Growing direction of two stacks implementing similar array - Data Structure

Discussion

• Nirja Shah   -Posted on 18 Nov 15
  If we implement two stacks in opposite direction with same array, then overflow chance will be reduce.
  Example:
  Suppose that array name is arr [ ]. We will consider the same array as two stack, stack1 and stack2. stack1 starts from the leftmost element, and stack2 starts from the rightmost corner. The first element in stack1 is pushed at index 0 and the first element in stack2 is pushed at index (n-1).
  class twoStacksTest
  {
      int ∗arr;
      int size;
      int top1, top2;
      int item, i,j;
  public:
     twoStacksTest() // constructor
  {
      cout << "\nEnter the size of stack" << endl;
      cin >>size;
      arr = new int[size];
      top1 = -1;
      top2 = size;
  }
  
  // Method to push an element to stack1
  void push1()
  {
      if (top1 < top2 - 1)
    {
      cout << "\nEnter the itemt\t";
       cin >>item;
       top1++;
       arr[top1]=item;
  
    }
  else
  {
       cout << "Stack Overflow\n\n";
  
  }
  }
  
  // Method to push an element to stack2
  void push2()
  {
  
  if (top1 < top2 - 1)
   {
       cout <<"Enter the itemt";
       cin >>item;
       top2--;
       arr[top2] = item;
   }
  else
   {
       cout << "Stack Overflow\n\n";
    }
  }
  
  // Method to pop an element from first stack
  void pop1()
    {
       if (top1 >= 0 )
    {
       cout <<"The deleted item is\t"<<arr[top1];
       top1--;
    }
  else
     {
       cout << "Stack UnderFlow\n\n";
  
     }
   }
  
  // Method to pop an element from second stack
  void pop2()
  {
    if (top2 < size)
   {
       cout <<"The deleted item is\t" <<arr[top2];
       top2++;
  
    }
  else
   {
      cout << "Stack UnderFlow\n\n";
  
    }
  }
  void display()
    {
      for(i=top1;i>=0;i--)
     {
       cout <<arr[i]<<"\t\t";
     }
       for(i=top2;i<size;i++)
     {
       cout<<arr[i]<<"\t\t";
     }
    }
  };
  int main()
    {
  
       twoStacksTest obj;
       int choice;
    do
      {
       cout <<"Enter 1. for Push1\n";
       cout <<"Enter 2. for Push2\n";
       cout <<"Enter 3. for Pop1\n";
       cout <<"Enter 4. for Pop2\n";
       cout <<"Enter 5. for Display\n";
       cout <<"Enter 6. for Exit\n";
       cout <<"Enter you choice\n";
       cin >>choice;
       switch(choice)
     {
      case 1:
        obj.push1();
        break;
       case 2:
        obj.push2();
        break;
       case 3:
        obj.pop1();
        break;
       case 4:
        obj.pop2();
        break;
       case 5:
        obj.display();
        break;
       case 6:
        break;
        default:
        cout<<"Wrong choice\n";
         break;
  
     }
  }
        while(choice <6);
        return 0;
     }
  

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