<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>

class Solution:
    def findMaximumLength(self, nums: List[int]) -> int:
        '''
            Intuition : Binary search with monotonic stack

            n, acc, total = 0,0,0
            dq = deque() # (n, total, need)
            for v in nums:
                total+=v
                while dq and dq[0][2]<=total: n,acc,_ = dq.popleft()
                while dq and dq[-1][2]>=2*total-acc: dq.pop()
                dq.append((n+1, total, 2*total-acc))
            return n+1
        '''
        # can do in one run with loops inside without searching via base conditions 
        # track max length 
        max_length = 0
        # track accumulated total so far separately from maximum total so far 
        accumulated_total = 0
        total = 0
        # use a deque for your monotonic stack 
        monotonic_stack = collections.deque()
        # loop nums 
        for num in nums:
            # add to total each time  
            total += num 
            # then, while you have a monotonic stack and the first items combined totals is less than or equal to total 
            while monotonic_stack and monotonic_stack[0][2] <= total: 
                # remove it, and update your max length and the accumulated total. No need to track the combined totals. 
                max_length, accumulated_total, _ = monotonic_stack.popleft()
            # then, while you have a monotonic stack, and the last items combined totals is gte to the next combined total
            while monotonic_stack and monotonic_stack[-1][2] >= (2 * total - accumulated_total) : 
                # remove it entirely as it is not satisfactory 
                monotonic_stack.pop() 
            # when done on this pass, append the increase in max length, the total, and the new combined total 
            monotonic_stack.append((max_length + 1, total, 2 * total - accumulated_total))
        # return max length + 1 when done to account for 1 indexing in the problem space 
        return max_length + 1 

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