feat: refactored docs structure

apps/docs/content/(05-decision-space)/greedy/(problems)/53_maximum-subarray.mdx

@@ -1,54 +0,0 @@
----
-title: '53. Maximum Subarray'
-description: Given an integer array nums, find the subarray with the largest sum, and return its sum
-icon: dot
-topics:
-  - { name: "Array", slug: "array" }
-  - { name: "Divide and Conquer", slug: "divide-and-conquer" }
-  - { name: "Dynamic Programming", slug: "dynamic-programming" }
-issue: "https://github.com/prdlk/leetcode/issues/174"
-sidebar:
-  label: 'Maximum Subarray'
-  badge: 'Medium'
----
-
-::::warning
-If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
-::::
-
-### Example 1:
-- Input: `nums = [-2,1,-3,4,-1,2,1,-5,4]`
-- Output: `6`
-- Explanation: The subarray `[4,-1,2,1]` has the largest sum `6`.
-
-### Example 2:
-- Input: `nums = [1]`
-- Output: `1`
-- Explanation: The subarray `[1]` has the largest sum `1`.
-
-### Example 3:
-- Input: `nums = [5,4,-1,7,8]`
-- Output: `23`
-- Explanation: The subarray `[5,4,-1,7,8]` has the largest sum `23`.
-
-### Constraints:
-
-- `1 <= nums.length <= 10^5`
-- `-10^4 <= nums[i] <= 10^4`
-
-## Solution
-
-```py
-class Solution:
-    def maxSubArray(self, nums: list[int]) -> int:
-        best, total = nums[0], nums[0]
-
-        for i in range(1, len(nums)):
-            if total < 0:
-                total = nums[i]
-            else:
-                total = total + nums[i]
-            best = max(best, total)
-
-        return best
-```