们要找这个index之后的最大值可以小于等于这个index的数（# dont use sorted(..., reverse = True) because we want the first index if there is a tie）

note: 因为不是离index最近的最小或者最大，所以单纯monotonic stack 不行，因为是最小或者最大，所以需要sort以下，然后再用类似于monotonic stack的方式来得到next_lower和next_higher

 

3. 利用dynamic programming，从后往前推

4. function： higher[index] = lower[next_higher[index]]

                     lower[index] = higher[next_lower[index]]

                    因为跳跃是higher（odd），lower(even)交替进行

5. 最后返回sum(higher) 因为我们从odd跳开始，也就是说最开始要找higher

 

 

Code:

class Solution:
    def oddEvenJumps(self, arr: List[int]) -> int:
        next_higher = self.makeStack(sorted([(num, index) for index, num in enumerate(arr)]))
        next_lower = self.makeStack(sorted([(-num, index) for index, num in enumerate(arr)])) # dont use sorted(..., reverse = True) because we want the first index if there is a tie
        n = len(arr)
        higher, lower = [False] * n, [False] * n # higher means odd, lower means even
        higher[-1], lower[-1] = True, True
        for i in range(n - 1, -1, -1):
            if next_higher[i] is not None:
                higher[i] = lower[next_higher[i]]
            if next_lower[i] is not None:
                lower[i] = higher[next_lower[i]]
        return sum(higher)
        
        
    
    
    def makeStack(self, s_indexes):
        res = [None] * (len(s_indexes))
        stack = []
        for _, index in s_indexes:
            while stack and index > stack[-1]:
                res[stack.pop()] = index
            stack.append(index)
        return res