feat: refactored docs structure

apps/docs/content/(01-linear-structures)/stack/(problems)/155_min-stack.mdx

@@ -1,82 +0,0 @@
----
-title: '155. Min Stack'
-description: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time
-icon: dot
-topics:
-  - { name: "Stack", slug: "stack" }
-  - { name: "Design", slug: "design" }
-video: "qkLl7nAwDPo"
-issue: "https://github.com/prdlk/leetcode/issues/68"
-sidebar:
-  label: 'Min Stack'
-  badge: 'Medium'
----
-
-::::warning
-You must implement a solution with O(1) time complexity for each function.
-::::
-
-### Example 1:
-- Input: ``
-- Output: ``
-- Explanation: Input `["MinStack","push","push","push","getMin","pop","top","getMin"]` [[],[-2],[0],[-3],[],[],[],[]] Output `[null,null,null,null,-3,null,0,-2]` Explanation MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); // return `-3` minStack.pop(); minStack.top();    // return `0` minStack.getMin(); // return `-2`
-
-### Constraints:
-
-- `-2^31 <= val <= 2^31 - 1`
-- Methods pop, top and getMin operations will always be called on non-empty stacks.
-- At most 3 * 10^4 calls will be made to push, pop, top, and getMin.
-
-## Approach
-
-```mermaid
-flowchart TD
-  subgraph state["two parallel stacks — same depth, always"]
-    A["stack — the values"]
-    B["minStack — the minimum as of that depth"]
-  end
-  state --> P
-  subgraph P["push(value)"]
-    P1["stack.append(value)"] --> P2["minStack.append(min(value, minStack[-1]))"]
-  end
-  state --> O
-  subgraph O["pop()"]
-    O1["pop both stacks together"]
-  end
-  state --> R
-  subgraph R["top() / getMin() — O(1)"]
-    R1(["stack[-1] / minStack[-1]"])
-  end
-```
-
-## Solution
-
-```py
-class MinStack:
-    def __init__(self):
-        self.stack = []
-        self.minStack = []
-
-    def push(self, value: int) -> None:
-        self.stack.append(value)
-        value = min(value, self.minStack[-1] if self.minStack else value)
-        self.minStack.append(value)
-
-    def pop(self) -> None:
-        self.stack.pop()
-        self.minStack.pop()
-
-    def top(self) -> int:
-        return self.stack[-1]
-
-    def getMin(self) -> int:
-        return self.minStack[-1]
-
-
-# Your MinStack object will be instantiated and called as such:
-# obj = MinStack()
-# obj.push(value)
-# obj.pop()
-# param_3 = obj.top()
-# param_4 = obj.getMin()
-```