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Odd Even Jump

Difficulty: Hard

Problem Description

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.

Key Insights

  • Understanding the rules for odd and even jumps is crucial for determining jump possibilities.
  • Utilizing a greedy approach can help efficiently find the next valid jump indices.
  • A monotonic stack can be used to keep track of the next jump indices for both odd and even jumps.
  • Dynamic programming can be employed to evaluate which starting indices can lead to reaching the end of the array.

Space and Time Complexity

Time Complexity: O(n log n)
Space Complexity: O(n)

Solution

To solve this problem, we will use a combination of dynamic programming and a monotonic stack. The idea is to maintain two boolean arrays, odd and even, where odd[i] indicates if we can reach the end of the array starting from index i with an odd-numbered jump, and even[i] indicates the same for even-numbered jumps.

  1. Initialization: Set the last index to true for both odd and even since we are already at the end.
  2. Monotonic Stack: Iterate backward through the array and use a stack to help find the next valid jump indices:
    • For odd jumps, find the smallest index j where arr[j] >= arr[i].
    • For even jumps, find the largest index j where arr[j] <= arr[i].
  3. Dynamic Programming Update: Update the odd and even arrays using the found jump indices.
  4. Count Good Indices: Finally, count how many starting indices are marked as good.

Code Solutions

def oddEvenJumps(arr):
    n = len(arr)
    if n == 1:
        return 1

    odd = [False] * n
    even = [False] * n
    odd[-1] = even[-1] = True

    # Create a sorted list of indices based on values in arr
    sorted_indices_odd = sorted(range(n), key=lambda i: (arr[i], i))
    sorted_indices_even = sorted(range(n), key=lambda i: (-arr[i], i))

    # Fill odd jumps
    stack = []
    for i in sorted_indices_odd:
        while stack and stack[-1] < i:
            odd[i] = odd[i] or even[stack.pop()]
        stack.append(i)

    # Fill even jumps
    stack = []
    for i in sorted_indices_even:
        while stack and stack[-1] < i:
            even[i] = even[i] or odd[stack.pop()]
        stack.append(i)

    return sum(odd)
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