Problem Description

In this problem, we have an array nums that represents the contents of a pile with an index starting at 0. The element at index 0 (nums[0]) is at the top of the pile. There are two types of moves we can perform on this pile:

1. Remove the topmost element of the pile, if the pile is not empty.
2. Add any one of the previously removed elements back onto the top of the pile, assuming there's at least one removed element available.

We are also given an integer k, which represents the total number of moves that have to be made.

Objective: Our task is to find out the maximum possible value of the element at the top of the pile after making exactly k moves. If it's not possible to have any elements left in the pile after k moves, we should return -1.

This problem requires us to manipulate the topmost element of a pile with some constraints and to plan the moves to maximize the value of the topmost element taking into consideration the number of available moves.

Intuition

To find the solution, we need to consider the number of elements in the pile and the number of available moves, k. We can break this down into different scenarios:

1. If k is 0, we do not need to make any moves, and we simply return the current topmost element, nums[0].

2. If there is only one element in the pile (len(nums) == 1) and k is odd, we will inevitably end up with an empty pile after these moves since each odd move removes the last element and each even move can't put it back (as there's no other to add). Hence, in this case, the answer is -1.

3. If k is less than the number of elements in the pile (k < len(nums)), we need to get the maximum between two potential topmost elements:

   • The largest element from the range nums[0] to nums[k-1], acknowledging that we can make k-1 removals and then put the maximum number back on the pile.
   • The element at index k (nums[k]), since it can be the top of the pile if we remove k elements (one of them being this nums[k]) and then put nums[k] back.

4. If k is greater or equal to the number of elements (k >= len(nums)), we only look at the maximum element within the bounds of nums in the first k-1 moves. The k-th move could be adding an element back if there are still remaining moves.

By considering these scenarios, we can design the solution which ensures that we return the maximum topmost value after k moves, or -1 if it's not possible to perform k moves and still have elements left in the pile.

Learn more about Greedy patterns.

Solution Approach

The provided solution is straightforward and makes use of conditional checks and the max function to compare elements within the nums array based on different scenarios discussed in the intuition section.

Here's a step-by-step explanation of the implementation:

1. Check for No Moves (k == 0): If the number of moves k is 0, we simply return the topmost element as no moves are made.

   if k == 0:
       return nums[0]

2. Check for Single Element Piles (n == 1): If the pile has only one element, we need to check if k is an odd number. If k is odd, we will end up with no element at the top, so return -1. If k is even, we can keep removing and adding the same element, effectively making no changes, and we return the topmost element.

   n = len(nums)
   if n == 1:
       if k % 2:
           return -1
       return nums[0]

3. Find Maximum with Constraints: We then find the maximum value from the array up to the k-1 index. This is because after k-1 removals, we could replace the topmost element with one of the removed elements. We use the max function over the slice nums[: k - 1], and use default=-1 to handle the edge case where the slice could be empty.

   ans = max(nums[: k - 1], default=-1)

4. Consider Element at Index k: If k is less than the length of the array (k < n), it means the element at the index k could be a candidate for the maximum topmost element after k moves. We should choose the larger between the current ans and nums[k].

   if k < n:
       ans = max(ans, nums[k])

5. Return the Result: Finally, we return ans as the maximum possible value for the top of the pile after exactly k moves.

In terms of data structures, the solution simply uses the list (array) given, without the need for additional data structures. The key algorithmic pattern here is the greedy approach, selecting the largest value at each step with consideration to the constraints given by k and the number of elements n. The implementation uses basic array operations and conditions to achieve the desired outcome.

Example Walkthrough

Let's take a small example to illustrate the solution approach. Suppose we are given the following array of integers representing our pile:

nums = [4, 1, 5, 6]

and we have k = 3 moves to make.

Following the solution steps:

1. Since k is not 0, we do not return nums[0] right away. We proceed with the next steps.

2. There is more than one element in the pile (len(nums) = 4), so we do not need to consider the scenario where k is odd, and we only have one element.

3. We find the maximum value from the array up to the k-1 index, which is 2 in this case, so we consider nums up to and including index 1: nums[:k-1] = [4, 1]. We find the maximum value in this range:

   ans = max(nums[:2])  # The maximum value from [4, 1] which is 4.

   So, ans is now 4.

4. Since k is less than the length of the array (k < n), we consider the element at index k, which is nums[3] = 6. We then compare it with our current ans and select the larger value:

   ans = max(ans, nums[3])  # We compare 4 and 6.

   So, ans is updated to 6.

5. After evaluating the given conditions, we have determined that the maximum value we can have at the top of the pile after exactly 3 moves is 6.

We can now visualize the moves:

• First move: we remove the topmost element 4 from the pile, now the pile is [1, 5, 6].
• Second move: we remove the topmost element 1 from the pile, now the pile is [5, 6].
• Third move: we add an element back to the pile, and we have the option to add either 1 or 4. We choose 4 as it is the larger value. However, we also realize that the remaining element 6 can be placed on top as it is the largest of all the elements we have, even larger than 4. Thus, we opt to place 6 on top instead, leading to our answer.

Therefore, the maximum possible value for the top of the pile after exactly 3 moves is 6.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(k), where k is the number of operations to be performed on the nums array. This is because the code potentially examines elements up to the k-1 position in the nums array to find the maximum value, and the number of elements looked at is limited by k. When k < n, it also compares the value at the k-th position once, which does not affect the overall time complexity.

The space complexity of the code is O(1) since the code only uses a constant amount of additional space. The variables n and ans are used to store the length of the nums array and the interim maximum values, respectively, and they do not depend on the size of the input array.

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