Maximize the Topmost Element After K Moves

Medium

American Express

0 views

Topics:

ArraysGreedy Algorithms

You are given a 0-indexed integer array nums representing the contents of a pile, where nums[0] is the topmost element of the pile.

In one move, you can perform either of the following:

If the pile is not empty, remove the topmost element of the pile.
If there are one or more removed elements, add any one of them back onto the pile. This element becomes the new topmost element.

You are also given an integer k, which denotes the total number of moves to be made.

Return the maximum value of the topmost element of the pile possible after exactly k moves. In case it is not possible to obtain a non-empty pile after k moves, return -1.

Example 1:

Input: nums = [5,2,2,4,0,6], k = 4
Output: 5
Explanation:
One of the ways we can end with 5 at the top of the pile after 4 moves is as follows:
- Step 1: Remove the topmost element = 5. The pile becomes [2,2,4,0,6].
- Step 2: Remove the topmost element = 2. The pile becomes [2,4,0,6].
- Step 3: Remove the topmost element = 2. The pile becomes [4,0,6].
- Step 4: Add 5 back onto the pile. The pile becomes [5,4,0,6].
Note that this is not the only way to end with 5 at the top of the pile. It can be shown that 5 is the largest answer possible after 4 moves.

Example 2:

Input: nums = [2], k = 1
Output: -1
Explanation: 
In the first move, our only option is to pop the topmost element of the pile.
Since it is not possible to obtain a non-empty pile after one move, we return -1.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i], k <= 10⁹

Try coding it in LeetCode

Maximize the Topmost Element After K Moves

Medium

American Express

0 views

Topics:

ArraysGreedy Algorithms

You are given a 0-indexed integer array nums representing the contents of a pile, where nums[0] is the topmost element of the pile.

In one move, you can perform either of the following:

If the pile is not empty, remove the topmost element of the pile.
If there are one or more removed elements, add any one of them back onto the pile. This element becomes the new topmost element.

You are also given an integer k, which denotes the total number of moves to be made.

Return the maximum value of the topmost element of the pile possible after exactly k moves. In case it is not possible to obtain a non-empty pile after k moves, return -1.

Example 1:

Input: nums = [5,2,2,4,0,6], k = 4
Output: 5
Explanation:
One of the ways we can end with 5 at the top of the pile after 4 moves is as follows:
- Step 1: Remove the topmost element = 5. The pile becomes [2,2,4,0,6].
- Step 2: Remove the topmost element = 2. The pile becomes [2,4,0,6].
- Step 3: Remove the topmost element = 2. The pile becomes [4,0,6].
- Step 4: Add 5 back onto the pile. The pile becomes [5,4,0,6].
Note that this is not the only way to end with 5 at the top of the pile. It can be shown that 5 is the largest answer possible after 4 moves.

Example 2:

Input: nums = [2], k = 1
Output: -1
Explanation: 
In the first move, our only option is to pop the topmost element of the pile.
Since it is not possible to obtain a non-empty pile after one move, we return -1.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i], k <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums` and the value of `k`?
Can the elements in the `nums` array be negative, zero, or non-integer values?
If `k` is greater than the length of `nums`, what should the function return?
If there are multiple elements that could be the maximum after k moves, is any one of them acceptable, or is there a specific criterion for choosing among them?
Is `nums` guaranteed to be non-empty, or should I handle the case where `nums` is empty?

Brute Force Solution

Approach

The brute force way to solve this is to try out every possible combination of moves, keeping track of the topmost element after each set of moves. This means we will explore every possible path of removing and potentially adding elements back until we've performed the allowed number of moves.

Here's how the algorithm would work step-by-step:

Consider all the different things you could do for the first move: remove the first element, or if possible, remove the first two elements, and so on.
For each of those initial moves, consider all the possible moves you can make next. This could involve removing elements or, if you've removed all the elements, putting back one of the removed elements on top.
Continue making moves until you have made the exact number of moves allowed.
After each set of moves, check what the topmost element is.
Remember the largest topmost element you've seen after any of these sets of moves.
After trying every possible set of moves, return the largest topmost element that you remembered.

Code Implementation

def maximize_topmost_element_after_k_moves_brute_force(numbers, maximum_moves):
    maximum_topmost = -1

    def explore_moves(current_index, remaining_moves, current_stack, removed_elements):
        nonlocal maximum_topmost

        # Base case: If we've made all our moves, check the topmost element
        if remaining_moves == 0:
            if current_stack:
                maximum_topmost = max(maximum_topmost, current_stack[-1])
            else:
                maximum_topmost = max(maximum_topmost, -1)
            return

        # Case 1: Remove an element
        if current_index < len(numbers):
            # Recursive call after removing element
            explore_moves(current_index + 1, remaining_moves - 1, current_stack, removed_elements + [numbers[current_index]])

        # Case 2: Add an element back, if possible
        if removed_elements:
            # Ensure moves are made and elements are available to add
            # Recursive call after adding element
            explore_moves(current_index, remaining_moves - 1, current_stack + [removed_elements[-1]], removed_elements[:-1])

    # Start exploring the moves from the beginning
    explore_moves(0, maximum_moves, [], [])
    return maximum_topmost

Big(O) Analysis

Time Complexity

O(n^k) – The described brute force solution explores all possible combinations of removing and potentially adding back elements up to k moves. In the worst case, for each move, we might have up to n choices (either remove an element or put one back), where n is the number of elements in the array. Since we are making k moves, this leads to approximately n * n * ... * n (k times), which is n^k. Therefore, the time complexity is O(n^k).

Space Complexity

O(K) – The brute force approach, as described, explores all possible move combinations, which can be visualized as a tree. The depth of this tree is K (the number of moves allowed). In the worst-case, recursion might be used to explore these combinations. Therefore, the maximum depth of the recursion stack could be K, leading to O(K) space for the call stack. No other auxiliary data structures of significant size are mentioned.

Optimal Solution

Approach

The goal is to figure out how to get the largest number to the top of a pile of numbers after a certain number of moves. The most efficient strategy focuses on carefully picking and placing the largest number while being mindful of move constraints.

Here's how the algorithm would work step-by-step:

First, consider a simple case where you only have one or two numbers. If you have one number and you have to make more than zero moves, you can't do anything.
If the number of moves is odd and you only have one number, you can't put anything back on top.
If you have multiple numbers, look at how many moves you're allowed to make compared to the number of numbers you have.
If you have enough moves to remove all numbers except one and then put some back, find the biggest number in your starting set, excluding the last one.
If you don't have enough moves to remove all but one number, carefully consider your options. If you can remove all the numbers, put the second to last one on the top.
If you're allowed an even number of moves, select the largest number from the beginning section of the pile up to the move count.
If you have more moves than numbers, just grab the largest one from the beginning.
In short, strategically pick the largest number you can reach within your move limit and put it on top.

Code Implementation

def maximize_topmost(numbers, moves):
    number_of_numbers = len(numbers)
    
    if number_of_numbers == 1:
        if moves % 2 != 0:
            return -1
        else:
            return numbers[0]

    if moves > number_of_numbers:
        # If we have enough moves, we remove and replace elements freely.
        return max(numbers)

    if moves == number_of_numbers:
        # Removing all but one, place the second to last on top.
        return max(numbers[:number_of_numbers-1])

    # Determine the reachable maximum.
    maximum_reachable = -1
    for i in range(min(number_of_numbers, moves - 1) + 1):

        #We are removing the first 'i' elements.
        if i == moves - 1:

            #After the moves, the next element will be at the top.
            maximum_reachable = max(maximum_reachable, numbers[i])

        if i == moves and i < number_of_numbers:
            maximum_reachable = max(maximum_reachable, numbers[i])

    return maximum_reachable

Big(O) Analysis

Time Complexity

O(n) – The algorithm identifies the largest element among the first k or n-1 elements of the input array nums, where n is the size of nums and k is the number of allowed moves. Finding the maximum element in a subset of the array requires iterating through that subset once. Since the size of this subset is bounded by either k or n-1, the time complexity is O(min(k, n)). In the worst case, k could be greater than or equal to n, so min(k,n) is essentially O(n), resulting in an overall time complexity of O(n).

Space Complexity

O(1) – The provided plain English explanation does not describe the use of any auxiliary data structures like arrays, hash maps, or trees. The strategy focuses on comparing elements within the input array and determining the largest element that can be placed on top within the given move constraints. Therefore, the algorithm's space usage remains constant regardless of the input size (N, the number of elements in the pile), implying O(1) auxiliary space complexity.

Edge Cases

Case	How to Handle
Empty input array	Return -1 because no element can be at the top.
k is 0	Return the first element of the array (nums[0]) because no moves are made.
k is greater than the array length	If k is greater than array length by more than one, return -1; otherwise, consider all elements for the maximum.
Array length is 1 and k is odd	Return -1 because we effectively remove the single element then try to put it back, but can't due to the number of moves.
Array length is 1 and k is even	Return the only element in the array.
k equals array length	Return the maximum of the entire array excluding the last element, as the last element cannot be placed at the top in this scenario.
k is slightly smaller than the array length	Iterate and track the maximum among the first k elements, if any remain after k moves, the kth element has to be considered as well.
Large array size to consider efficiency	Use a single pass approach to find the max element up to k-1 to avoid unnecessary computations.