feat: refactored docs structure

docs/(01-linear-structures)/monotonic-stack/(problems)/739_daily-temperatures.mdx

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+---
+title: '739. Daily Temperatures'
+description: Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the i^th day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Stack", slug: "stack" }
+  - { name: "Monotonic Stack", slug: "monotonic-stack" }
+video: "cTBiBSnjO3c"
+issue: "https://github.com/prdlk/leetcode/issues/72"
+sidebar:
+  label: 'Daily Temperatures'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `temperatures = [73,74,75,71,69,72,76,73]`
+- Output: `[1,1,4,2,1,1,0,0]`
+
+### Example 2:
+- Input: `temperatures = [30,40,50,60]`
+- Output: `[1,1,1,0]`
+
+### Example 3:
+- Input: `temperatures = [30,60,90]`
+- Output: `[1,1,0]`
+
+### Constraints:
+
+- `1 <= temperatures.length <= 10^5`
+- `30 <= temperatures[i] <= 100`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["dailyTemperatures(temperatures)"]) --> I["ans = zeros(n), stack = [] — indices of days still waiting"]
+  I --> L{"more (i, temp)?"}
+  L -- no --> E(["return ans"])
+  L -- yes --> W{"stack non-empty and temperatures[stack[-1]] < temp?"}
+  W -- yes --> P["prev = stack.pop(); ans[prev] = i - prev — today resolves that day"]
+  P --> W
+  W -- no --> A["stack.append(i)"]
+  A --> L
+```
+
+## Solution
+
+```py
+class Solution:
+    def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
+        ans = [0] * len(temperatures)
+        stack = []
+        for i, temp in enumerate(temperatures):
+            while stack and temperatures[stack[-1]] < temp:
+                prev = stack.pop()
+                ans[prev] = i - prev
+            stack.append(i)
+        return ans
+```

docs/(01-linear-structures)/monotonic-stack/(problems)/853_car-fleet.mdx

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+---
+title: '853. Car Fleet'
+description: There are n cars at given miles away from the starting mile 0, traveling to reach the mile target
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Stack", slug: "stack" }
+  - { name: "Sorting", slug: "sorting" }
+  - { name: "Monotonic Stack", slug: "monotonic-stack" }
+video: "Pr6T-3yB9RM"
+issue: "https://github.com/prdlk/leetcode/issues/73"
+sidebar:
+  label: 'Car Fleet'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]`
+- Output: `3`
+- Explanation: • The cars starting at `10 (speed 2`) and `8 (speed 4`) become a fleet, meeting each other at `12`. The fleet forms at `target`. • The car starting at `0 (speed 1`) does not catch up to any other car, so it is a fleet by itself. • The cars starting at `5 (speed 1`) and `3 (speed 3`) become a fleet, meeting each other at `6`. The fleet moves at `speed 1` until it reaches `target`.
+
+### Example 2:
+- Input: `target = 10, position = [3], speed = [3]`
+- Output: `1`
+- Explanation: There is only one car, hence there is only one fleet.
+
+### Example 3:
+- Input: `target = 100, position = [0,2,4], speed = [4,2,1]`
+- Output: `1`
+- Explanation: • The cars starting at `0 (speed 4`) and `2 (speed 2`) become a fleet, meeting each other at `4`. The car starting at `4 (speed 1`) travels to `5`. • Then, the fleet at `4 (speed 2`) and the car at `position 5 (speed 1`) become one fleet, meeting each other at `6`. The fleet moves at `speed 1` until it reaches `target`.
+
+### Constraints:
+
+- `n == position.length == speed.length`
+- `1 <= n <= 10^5`
+- `0 < target <= 10^6`
+- `0 <= position[i] < target`
+- All the values of `position` are unique.
+- `0 < speed[i] <= 10^6`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["carFleet(target, position, speed)"]) --> O["zip and sort by position, descending — nearest the target first"]
+  O --> I["stack = [] — arrival time of each fleet's leader"]
+  I --> L{"more (pos, spd)?"}
+  L -- no --> E(["return len(stack) — one entry per fleet"])
+  L -- yes --> T["time = (target - pos) / spd"]
+  T --> Q{"stack empty or time > stack[-1]?"}
+  Q -- yes --> A["stack.append(time) — slower, so it starts a new fleet"]
+  Q -- no --> B["discard — it catches the car ahead and joins that fleet"]
+  A --> L
+  B --> L
+```
+
+## Solution
+
+```py
+class Solution:
+    def carFleet(self, target: int, position: List[int], speed: List[int]) -> int:
+        stack = []
+        for pos, spd in sorted(zip(position, speed), reverse=True):
+            time = (target - pos) / spd
+            if not stack or time > stack[-1]:
+                stack.append(time)
+
+        return len(stack)
+```

docs/(01-linear-structures)/monotonic-stack/index.mdx

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+---
+title: Monotonic Stack
+description: 'A stack keeps its items in sorted order, which finds the next larger or smaller item in one pass.'
+hidden: true
+---