<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>.</p>

<p>You can perform any number of operations, where each operation involves selecting a <strong>subarray</strong> of the array and replacing it with the <strong>sum</strong> of its elements. For example, if the given array is <code>[1,3,5,6]</code> and you select subarray <code>[3,5]</code> the array will convert to <code>[1,8,6]</code>.</p>

<p>Return <em>the </em><strong><em>maximum</em></strong><em> length of a </em><strong><em>non-decreasing</em></strong><em> array that can be made after applying operations.</em></p>

<p>A <strong>subarray</strong> is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,2,2]
<strong>Output:</strong> 1
<strong>Explanation:</strong> This array with length 3 is not non-decreasing.
We have two ways to make the array length two.
First, choosing subarray [2,2] converts the array to [5,4].
Second, choosing subarray [5,2] converts the array to [7,2].
In these two ways the array is not non-decreasing.
And if we choose subarray [5,2,2] and replace it with [9] it becomes non-decreasing. 
So the answer is 1.
</pre>

<p><strong>Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The array is non-decreasing. So the answer is 4.
</pre>

<p><strong>Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,3,2,6]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Replacing [3,2] with [5] converts the given array to [4,5,6] that is non-decreasing.
Because the given array is not non-decreasing, the maximum<!-- notionvc: 3447a505-d1ee-4411-8cae-e52162f53a55 --> possible answer is 3.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

from itertools import accumulate
from bisect import bisect_left
from typing import List

class Solution:
    def findMaximumLength(self, loads: List[int]) -> int:
        """
        Optimizes the distribution of server loads by summing subarrays of servers and replacing them with 
        their summed values to create the maximum length of non-decreasing server loads.
        
        Args:
            loads (List[int]): A list of integers representing the load on each server.

        Returns:
            int: The maximum number of servers that can have non-decreasing loads after the optimization.
        """
        
        # Step 1: Calculate the length of the loads array
        n = len(loads)
        
        # Step 2: Compute the cumulative sum of the loads array using accumulate.
        #         'acc' stores the cumulative sum starting with 0.
        acc = list(accumulate(loads, initial=0))
        
        # Step 3: Initialize arrays to track the optimization process.
        #         'pre' will track the previous index where the operation was performed for optimization.
        #         'dp' will store the dynamic programming states for the length of non-decreasing subarrays.
        pre = [0] * (n + 2)  # One extra element for boundary conditions.
        dp = [0] * (n + 1)   # One extra element for easier index handling in the dp array.
        
        # 'i' tracks the index of the last operation or subarray merging that preserves the non-decreasing order.
        i = 0
        
        # Step 4: Loop through the loads array to process each server's load and optimize.
        for j, load in enumerate(loads, 1):
            # Update 'i' to ensure we are working with the most recent subarray that preserves order.
            i = max(i, pre[j])
            
            # Update the dp array for the current index j. dp[i] holds the best non-decreasing length at i.
            dp[j] = dp[i] + 1
            
            # Calculate the index 'k' where the sum of subarray [i, j] maintains the non-decreasing property.
            # We use bisect_left to find the position in the cumulative sum array.
            k = bisect_left(acc, acc[j] * 2 - acc[i])
            
            # Update the pre array at index k with the current index j.
            pre[k] = j
        
        # Step 5: Return the length of the maximum non-decreasing subarray, which is stored in dp[n].
        return dp[n]
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