Min Stack

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. Implement the MinStack class with the following operations: - MinStack() initializes the stack object - void push(int val) pushes the element val onto the stack - void pop() removes the element on the top of the stack - int top() gets the top element of the stack - int getMin() retrieves the minimum element in the stack

Example 1

  
Input: ["MinStack", "push", "push", "push", "getMin", "pop", "top", "getMin"] [[], [-2], [0], [-3], [], [], [], []]
  
Output: [null, null, null, null, -3, null, 0, -2]
  
Explanation:

 Step 1: MinStack minStack = new MinStack();

 Step 2: minStack.push(-2);

 Step 3: minStack.push(0);

 Step 4: minStack.push(-3);

 Step 5: minStack.getMin() returns -3 (minimum element in stack)

 Step 6: minStack.pop() removes the top element (-3)

 Step 7: minStack.top() returns 0 (current top element)

 Step 8: minStack.getMin() returns -2 (current minimum element)

Example 2

  
Input: ["MinStack", "push", "push", "getMin", "push", "getMin", "pop", "getMin"] [[], [5], [7], [], [2], [], [], []]
  
Output: [null, null, null, 5, null, 2, null, 5]
  
Explanation:

 Step 1: MinStack minStack = new MinStack();

 Step 2: minStack.push(5);

 Step 3: minStack.push(7);

 Step 4: minStack.getMin() returns 5 (minimum element in stack)

 Step 5: minStack.push(2);

 Step 6: minStack.getMin() returns 2 (new minimum element)

 Step 7: minStack.pop() removes the top element (2)

 Step 8: minStack.getMin() returns 5 (minimum element after pop)

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
  
Time Complexity: O(1) for all operations
  
Space Complexity: O(n) where n is the number of elements in the stack