MinStack.java

package org.example;

import java.util.*;

// Time Complexity : O(1)
// Space Complexity : O(1)
// Did this code successfully run on Leetcode : Yes
// Any problem you faced while coding this : No


// Your code here along with comments explaining your approach
// The basic idea is to use a single stack and initially the min is assumed to be infinity. When a value less the current minimum is identified,
// then push the previous minimum followed by the new value. And when popping, pop the top of the stack and the next element below it if the
// popped value is currently the minimum, then the lastly popped element will be the new minimum

// Using one stack
class MinStack {
    Stack<Integer> stack;
    int min;

    public MinStack() {
        this.min = Integer.MAX_VALUE;
        this.stack = new Stack<>();
    }

    public void push(int val) {
        if(min>= val){
            stack.push(min);
            min = val;
        }
        stack.push(val);
    }

    public void pop() {
        if(stack.pop() == min){
            min = stack.pop();
        }

    }

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

    public int getMin() {
        return min;
    }
}

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