feat: refactored docs structure

docs/(01-linear-structures)/binary-search/(problems)/35_search-insert-position.mdx

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+---
+title: '35. Search Insert Position'
+description: Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Binary Search", slug: "binary-search" }
+video: "K-RYzDZkzCI"
+issue: "https://github.com/prdlk/leetcode/issues/63"
+sidebar:
+  label: 'Search Insert Position'
+  badge: 'Easy'
+---
+
+::::warning
+You must write an algorithm with O(log n) runtime complexity.
+::::
+
+### Example 1:
+- Input: `nums = [1,3,5,6], target = 5`
+- Output: `2`
+
+### Example 2:
+- Input: `nums = [1,3,5,6], target = 2`
+- Output: `1`
+
+### Example 3:
+- Input: `nums = [1,3,5,6], target = 7`
+- Output: `4`
+
+### Constraints:
+
+- `1 <= nums.length <= 10^4`
+- `-10^4 <= nums[i] <= 10^4`
+- `nums` contains distinct values sorted in ascending order.
+- `-10^4 <= target <= 10^4`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["searchInsert(nums, target)"]) --> I["l = 0, r = n - 1"]
+  I --> W{"l <= r?"}
+  W -- yes --> M["m = (l + r) // 2"]
+  M --> Q{"nums[m] vs target"}
+  Q -- equal --> R(["return m"])
+  Q -- "nums[m] < target" --> L["l = m + 1"]
+  L --> W
+  Q -- "nums[m] > target" --> H["r = m - 1"]
+  H --> W
+  W -- no --> E(["return (l + r) // 2 + 1 — the insert slot"])
+```
+
+## Solution
+
+```py
+class Solution:
+    def searchInsert(self, nums: List[int], target: int) -> int:
+        l, r = 0, len(nums) - 1
+
+        while l <= r:
+            m = (l + r) // 2
+            if nums[m] == target:
+                return m
+            elif nums[m] < target:
+                l = m + 1
+            elif nums[m] > target:
+                r = m - 1
+
+        return (l + r) // 2 + 1
+```