Optimizing Subarray Lengths: Understanding a Dynamic Programming Approach
Intuition\n Describe your first thoughts on how to solve this problem. \nThe code seems to be addressing a problem related to finding the maximum length of a su
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code seems to be addressing a problem related to finding the maximum length of a subarray with a specific property. It involves dynamic programming and maintains a deque (double-ended queue) to efficiently track the relevant information for each element in the input array.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. The code initializes a dynamic programming array `dp` to store the maximum length of subarrays meeting the criteria.\n2. It also creates a list `prefix_sums` to store the cumulative sums of the input array `nums`.\n3. A deque named `queue` is used to keep track of potential candidates for the starting index of the subarrays.\n4. The code iterates through the elements of the input array `nums` using a for loop.\n5. It updates the deque by removing elements from the front until it finds a valid starting index.\n6. Computes the length of the current subarray using the difference between the current and previous starting indices.\n7. Updates the dynamic programming array `dp` with the calculated length.\n8. Adjusts the deque by removing elements from the back until the deque is in decreasing order of costs.\n9. Appends the current element along with its cost to the deque.\n10. The final result is the value stored in `dp[-1]`, representing the maximum length of subarrays meeting the criteria.\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def findMaximumLength(self, nums: List[int]) -> int:\n        dp = [0] * (len(nums) + 1)\n        prefix_sums = list(accumulate(nums, initial = 0))\n        queue = deque()\n        prev = 0\n        for i, val in enumerate(nums, 1):\n            while queue and queue[0][0] >= -prefix_sums[i]:\n                _, prev = queue.popleft()\n            dp[i] = dp[prev] + 1\n            cost = -prefix_sums[i] - (prefix_sums[i] - prefix_sums[prev])\n            while queue and queue[-1][0] <= cost:\n                queue.pop()\n            queue.append([cost, i])\n        return dp[-1]\n```
['Dynamic Programming', 'Python3']