Fasted solution using prefix sum and dequeue
Intuition\n Describe your first thoughts on how to solve this problem. \nInitialize with a prefix sum of the array.\n\n\n# Approach\n Describe your approach to 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nInitialize with a prefix sum of the array.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe second while loop in your function, let\'s consider an example and walk through the loop\'s behavior. This loop is crucial for maintaining the deque (dq) in a way that it only contains elements that are potentially useful for finding the maximum length of a non-decreasing array.\n\nThe condition `2*prefix_sum[i] - cur` is crucial. It ensures that the elements in the deque represent possible subarray sums that could form a non-decreasing sequence when summed with future elements.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n> The prefix sum of the array A is computed using accumulate, which takes \n$O(n)$ time, where `n` is the length of the array. Each element is added to the deque only once and removed from the deque at most once. Therefore, across all iterations of the main loop, the total number of deque operations (append and pop) is $O(n)$.\n\nCombining these, the total time complexity of the function is \n$O(n)$. Despite the nested loops, the *deque* ensures that each element is processed in constant time on average.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n> The space complexity is influenced by the prefix_sum array and the deque (dq).\n- prefix_sum of size n takes $O(n)$ space\n- The `deque` can store upto n elements so takes $O(n)$ space\n\nThus, total space complexity is $O(n)$\n\n\n# Code\n```\n\nclass Solution:\n    def findMaximumLength(self, A: List[int]) -> int:\n        n = len(A)\n        if n == 1: return 1\n        prefix_sum = list(accumulate(A))\n        dq, cur, r = deque(), 0, 0  \n        for i in range(n):\n            while dq and dq[0][0] <= prefix_sum[i]:\n                (_, r, cur) = dq.popleft()\n            minv = 2*prefix_sum[i] - cur\n            while dq and dq[-1][0] >= minv: \n                dq.pop()\n            dq.append((minv, r+1, prefix_sum[i]))\n        return r+1\n\n    \n    \n\n   \n```
['Python3']