<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:
        
        prefix = list(itertools.accumulate(nums, initial=0))
        n = len(nums)
        bestleft = [0] * (n + 2)
        # bestleft[i]: looking at first i elements
        # in the optimal solution, the last block is nums[l, i) where l = bestleft[i]
        dp = [0] * (n + 1)
        # dp[i]: looking at first i elements, max length after transformation

        dp[1] = 1
        for i in range(1, n + 1):

            bestleft[i] = max(bestleft[i], bestleft[i - 1])
            left = bestleft[i]
            right = bisect.bisect_left(prefix, 2 * prefix[i] - prefix[left])
            bestleft[right] = max(bestleft[right], i)
            dp[i] = dp[left] + 1
        
        return dp[n]








        # solution 1: time limit exceeds
        #n = len(nums)
        #presum = [None] * n
        #presum[0] = nums[0]
        #for i in range(1, n):
        #    presum[i] = presum[i - 1] + nums[i]
        #
        #memo = {}
        #def dp(prev, idx):
        #    # prev: last element of previous adjusted array
        #    # idx: only look from index idx and forwards
        #    # return max length of the adjusted array starting from idx to the end
        #    if idx == n:
        #        return 0
        #
        #    if (prev, idx) in memo:
        #        return memo[(prev, idx)]
        #
        #    res = 0
        #    if nums[idx] >= prev:
        #        res = max(res, 1 + dp(nums[idx], idx + 1))
        #
        #    for j in range(idx + 1, n):
        #        if idx == 0:
        #            curr = presum[j]
        #        else:
        #            curr = presum[j] - presum[idx - 1]
        #        if curr >= prev:
        #            res = max(res, 1 + dp(curr, j + 1))
        #
        #    memo[(prev, idx)] = res
        #    return res
        #
        #return dp(float('-inf'), 0)

            
            

35000