Solutions with two stacks

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class MinStack {
    Stack<Integer> nums = new Stack<Integer>();
    Stack<Integer> min = new Stack<Integer>();
    
    public void push(int x) {
        nums.push(x);
        if (!min.empty() && min.peek() < x){
            min.push(min.peek());
        }else{
            min.push(x);
        }
    }

    public void pop() {
        if (!nums.empty() && !min.empty()){
            nums.pop();
            min.pop();
        }
    }

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

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

Solutions with one stacks

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public class MinStack {
    long min;
    Stack<Long> stack;

    /** initialize your data structure here. */
    public MinStack() {
        stack = new Stack<Long>();
    }
    
    public void push(int x) {
        if (stack.empty()) {
            stack.push(0L);
            min = x;
        } else {
            stack.push(x - min); // the max can be Integer.max_value - Integer.min_value, which will overflow, that's why we need long
            min = Math.min(x, min);
        }
    }
    
    public void pop() {
        if (!stack.empty()) {
            Long pop = stack.pop();
            if (pop < 0) {
                min -= pop;
            }
        }
    }
    
    public int top() {
        Long peek = stack.peek();
        if (peek < 0) {
            return (int) min;
        } else {
            return (int) (min + peek);
        }
    }
    
    public int getMin() {
        return (int) min;
    }
}

/**
 * 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();
 */