Problem Description

The given problem presents us with a particular type of jumping puzzle. You have an array called nums which consists of n integers, indexed from 0 to n-1. You also have a value called target. Starting at the first element of the array (index 0), you can jump to any later element in the array (from index i to index j, where i < j) as long as the absolute difference between the values of nums[i] and nums[j] is less than or equal to target. The question is to determine the maximum number of jumps that can be made to reach the last element of the array (n-1 index).

If, at any point, there are no legal jumps to make, which means you cannot reach the last index of the array, the function should return -1.

In essence, the problem is asking for the furthest reach in the array through a series of legal jumps, where each jump abides by the rule concerning the allowable difference defined by the target value.

Intuition

The solution approach relies on the idea of recursion and dynamic programming. We can define a recursive function, say dfs(i), that computes the maximum number of jumps needed to reach the end of the array starting from the current index i.

Initially, we might consider simply iterating through the array, and at each step, trying each possible legal jump. However, this brute-force approach can be highly inefficient as it involves many repeated calculations.

Instead, we can use a technique called memoization, which is a strategy to store the results of expensive function calls and return the cached result when the same inputs occur again. Thus, for each index i, we remember the maximum number of jumps we can make. If we revisit the same index i, we don't recalculate; we simply use the stored value.

Here's how the thought process goes for the given solution:

1. If we are at the last index (n - 1), return 0 because we don't need any more jumps;
2. If we are at any other index i, look at all potential jumps, i.e., loop from i + 1 to n - 1 and find indexes j to jump to where |nums[i] - nums[j]| <= target;
3. Use our recursive function dfs(j) to compute the maximum number of jumps from index j to the end. The answer for our current position i would then be the maximum value of 1 + dfs(j) over all legal j's plus one (for the jump we are currently considering);
4. If we cannot jump anywhere from i, we use negative infinity to mark that we can't reach the end from this index;
5. We use memoization (@cache decorator) to avoid re-computation for indexes we have already visited.
6. Finally, we initiate our recursive calls with dfs(0) to start the process from the first index. The result is either the maximum number of jumps or -1 if the end is unreachable (ans < 0).

This dynamic programming approach, combined with memoization (caching), provides an efficient solution to what could otherwise be a very time-consuming problem if solved with plain recursion or brute-forcing.

Learn more about Dynamic Programming patterns.

Solution Approach

The implementation uses a recursive depth-first search (DFS) approach in combination with memoization. The core of this approach is the dfs function, which solves the problem for a given index i and provides the maximum number of jumps that can be made from that index to the end. To prevent re-computation of results for each index i, memoization is used via the @cache decorator provided by Python's standard library. Here's an outline of how the algorithm works:

• A helper function dfs(i) is defined:
  • If i is the last index (n - 1), return 0 because no more jumps are necessary.
  • If not, initialize a variable ans to negative infinity, symbolizing that the end is not yet reachable from i.
  • Loop through each potential target index j (where j goes from i + 1 to n - 1), and for each target index, check if the jump from i to j is legal; that is, if abs(nums[i] - nums[j]) <= target.
  • For every legal jump, use the dfs function to compute the maximum number of jumps from index j to the end. The value of ans is updated to be the maximum of the current ans and 1 + dfs(j), which represents making one jump to index j plus the maximum jumps from j to the end.
• The global scope begins by determining n, the length of the nums array.
• It then calls the helper function dfs(0) to initiate the recursive jumping process from the first element of the array.
• Lastly, the function returns -1 if ans < 0 indicating that the last index is unreachable or ans if the end can be reached, providing the maximum number of jumps to reach the end.

Data structures used:

• The nums array holds the input sequence of integers.
• An implicit call stack for recursive function calls.
• An internal cache for memoization, which is abstracted away by the @cache decorator but essentially behaves as a hash map storing computed results keyed by function arguments.

Algorithms and patterns used:

• DFS (Depth-First Search): This recursive strategy is used to explore all possible jumping paths.
• Memoization: This pattern avoids recalculating dfs(i) for any index by caching the result. When dfs(i) is called multiple times with the same i, it returns the cached result instead of recalculating.
• Dynamic Programming (Top-Down Approach): The problem is broken down into smaller subproblems (dfs(j) for each j) and solved in a recursive manner. The overlapping subproblems are handled efficiently using memoization.

In summary, the solution employs a combination of recursive DFS and memoization in a top-down dynamic programming framework to efficiently compute the maximum number of jumps to the end of the array.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have the following input:

• nums = [2, 3, 1, 1, 4]
• target = 1

We want to find the maximum number of jumps to reach the end of the array.

1. We call dfs(0), starting at index 0, where the value is 2. Our first action is to look for all indices we can jump to from 0.

2. We can only jump to an index i if abs(nums[0] - nums[i]) <= target. In this case, we can jump from index 0 (nums[0] = 2) to index 1 (nums[1] = 3) because abs(2 - 3) = 1 <= target.

3. Now at index 1, we call dfs(1) and look for a jump. We can jump from 1 to 2 and 1 to 3 since both satisfy the condition (abs(3 - 1) <= 1 and abs(3 - 1) <= 1).

4. At index 2, calling dfs(2) returns 0, as we can jump directly to the end (index 4) from here (abs(1 - 4) <= 1).

5. At index 3, calling dfs(3) also returns 0, for the same reason as dfs(2).

6. Backtracking, we can see that from dfs(1), we have two possible destinations to consider: index 2 and index 3. Choosing index 2 lets us reach the end in one jump (the same for index 3), so we update our maximum jumps from 1 to 2.

7. Finally, considering jumps from dfs(0) to dfs(1), we have now found that from dfs(1) we can reach the end with 2 jumps, hence we add one more jump (the jump from 0 to 1) and update our answer to 3.

8. The recursive function will use memoization to ensure that if we call dfs(2) or dfs(3) again during exploration, it will return the cached result without recalculating it.

9. If there were no possible jumps at any point, dfs(i) would return negative infinity to indicate that it's impossible to proceed. But in this case, we can reach the end, and the maximum number of jumps is 3.

Hence, dfs(0) will finally return 3 as the maximum number of jumps needed to reach the end of the array from the first element.

Using this approach, the solution capitalizes on the efficiency of memoization to avoid re-computing the same values, thereby reducing the time complexity significantly from what would be seen in a brute-force approach.

Solution Implementation

Time and Space Complexity

The given Python code defines a function maximumJumps that calculates the maximum number of jumps you can make in a list of integers, where a jump from index i to index j is valid if the absolute difference between nums[i] and nums[j] is less than or equal to a given target. The analysis of the time complexity and space complexity is as follows:

The time complexity of the maximumJumps function is O(n^2) where n is the length of the input array nums. This quadratic time complexity arises because, in the worst case, the recursive function dfs is called for every pair of indices i and j in the array. The outer loop runs once for each of the n elements, and the inner loop runs up to n-1 times for each iteration of the outer loop, hence the n * (n - 1) which simplifies to O(n^2).

The space complexity of the function is O(n) which is attributable to two factors:

• The recursion depth can go up to n in the worst case, where each function call adds a new frame to the call stack.
• The use of the @cache decorator on the dfs function adds memoization, storing the results of subproblems to prevent re-computation. As there are n possible starting positions for jumps, the cache could potentially hold n entries, one for each subproblem.

Therefore, the space used by the call stack and memoization dictates the space complexity of O(n).

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