Odd Even Jump Python Solution with Tests

lcpy gen -n 975   # by problem number
lcpy gen -s odd_even_jumps   # by problem name

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.

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.

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

class Solution:
    # Time: O(n log n) sorting plus linear DP
    # Space: O(n) for the two jump maps
    def odd_even_jumps(self, arr: list[int]) -> int:
        n = len(arr)

        def make_next(indices: list[int]) -> list[int | None]:
            # For each index j, the next index (in sorted order) greater than j:
            # its first jump target, honoring the smallest-index tie-break.
            nxt: list[int | None] = [None] * n
            stack: list[int] = []
            for i in indices:
                while stack and i > stack[-1]:
                    nxt[stack.pop()] = i
                stack.append(i)
            return nxt

        odd_next = make_next(sorted(range(n), key=lambda i: arr[i]))
        even_next = make_next(sorted(range(n), key=lambda i: -arr[i]))

        # higher[i]: a good end is reachable from i when the next jump is odd-numbered
        higher = [False] * n
        lower = [False] * n
        higher[n - 1] = lower[n - 1] = True
        for i in range(n - 2, -1, -1):
            odd_target = odd_next[i]
            if odd_target is not None:
                higher[i] = lower[odd_target]
            even_target = even_next[i]
            if even_target is not None:
                lower[i] = higher[even_target]

        return sum(higher)