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 subarray [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 operations.
\n\nA subarray is a contiguous non-empty sequence of elements within an array.
\n\n\n
Example 1:
\n\n\nInput: nums = [5,2,2]\nOutput: 1\nExplanation: This array with length 3 is not non-decreasing.\nWe have two ways to make the array length two.\nFirst, choosing subarray [2,2] converts the array to [5,4].\nSecond, choosing subarray [5,2] converts the array to [7,2].\nIn these two ways the array is not non-decreasing.\nAnd if we choose subarray [5,2,2] and replace it with [9] it becomes non-decreasing. \nSo the answer is 1.\n\n\n
Example 2:
\n\n\nInput: nums = [1,2,3,4]\nOutput: 4\nExplanation: The array is non-decreasing. So the answer is 4.\n\n\n
Example 3:
\n\n\nInput: nums = [4,3,2,6]\nOutput: 3\nExplanation: Replacing [3,2] with [5] converts the given array to [4,5,6] that is non-decreasing.\nBecause the given array is not non-decreasing, the maximum possible answer is 3.\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 105dp[i] be the maximum number of elements in the increasing sequence after processing the first i elements of the original array.",
"We have dp[0] = 0. dp[i + 1] >= dp[i] (since if we have the solution for the first i elements, we can always merge the last one of the first i + 1 elements which is nums[i] into the solution of the first i elements.",
"For i > 0, we want to dp[i] = max(dp[j] + 1) where sum(nums[i - 1] + nums[i - 2] +… + nums[j]) >= v[j] and v[j] is the last element of the solution ending with nums[j - 1]."
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