<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>


# @lc code=start
class Solution:

    def findMaximumLength(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [0] * (n + 1)  # 前 i 个元素，最长非递减序列长度
        last = [0] * (n + 1)  # 第 i 个元素，取dp[i] 时最大值的最小值
        # dp[i+1] >= dp[i]
        # 如果 sum([j+1...i]) >= last[j]，则 j 可以扩展到 i: pre_sum[i+1]-pre_sum[j+1]
        #   dp[i] = max(dp[i], dp[j] + 1)
        # last[i]越小，对结果越有利
        # 满足 s[i] >= s[j] + last[j] 的前提下，j越大，s[j]就越大，所以last[j]就越小，为了获得更大的j，要弹出队首
        # 转移之后，队列的后半部分，不一定满足 s[j] + last[j] <= s[i] + last[i]，为了保证队列的单调性，需要弹出队尾之后再将i入队
        q = [0]
        pre_sum = [0]
        for i, num in enumerate(nums):
            pre_sum.append(pre_sum[-1] + num)
            # 1. 去掉队首无用数据（计算转移时，直接取队首）
            while len(q) > 1 and pre_sum[q[1]] + last[q[1]] <= pre_sum[i + 1]:
                q.pop(0)
            # 2. 计算转移
            dp[i + 1] = dp[q[0]] + 1
            last[i + 1] = pre_sum[i + 1] - pre_sum[q[0]]
            # 3. 去掉队尾无用数据
            while q and pre_sum[q[-1]] + last[q[-1]] >= pre_sum[i + 1] + last[i + 1]:
                q.pop()
            q.append(i + 1)

        return dp[-1]

35900