Maximize the Topmost Element After K Moves

• Define a recursive function solve(moves_left, current_pile, removed_elements).
• Base Case: If moves_left == 0:
  • If current_pile is empty, return -1.
  • Otherwise, return the top element of current_pile.
• Initialize max_top = -1.
• Recursive Step 1 (Remove):
  • If current_pile is not empty:
    • Simulate removing the top element x.
    • Update max_top = max(max_top, solve(moves_left - 1, new_pile, removed_elements + {x})).
• Recursive Step 2 (Add):
  • For each element y in removed_elements:
    • Simulate adding y to the top of the pile.
    • Update max_top = max(max_top, solve(moves_left - 1, new_pile_with_y, removed_elements)).
• Return max_top.

The brute-force strategy involves a recursive exploration of every possible move at each step. We can define a function, say findMax(k, pile, removed_elements), that represents the state of the problem.

The base case for the recursion is when k (moves remaining) becomes 0. At this point, if the pile is not empty, we have a potential answer, which is the top element. If the pile is empty, this path is invalid.

In the recursive step (when k > 0), we branch out based on the allowed moves:

1. Remove operation: If the pile is not empty, we simulate removing the top element. We then make a recursive call with k-1 moves, the updated pile, and the newly removed element added to our set of removed elements.
2. Add operation: If there are any removed elements, we can add any one of them back. We iterate through each removed element, simulate adding it to the top of the pile, and make a recursive call with k-1 moves.

The function returns the maximum value obtained from all valid terminal states.

Due to the massive state space ((pile_state, removed_elements_state, k)) and the branching factor, this approach is computationally prohibitive.

1// This is a conceptual representation. A real implementation would be too slow2// and complex for the given constraints, likely causing a Time Limit Exceeded error.3class Solution {4    public int maximumTop(int[] nums, int k) {5        // The state space is too large for a direct recursive solution.6        // The number of branches at each step can be up to nums.length.7        // The depth of recursion is k.8        // This leads to a complexity roughly O((n)^k), which is not feasible.9        // A practical implementation would require memoization with a complex state,10        // but the state space itself is too large.11        return -1; // Placeholder for this infeasible approach.12    }13}

• Let n be the length of nums.
• Handle the edge case where the pile has only one element (n == 1). If k is odd, it's impossible to have a non-empty pile, so return -1. If k is even, we can remove and add the element back, so return nums[0].
• Initialize a variable max_candidate to 0 (or a suitable minimum).
• Candidate Set 1: Consider elements that are removed and then added back. To do this, we can use k-1 moves to manipulate the pile and the last move to add an element. The pool of elements we can add back are the ones we removed. To maximize our chances, we should consider the maximum among the first k-1 elements. We iterate from i = 0 to min(n, k - 1) - 1 and update max_candidate with nums[i] if it's larger.
• Candidate Set 2: Consider the element that is left on top after exactly k removals. This would be nums[k]. This is a valid candidate only if the pile is large enough, i.e., k < n. If so, we update max_candidate = max(max_candidate, nums[k]).
• The final answer is the max_candidate found.

Instead of exploring all move sequences, we can deduce the possible candidates for the maximum top element. We must first handle a key edge case:

• If nums.length == 1: With a single element, we can only alternate between removing it and adding it back. If k is odd, the pile will end up empty, so we return -1. If k is even, the pile will end up with nums[0] on top.

For the general case, we identify two main scenarios for the final top element:

1. The top element is one of the first k-1 elements: We can use k-1 moves to remove some number of elements from the top. With the final k-th move, we can add any of the removed elements back. To get the maximum value, we would choose to add back the largest element among those removed. The largest possible pool of removed elements we can choose from are nums[0], ..., nums[k-2]. So, the maximum of these elements is a candidate for the answer. This covers cases where k > n as well; if we remove all n elements, we can add back the maximum of the entire array.

2. The top element is nums[k]: This can be achieved by simply performing k consecutive removal operations. This is only possible if the pile has at least k+1 elements, i.e., k < nums.length.

By finding the maximum of all candidates from these two scenarios, we arrive at the answer. The algorithm iterates up to min(n, k-1) to find the first type of candidate and then checks nums[k] for the second type.

1class Solution {2    public int maximumTop(int[] nums, int k) {3        int n = nums.length;4 5        // If k is 0, no moves are made. The top is nums[0].6        if (k == 0) {7            return n > 0 ? nums[0] : -1;8        }9 10        // If there's only one element, the pile is non-empty after k moves only if k is even.11        if (n == 1) {12            return (k % 2 == 1) ? -1 : nums[0];13        }14 15        int maxVal = 0;16        // Candidate 1: Max of the first k-1 elements.17        // We can remove these and use the last move to add the max one back.18        // We can access at most n elements or k-1 elements.19        int limit = Math.min(n, k - 1);20        for (int i = 0; i < limit; i++) {21            maxVal = Math.max(maxVal, nums[i]);22        }23 24        // Candidate 2: The element at index k after k removals.25        // This is only possible if k is less than the number of elements.26        if (k < n) {27            maxVal = Math.max(maxVal, nums[k]);28        }29 30        return maxVal;31    }32}