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+# 2945. Find Maximum Non-decreasing Array Length
+from itertools import accumulate
+from bisect import bisect_left
+
+"""
+you are given a 0-indexed integer array 'nums'. you can perform any number of
+operations, where each operation involves selecting a subarray of the array
+and replacing it with the sum of its elements. for example, if the given
+array is '[1, 3, 5, 6]' and you select '[3, 5]', the array will convert to
+'[1, 8, 6]'. return the maximum length of a non-decreasing array that can be
+made after applying the aforementioned operations. a subarray is a contiguous
+non-empty sequence of elements within an array
+"""
+
+
+class Solution(object):
+ def findMaximumLength(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: int
+ """
+ n = len(nums)
+ acc = list(accumulate(nums, initial=0))
+ pre = [0] * (n + 2)
+ dp = [0] * (n + 1)
+ i = 0
+ for j, k in enumerate(nums, 1):
+ i = max(i, pre[j])
+ dp[j] = dp[i] + 1
+ l = bisect_left(acc, acc[j] * 2 - acc[i])
+ pre[l] = j
+ return dp[n]
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.findMaximumLength(nums=[5, 2, 2]))
+ print(obj.findMaximumLength(nums=[1, 2, 3, 4]))
+ print(obj.findMaximumLength(nums=[4, 3, 2, 6]))