Problem Description

This problem asks us to count "good" starting indices in an array arr. A good starting index is defined as an index from which it is possible to reach the end of the array by making a series of jumps following specific rules. These jumps are of two types: odd-numbered jumps and even-numbered jumps. For odd-numbered jumps, you look for the smallest value in the array that is greater than or equal to the value at the current index and jump to the first occurrence of this smallest value. For even-numbered jumps, you look for the largest value that is less than or equal to the value at the current index and jump to the first occurrence of this largest value. If there are no valid jumps possible from an index, you cannot proceed from that point. The goal is to determine the count of indices from which reaching the end of the array is possible.

Intuition

The intuition behind this solution lies in dynamic programming and the use of efficient data structures to keep track of legal jumps.

Since the jumps need us to look up the smallest larger (or largest smaller) value, we can leverage a sorted data structure like SortedDict from the sortedcontainers module in Python. This allows us to efficiently locate the next jump index using binary search operations (in O(log n) time complexity).

With dynamic programming, we can avoid recalculating whether it's possible to reach the end from a given index by storing the results of our calculations. We define a recursive function dfs which tries to determine if it is possible to reach the end of the array starting at index i and making a specific type of jump (odd or even). We then apply memoization using the cache decorator to store the results and avoid redundant computations.

For the overall solution, we iterate backward through the array. At each index, we use the SortedDict structure to find the correct jump indices for both odd and even jumps. Then, we attempt to find out if it's possible to reach the end from each index making an odd jump first (since we must start with an odd-numbered jump). The total count of "good" starting indices is the sum of all indices from which the end can be reached by following the jumping rules.

The crux of the solution is to carefully build a graph-like structure that maps each index to the next index for an odd or even jump. And then, use a depth-first search (DFS) strategy to explore possible paths from each potential starting index to the last index, using memoization to speed up the process significantly.

Learn more about Stack, Dynamic Programming and Monotonic Stack patterns.

Solution Approach

The given Python solution follows a bottom-up dynamic programming strategy with memoization. Here is a step-by-step breakdown of the implementation details:

1. Memoization Decorator: The @cache decorator is applied to the dfs function. This is Python's built-in way to store results of expensive function calls and return the cached result when the same inputs occur again. It helps to avoid repeated computation for the same starting index and jump type.

2. Helper Function dfs: This is the function that attempts to find if the current index (i) can reach the end of the array with a sequence of odd and even jumps. The parameter k indicates the jump type, 0 for even and 1 for odd.

   • Base Case: If the current index is the last in the array (i == n - 1), the function returns True as we've successfully reached the end.
   • Recursive Case: If a jump is possible (g[i][k] is not -1), the function recursively calls itself, toggling the jump type (k ^ 1) to simulate the alternation of jumps.
   • Jump Not Possible: If no jump is possible from the current index (g[i][k] is -1), the function returns False.

3. Graph g: A 2D list g is initialized to keep track of next indices for odd and even jumps from each index. The first dimension corresponds to the index in arr and the second dimension (0 or 1) corresponds to the type of jump.

4. SortedDict sd: A SortedDict is used to efficiently find the next indices for jumps. As we iterate backward through the array (i going from n - 1 to 0), this structure allows us to:

   • Odd-numbered jumps: Use sd.bisect_left(arr[i]) to find the smallest index where arr[j] is greater than or equal to arr[i].
   • Even-numbered jumps: Use sd.bisect_right(arr[i]) - 1 to find the largest index where arr[j] is smaller than or equal to arr[i].

5. Building the Graph g: For each index i, we update g[i][1] for odd jumps and g[i][0] for even jumps, or assign -1 if no jump is possible. Here, sd.values() returns the list of indices in the order of their corresponding values in sd.

6. Calculating Good Starting Indices: Finally, the sum of all indices from which the end of the array can be reached by starting with an odd jump is computed. We do this by invoking dfs(i, 1) for all indices from 0 to n - 1, and then taking the sum of these boolean results.

The solution smartly combines memoized recursion for dynamic programming and binary search within a sorted dictionary to maintain a balance between time complexity and the practical execution speed of the algorithm.

Example Walkthrough

Let's consider a small array arr = [5, 1, 3, 4, 2] to illustrate the solution approach:

1. Initialization: We prepare a 2D list g where g[i][0] and g[i][1] will store the next index for an even and an odd jump from the index i, respectively. We also create a SortedDict named sd.

2. Building the Graph: Start iterating from the end (i = 4) to the start (i = 0) of the array:

   • For i = 4 (value 2), sd is empty, so no jumps are possible. g[4] = [-1, -1].
   • Insert (2, 4) into sd.
   • For i = 3 (value 4), the next odd jump goes to the smallest larger or equal value in sd (bisect_left(4)), but there is none, so g[3][1] = -1. For even jumps, the next jump is to the largest smaller or equal value in sd (bisect_right(4) - 1), which is the index of value 2, so g[3][0] = 4.
   • Insert (4, 3) into sd.
   • For i = 2 (value 3), the next odd jump is to 4 and the next even jump is to 2. So g[2] = [4, 3].
   • Insert (3, 2) into sd.
   • For i = 1 (value 1), the next odd jump is to 3 and there is no even jump, so g[1] = [-1, 2].
   • Insert (1, 1) into sd.
   • For i = 0 (value 5), there are no larger or smaller values than 5 in sd, so g[0] = [-1, -1].

3. Calculating Good Starting Indices: We create a dfs function with memoization using the @cache decorator:

   • Start with index 0 and an odd jump (dfs(0, 1)). This function call returns False because g[0][1] = -1.
   • dfs(1, 1) returns True because g[1][1] = 2 and from index 2, an even jump leads to the end (dfs(2, 0) returns True).
   • dfs(3, 1) returns False since no odd jump is possible from 3.
   • dfs(4, 1) also returns False since no jump is possible from 4.

4. Result: Count all True results from dfs(i, 1) for i in [0, 1, 2, 3, 4] to get the total number of "good" starting indices. In our example, the count is 1 (only index 1 leads to the end of the array by following jump rules).

Hence, for our array arr = [5, 1, 3, 4, 2], there is only one "good" starting index, which is the index 1.

Solution Implementation

Time and Space Complexity

Time Complexity

The function oddEvenJumps involves both a dynamic programming approach (through memoization with @cache) and the use of a balanced binary search tree (BST) through SortedDict.

• The loop that constructs the graph g runs for n iterations, where n is the length of arr. In each iteration, two bisect operations are performed on SortedDict to find the next indices for odd and even jumps.

  • The bisect_left and bisect_right operations on a BST have a time complexity of O(log m), where m is the number of elements in the BST at the time of the query. In the worst case, m can be as large as n.
  • Inserting an element into the SortedDict also has a time complexity of O(log n).

• The DFS function dfs could potentially be called for every starting position i with every jump type k. As memoization is used, each pair (i, k) will be computed at most once. Therefore, we can consider it to have a constant time complexity O(1) for each call, after the initial computation.

• The final sum calls dfs for each starting position i with the initial jump type 1 (odd jump), totaling n calls to dfs. However, since dfs is memoized, the actual complexity of all these calls altogether is O(n).

Combining these parts, the overall time complexity is O(n log n) due to the combination of the loop and the operations on the SortedDict within the loop, which dominates over the linear time complexity of the final sum.

Space Complexity

• The space used by the memoization in dfs function will require O(n * 2) space since it stores results for each element and jump type (odd or even). This simplifies to O(n).

• The graph g is a 2D array with dimensions n * 2, also taking O(n) space.

• The SortedDict can hold up to n elements, so it also consumes O(n) space.

Overall, combining the space for memoization, the graph g, and the SortedDict, the total space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.