Question

Design a stack in C++ that in addition to push, pop, and top functions, also has a min function that returns the minimum element in the entire stack. The stack can contain the minimum element in the top.

The purpose of having a min function is for run time purposes, such that the minimum element can be retrieved quickly when needed (as opposed to traversing an entire linked list from start to end just to find the minimum object).

You can implement this by modifying the push function, so that each time a new element is pushed it is checked if the new element is less than the top or greater than the top (if the new element is greater than the top then the top and new are swapped).

Answer 1

Code

#include <iostream>
using namespace std;
int stack[100], n=100, top=-1;
void push(int val) {
   if(top>=n-1)
cout<<"Stack Overflow"<<endl;
   else
   {
       if(top==-1)
       {
           top++;
           stack[top]=val;
       }
       else
       {
           int topElement=stack[top];
           if(val>topElement)
           {
               stack[top]=val;
               top++;
               stack[top]=topElement;
           }
           else
           {
               top++;
               stack[top]=val;
           }
       }
   }
}
int min()
{
   return stack[top];
}
void pop() {
if(top<=-1)
cout<<"Stack Underflow"<<endl;
else {
cout<<"The popped element is "<< stack[top] <<endl;
top--;
}
}
void display() {
if(top>=0) {
cout<<"Stack elements are:";
for(int i=top; i>=0; i--)
cout<<stack[i]<<" ";
cout<<endl;
} else
cout<<"Stack is empty";
}
int main()
{
   push(10);
   push(20);
   push(30);
   display();
   cout<<"Minimim element in stack is:"<<min()<<endl;
}

outptu

Enter the first rational number: 2/3 Enter the second rational number: 5/7 Two Rationla number are: 2/3 + 5/7-29/21 2/3 5/7--1/21 /35/7 -18/21 2/35/7 -14/15 2/3 is less than 5/7 Press any key to continue . .-- 12:18 O Type here to search 08-05-2019 2

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