# [LeetCode 155] Min Stack

### Question

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• getMin() -- Retrieve the minimum element in the stack.

sn’t look possible to go to left half or right half by doing constant number of comparisons at the mSolution

### Analysis

I’ve already cover this question in another post [Question] Min Stack.

### My Code

``````class MinStack {

Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> min = new Stack<Integer>();

public void push(int x) {
stack.push(x);
if (min.isEmpty() || min.peek() >= x) {
min.push(x);
}
}

public void pop() {
if (stack.isEmpty()) {
return;
}
int topNum = stack.pop();
if (topNum == min.peek()) {
min.pop();
}
}

public int top() {
if (stack.isEmpty()) {
return 0;
}
return stack.peek();
}

public int getMin() {
if (min.isEmpty()) {
return 0;
}
return min.peek();
}
}
``````