You are given an array of servers represented by their processing power. You want to join consecutive servers together to form larger, more powerful servers. You can only join server i with server i+1 if the processing power of server i+1 is strictly greater than the processing power of server i. This joining process can cascade, meaning that if you join i and i+1, and the new combined processing power is less than i+2, you can then join the combined server with i+2, and so on. The goal is to maximize the number of these larger servers.

For example:

Consider the array [1, 2, 3, 4, 5]. You can join 1 and 2 to form 3. Then you can join 3 and 3 to form 6. Then you can join 6 and 4 to form 10. Finally, you can join 10 and 5 to form 15, resulting in one server with processing power 15.

Consider the array [5, 4, 3, 2, 1]. You cannot join any servers because the processing power is always decreasing.

Consider the array [1, 3, 2, 4, 1, 5]. You can join 1 and 3 to get 4. You cannot join 4 and 2. You can join 2 and 4 to get 6. You cannot join 6 and 1. You can join 1 and 5 to get 6. The final servers will be [4, 6, 6].

Write a function that takes an array of integers representing server processing powers and returns the final array of server processing powers after performing all possible joins.

Joining Servers Together

Problem Description

For example:

Consider the array [5, 4, 3, 2, 1]. You cannot join any servers because the processing power is always decreasing.

Write a function that takes an array of integers representing server processing powers and returns the final array of server processing powers after performing all possible joins.

Naive Solution

A naive solution would involve iterating through the array and, for each element, checking if it can be joined with the next element. If it can, join them and continue the process. This can be implemented using a loop and a conditional statement.

def naive_join_servers(servers):
    i = 0
    while i < len(servers) - 1:
        if servers[i+1] > servers[i]:
            servers[i+1] = servers[i] + servers[i+1]
            servers.pop(i)
            i = 0  # Restart from the beginning
        else:
            i += 1
    return servers

This approach has a time complexity of O(n^2) in the worst case, where n is the number of servers. This is because in the worst case, we might have to restart the loop from the beginning multiple times.

Optimal Solution

A more efficient solution can be achieved by using a stack or a similar data structure to keep track of the servers that can be potentially joined. This allows us to avoid restarting the loop from the beginning every time we perform a join.

def join_servers(servers):
    result = []
    i = 0
    while i < len(servers):
        current = servers[i]
        while result and current > result[-1]:
            current += result[-1]
            result.pop()
        result.append(current)
        i += 1
    return result

Example Usage

servers1 = [1, 2, 3, 4, 5]
print(f"Input: {servers1}, Output: {join_servers(servers1)}")  # Output: [15]

servers2 = [5, 4, 3, 2, 1]
print(f"Input: {servers2}, Output: {join_servers(servers2)}")  # Output: [5, 4, 3, 2, 1]

servers3 = [1, 3, 2, 4, 1, 5]
print(f"Input: {servers3}, Output: {join_servers(servers3)}")  # Output: [4, 6, 6]

servers4 = [1, 2, 0, 3, 4]
print(f"Input: {servers4}, Output: {join_servers(servers4)}")  # Output: [3, 7]

Big(O) Run-time Analysis

The optimal solution has a time complexity of O(n), where n is the number of servers. This is because each server is visited and potentially added to the result list only once. The inner while loop, which performs the joins, also processes each element at most once, as each element that is popped from the result list is permanently removed.

Big(O) Space Usage Analysis

The space complexity of the optimal solution is O(n) in the worst case, where n is the number of servers. This is because, in the worst-case scenario (e.g., a decreasing sequence of server processing powers), all the servers might be added to the result list before any joins can occur. Therefore, the result list could potentially store all the elements of the input servers list.

Edge Cases

Empty input array: If the input array is empty, the function should return an empty array.
Array with a single element: If the input array has only one element, the function should return an array containing that single element.
Array with all elements in decreasing order: In this case, no joins are possible, and the function should return the original array.
Array with all elements in increasing order: In this case, all elements should be joined into a single element.
Array with zero or negative values: The algorithm should work correctly with zero or negative values. If the server[i+1] is greater than server[i], they should still be joined, regardless of whether the values are positive, negative, or zero.

Given an array representing server processing powers, merge consecutive servers if the next server's power is greater than the current. What is the final array of server powers after all possible merges are done?