<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 bisect import bisect_left
from typing import List

class Solution:
    def findMaximumLength(self, nums: List[int]) -> int:
        n = len(nums)
        # 前缀和数组，acc[i] 表示数组 nums 前 i 个元素的累积和
        acc = [0] * (n + 1)
        for i in range(1, n + 1):
            acc[i] = acc[i - 1] + nums[i - 1]
        
        # 辅助数组 pre 用于跟踪子数组结束位置
        pre = [0] * (n + 2)
        # 动态规划数组 dp，dp[i] 表示到第 i 个元素为止的最大长度
        dp = [0] * (n + 1)
        # 初始化 i 为 0
        i = 0
        
        for j in range(1, n + 1):
            # 确保子数组不重叠，更新 i 的位置
            i = max(i, pre[j])
            # 更新 dp 数组，以 j 结尾的子数组长度
            dp[j] = dp[i] + 1
            # 计算目标和，找到满足条件的子数组
            target = acc[j] * 2 - acc[i]
            # 二分查找目标和的位置
            k = bisect_left(acc, target)
            # 更新 pre 数组
            pre[k] = j
        
        # 返回满足条件的最长子数组长度
        return dp[n]

# 示例用法
solution = Solution()
nums = [1, 2, 3, 4, 5]
print(solution.findMaximumLength(nums))  # 输出应为满足条件的最大子数组长度

        
        
34400