feat: refactored docs structure

docs/(01-linear-structures)/binary-search/(problems)/162_find-peak-element.mdx

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+---
+title: '162. Find Peak Element'
+description: A peak element is an element that is strictly greater than its neighbors
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Binary Search", slug: "binary-search" }
+issue: "https://github.com/prdlk/leetcode/issues/64"
+sidebar:
+  label: 'Find Peak Element'
+  badge: 'Medium'
+---
+
+::::warning
+You must write an algorithm that runs in O(log n) time.
+::::
+
+### Example 1:
+- Input: `nums = [1,2,3,1]`
+- Output: `2`
+- Explanation: `3` is a peak element and your function should return the index number `2`.
+
+### Example 2:
+- Input: `nums = [1,2,1,3,5,6,4]`
+- Output: `5`
+- Explanation: Your function can return either index number `1` where the peak element is `2`, or index number `5` where the peak element is `6`.
+
+### Constraints:
+
+- `1 <= nums.length <= 1000`
+- `-2^31 <= nums[i] <= 2^31 - 1`
+- `nums[i] != nums[i + 1] for all valid i.`
+
+## Solution
+
+```py
+class Solution:
+    def findPeakElement(self, nums: list[int]) -> int:
+        left, right = 0, len(nums) - 1
+        while left < right:
+            mid = (left + right) // 2
+            if nums[mid] < nums[mid + 1]:
+                left = mid + 1
+            else:
+                right = mid
+
+        return left
+```