feat: refactored docs structure

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

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+---
+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()
+```