# 155 Min Stack – Easy

### Problem:

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.

### Thoughts:

The trick is to use an extra stack to keep the current min value.

### Solutions:

``````public class MinStack {
private Stack<Integer> data;
private Stack<Integer> mins;
/** initialize your data structure here. */
public MinStack() {
data = new Stack<Integer>();
mins = new Stack<Integer>();
}
public void push(int x) {
data.push(x);
if (mins.isEmpty()) {
mins.push(x);
}
else {
mins.push(Math.min(x, mins.peek()));
}
}
public void pop() {
data.pop();
mins.pop();

}

public int top() {
return data.peek();
}

public int getMin() {
return mins.peek();
}
}

/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
``````
``````public class MinStack {
private Stack<Integer> data;
private Stack<Integer> mins;
/** initialize your data structure here. */
public MinStack() {
data = new Stack<Integer>();
mins = new Stack<Integer>();
data.push(-1);
mins.push(Integer.MAX_VALUE);
}
public void push(int x) {
data.push(x);
mins.push(Math.min(x, mins.peek()));
}
public void pop() {
if (data.size() > 1) {
data.pop();
mins.pop();
}
}

public int top() {
return data.peek();
}

public int getMin() {
return mins.peek();
}
}

/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
``````