Skip to main content

Stack with find minimum function

Question : Implement a stack which finds the minimum of elements in the stack. All the three functions push(), pop() and findmin() must have time complexity of O(1)

This is one of the common questions asked in interviews.

The problem is finding minimum requires you to traverse through the stack, which is not possible without popping elements. So you can pop elements, find minimum and then push the elements back.

But the requirement says that findmin() function must have a constant complexity - O(1).

You can achieve this with the help of one more stack, to store minimum values.

Let us call this minstack. This is how the operations will work

push - 

  • push a value to stack
  • compare the value with top of minstack
    • if value is smaller than minstack top element push it to minstack
    • if not push minstack top element again to minstack

pop 

  • pop a value from stack
  • pop a value from minstack

findmin

  • get the value from minstack without popping it (peek)
 So this way we can have all three functions with O(1) time complexity. Here is our code


#include<stdio.h>  
#include<stdlib.h>
#define ERROR -1000
 struct node  
 {  
   int value;  
   struct node *next;   
 };  
 typedef struct node * NODEPTR;  
 
 NODEPTR create_node(int value)  
 {  
   NODEPTR temp = (NODEPTR) malloc(sizeof(struct node));  
    temp->next = NULL;  
   temp->value= value;  
   return temp;  
 } 
 NODEPTR push(NODEPTR top,int val)
 { 
     NODEPTR newnode = create_node(val);
     newnode->next = top;
     top = newnode;
     return top;
}
int isempty(NODEPTR top)
{
   return top==NULL;
}
int pop(NODEPTR *topPtr)
{
   if(isempty(*topPtr))
       return ERROR;
   int num = (*topPtr)->value;
   NODEPTR temp = *topPtr;
   *topPtr = (*topPtr)->next;
   free(temp);
   return num;
}
int peek(NODEPTR nd)
{
   return nd->value;
 }

void push1(NODEPTR*topptr,NODEPTR* mintopptr,int val)
{
     int minval =val;
     if(*mintopptr!=NULL)
        {
    int temp = peek(*mintopptr);
           if (temp<minval)
               minval = temp;
        }
     *topptr = push(*topptr,val);
     *mintopptr = push(*mintopptr,minval);
 }

 int pop1(NODEPTR* topptr,NODEPTR *mintopptr)
 {
   if(!isempty(*topptr))
   {
    pop(mintopptr);
    return pop(topptr);
   }else
       return -1;
 }
      
int findmin(NODEPTR mintop)
{
   if(mintop==NULL)
      return -1;
   return mintop->value;
}   

 int main()  
 {  
     NODEPTR top = NULL;//stack
     NODEPTR nd;
     NODEPTR mintop=NULL;//minstack

     while(1)
     {  
       int option,value; 
       printf("Enter 1- push 2- pop 3 - exit 4- find minimum"); 
       scanf("%d",&option);
       if(option==3)
          break;
       switch(option)
       {
           case 1:  printf("new value to push=");  
             scanf("%d",&value);   
                    push1(&top,&mintop,value);
                    break;
           case 2:  value = pop1(&top,&mintop);
                    if(value==ERROR)
                         printf("Stack empty\n");
                    else
    printf("Value popped is %d\n",value);
                    break; 
           case 4:  value=findmin(mintop);
                    if(value==-1)
                        printf("Stack empty\n");
                    else
                    printf("Minimum of stack is %d\n",value);
             
                    break;
        }  
    }     
 }

Notes
  1. We are implementing the stack using a linked list
  2. push1() function is used for pushing the value to both the stack and minstack. pop1 is used for popping values.
Labels:

Comments

Post a Comment

Popular posts from this blog

Delete a node from doubly linked list

Deletion operation in DLL is simpler when compared to SLL. Because we don't have to go in search of previous node of to-be-deleted node.  Here is how you delete a node Link previous node of node of to-be-deleted to next node. Link next node of node of to-be-deleted to previous node. Free the memory of node of to-be-deleted Simple, isn't it. The code can go like this. prevnode = delnode->prev; nextnode = delnode->next; prevnode->next = nextnode; nextnode->prev = prevnode; free(delnode); And that is it. The node delnode is deleted. But we should always consider boundary conditions. What happens if we are trying to delete the first node or last node? If first node is to be deleted, its previous node is NULL. Hence step 3 should not be used.  And also, once head is deleted, nextnode becomes head . Similarly if last node is to be deleted, nextnode is NULL. Hence step 4 is as strict NO NO. And we should set prevnode to tail. After we put these things together, we have...
Post a Comment
Read more

Program to delete a node from linked list

How do you remove a node from a linked list? If you have to delete a node, first you need to search the node. Then you should find its previous node. Then you should link the previous node to the next node. If node containing 8 has to be deleted, then n1 should be pointing to n2. Looks quite simple. Isn't it? But there are at least two special cases you have to consider. Of course, when the node is not found. If the node is first node of the list viz head. If the node to be deleted is head node, then if you delete, the list would be lost. You should avoid that and make the second node as the head node. So it becomes mandatory that you return the changed head node from the function.   Now let us have a look at the code. #include<stdio.h> #include<stdlib.h> struct node { int data; struct node * next; }; typedef struct node * NODEPTR; NODEPTR create_node ( int value) { NODEPTR temp = (NODEPTR) malloc( size...
3 comments
Read more

Program to create a Linked List in C

An array is a commonly used data structure in most of the languages. Because it is simple, it needs O(1) time for accessing elements. It is also compact. But an array has a serious drawback - it can not grow or shrink. You need to estimate the array size and define it during compile time. This drawback is not present a linked list. A linked list is a data structure which can grow or shrink dynamically.  A linked list has nodes each of which contain  contain  data and a link to next node . These nodes are dynamically allocated structures. If you need more nodes, you just need to allocate memory for these and link these nodes to the existing list. The nodes of a linked list have to be defined as self-referential structures in C. That is structures with data members and one member which is a pointer to the structure of same type.  This pointer will work as a link to next node. struct node { int data; struct node * next; //pointer to another node }...
1 comment
Read more