beats 100, with full explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\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    n = len(nums)\n    # prefix[i] := the sum of the first i nums\n    prefix = list(itertools.accumulate(nums, initial=0))\n    # dp[i] := the maximum number of elements in the increasing\n    # sequence after processing the first i nums\n    dp = [0] * (n + 1)\n    # bestLeft[i] := the index l s.t. merging nums[l..i) is the\n    # optimal strategy among processing the first i nums\n    bestLeft = [0] * (n + 2)\n\n    for i in range(1, n + 1):\n      bestLeft[i] = max(bestLeft[i], bestLeft[i - 1])\n      # When merging nums[l, i), consider the next segment as [i, r).\n      # Find the minimum `r` where the sum(nums[l, i)) <= sum(nums[i, r)).\n      # Equivalently, prefix[i] - prefix[l] <= prefix[r] - prefix[i].\n      #            => prefix[r] >= prefix[i] * 2 - prefix[l]\n      # Therefore, we can binary search `prefix` to find the minimum `r`.\n      l = bestLeft[i]\n      r = bisect.bisect_left(prefix, 2 * prefix[i] - prefix[l])\n      dp[i] = dp[l] + 1\n      bestLeft[r] = i\n\n    return dp[n]\n```
['Python3']