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Design and Implement Special Stack Data Structure | Added Space Optimized Version

  • Difficulty Level : Medium
  • Last Updated : 07 Jun, 2021

Question: Design a Data Structure SpecialStack that supports all the stack operations like push(), pop(), isEmpty(), isFull() and an additional operation getMin() which should return minimum element from the SpecialStack. All these operations of SpecialStack must be O(1). To implement SpecialStack, you should only use standard Stack data structure and no other data structure like arrays, list, . etc. 

Example: 

Consider the following SpecialStack
16  --> TOP
15
29
19
18

When getMin() is called it should 
return 15, which is the minimum 
element in the current stack. 

If we do pop two times on stack, 
the stack becomes
29  --> TOP
19
18

When getMin() is called, it should 
return 18 which is the minimum in 
the current stack.

Solution: Use two stacks: one to store actual stack elements and the other as an auxiliary stack to store minimum values. The idea is to do push() and pop() operations in such a way that the top of the auxiliary stack is always the minimum. Let us see how push() and pop() operations work.

Push(int x) // inserts an element x to Special Stack 
1) push x to the first stack (the stack with actual elements) 
2) compare x with the top element of the second stack (the auxiliary stack). Let the top element be y. 
…..a) If x is smaller than y then push x to the auxiliary stack. 
…..b) If x is greater than y then push y to the auxiliary stack.
int Pop() // removes an element from Special Stack and return the removed element 
1) pop the top element from the auxiliary stack. 
2) pop the top element from the actual stack and return it.
Step 1 is necessary to make sure that the auxiliary stack is also updated for future operations.
int getMin() // returns the minimum element from Special Stack 
1) Return the top element of the auxiliary stack.

We can see that all the above operations are O(1)
Let us see an example. Let us assume that both stacks are initially empty and 18, 19, 29, 15, and 16 are inserted to the SpecialStack. 



When we insert 18, both stacks change to following.
Actual Stack
18 <--- top     
Auxiliary Stack
18 <---- top

When 19 is inserted, both stacks change to following.
Actual Stack
19 <--- top     
18
Auxiliary Stack
18 <---- top
18

When 29 is inserted, both stacks change to following.
Actual Stack
29 <--- top     
19
18
Auxiliary Stack
18 <---- top
18
18

When 15 is inserted, both stacks change to following.
Actual Stack
15 <--- top     
29
19 
18
Auxiliary Stack
15 <---- top
18
18
18

When 16 is inserted, both stacks change to following.
Actual Stack
16 <--- top     
15
29
19 
18
Auxiliary Stack
15 <---- top
15
18
18
18

The following is the implementation for SpecialStack class. In the below implementation, SpecialStack inherits from Stack and has one Stack object min which works as an auxiliary stack.

C++




#include <iostream>
#include <stdlib.h>
 
using namespace std;
 
/* A simple stack class with
basic stack funtionalities */
class Stack {
private:
    static const int max = 100;
    int arr[max];
    int top;
 
public:
    Stack() { top = -1; }
    bool isEmpty();
    bool isFull();
    int pop();
    void push(int x);
};
 
/* Stack's member method to check
if the stack is empty */
bool Stack::isEmpty()
{
    if (top == -1)
        return true;
    return false;
}
 
/* Stack's member method to check
if the stack is full */
bool Stack::isFull()
{
    if (top == max - 1)
        return true;
    return false;
}
 
/* Stack's member method to remove
an element from it */
int Stack::pop()
{
    if (isEmpty()) {
        cout << "Stack Underflow";
        abort();
    }
    int x = arr[top];
    top--;
    return x;
}
 
/* Stack's member method to insert
an element to it */
void Stack::push(int x)
{
    if (isFull()) {
        cout << "Stack Overflow";
        abort();
    }
    top++;
    arr[top] = x;
}
 
/* A class that supports all the stack
operations and one additional
  operation getMin() that returns the
minimum element from stack at
  any time.  This class inherits from
the stack class and uses an
  auxiliary stack that holds minimum
elements */
class SpecialStack : public Stack {
    Stack min;
 
public:
    int pop();
    void push(int x);
    int getMin();
};
 
/* SpecialStack's member method to insert
 an element to it. This method
   makes sure that the min stack is also
updated with appropriate minimum
   values */
void SpecialStack::push(int x)
{
    if (isEmpty() == true) {
        Stack::push(x);
        min.push(x);
    }
    else {
        Stack::push(x);
        int y = min.pop();
        min.push(y);
        if (x < y)
            min.push(x);
        else
            min.push(y);
    }
}
 
/* SpecialStack's member method to
remove an element from it. This method
   removes top element from min stack also. */
int SpecialStack::pop()
{
    int x = Stack::pop();
    min.pop();
    return x;
}
 
/* SpecialStack's member method to get
 minimum element from it. */
int SpecialStack::getMin()
{
    int x = min.pop();
    min.push(x);
    return x;
}
 
/* Driver program to test SpecialStack
 methods */
int main()
{
    SpecialStack s;
    s.push(10);
    s.push(20);
    s.push(30);
    cout << s.getMin() << endl;
    s.push(5);
    cout << s.getMin();
    return 0;
}

Java




// Java implementation of SpecialStack
// Note : here we use Stack class for
// Stack implementation
 
import java.util.Stack;
 
/* A class that supports all the
stack operations and one additional
operation getMin() that returns
the minimum element from stack at
any time. This class inherits from
the stack class and uses an
auxiliary stack that holds minimum
 elements */
 
class SpecialStack extends Stack<Integer> {
    Stack<Integer> min = new Stack<>();
 
    /* SpecialStack's member method to
insert an element to it. This method
    makes sure that the min stack is
also updated with appropriate minimum
    values */
    void push(int x)
    {
        if (isEmpty() == true) {
            super.push(x);
            min.push(x);
        }
        else {
            super.push(x);
            int y = min.pop();
            min.push(y);
            if (x < y)
                min.push(x);
            else
                min.push(y);
        }
    }
 
    /* SpecialStack's member method to
insert an element to it. This method
    makes sure that the min stack is
also updated with appropriate minimum
    values */
    public Integer pop()
    {
        int x = super.pop();
        min.pop();
        return x;
    }
 
    /* SpecialStack's member method to get
minimum element from it. */
    int getMin()
    {
        int x = min.pop();
        min.push(x);
        return x;
    }
 
    /* Driver program to test SpecialStack
methods */
    public static void main(String[] args)
    {
        SpecialStack s = new SpecialStack();
        s.push(10);
        s.push(20);
        s.push(30);
        System.out.println(s.getMin());
        s.push(5);
        System.out.println(s.getMin());
    }
}
// This code is contributed by Sumit Ghosh

Python3




# Python3 program for the
# above approach
# A simple stack class with 
# basic stack funtionalities
class stack:
 
  def __init__(self):
 
    self.array = []
    self.top = -1
    self.max = 100 
 
  # Stack's member method to check
  # if the stack is empty
  def isEmpty(self):
 
    if self.top == -1:
      return True
    else:
      return False 
 
  # Stack's member method to check
  # if the stack is full 
  def isFull(self): 
     
    if self.top == self.max - 1:
      return True
    else:
      return False  
 
  # Stack's member method to
  # insert an element to it  
 
  def push(self, data):
 
    if self.isFull():
      print('Stack OverFlow')
      return
    else:
      self.top += 1
      self.array.append(data)    
 
  # Stack's member method to
  # remove an element from it
  def pop(self):
 
    if self.isEmpty():
      print('Stack UnderFlow')
      return
    else:
      self.top -= 1
      return self.array.pop()
 
# A class that supports all the stack 
# operations and one additional
# operation getMin() that returns the
# minimum element from stack at
# any time.  This class inherits from
# the stack class and uses an
# auxiliary stack that holds
# minimum elements 
class SpecialStack(stack):
 
  def __init__(self):
    super().__init__()
    self.Min = stack() 
 
  # SpecialStack's member method to
  # insert an element to it. This method
  # makes sure that the min stack is also
  # updated with appropriate minimum
  # values
  def push(self, x):
 
    if self.isEmpty():
      super().push(x)
      self.Min.push(x)
    else:
      super().push(x)
      y = self.Min.pop()
      self.Min.push(y)
      if x <= y:
        self.Min.push(x)
      else:
        self.Min.push(y) 
 
  # SpecialStack's member method to 
  # remove an element from it. This
  # method removes top element from
  # min stack also.
  def pop(self):
 
    x = super().pop()
    self.Min.pop()
    return
 
  # SpecialStack's member method
  # to get minimum element from it.
  def getmin(self):
 
    x = self.Min.pop()
    self.Min.push(x)
    return x
 
# Driver code
if __name__ == '__main__':
   
  s = SpecialStack()
  s.push(10)
  s.push(20)
  s.push(30)
  print(s.getmin())
  s.push(5)
  print(s.getmin())
 
# This code is contributed by rachitkatiyar99
Output
10
5

Complexity Analysis: 

  • Time Complexity: 
    1. For insert operation: O(1) (As insertion ‘push’ in a stack takes constant time)
    2. For delete operation: O(1) (As deletion ‘pop’ in a stack takes constant time)
    3. For ‘Get Min’ operation: O(1) (As we have used an auxiliary stack which has it’s top as the minimum element)
  • Auxiliary Space: O(n). 
    Use of auxiliary stack for storing values.

Space Optimized Version 
The above approach can be optimized. We can limit the number of elements in the auxiliary stack. We can push only when the incoming element of the main stack is smaller than or equal to the top of the auxiliary stack. Similarly during pop, if the pop-off element equal to the top of the auxiliary stack, remove the top element of the auxiliary stack. Following is the modified implementation of push() and pop(). 

C++




/* SpecialStack's member method to
   insert an element to it. This method
   makes sure that the min stack is
   also updated with appropriate minimum
   values */
void SpecialStack::push(int x)
{
    if (isEmpty() == true) {
        Stack::push(x);
        min.push(x);
    }
    else {
        Stack::push(x);
        int y = min.pop();
        min.push(y);
 
        /* push only when the incoming element
           of main stack is smaller
        than or equal to top of auxiliary stack */
        if (x <= y)
            min.push(x);
    }
}
 
/* SpecialStack's member method to
   remove an element from it. This method
   removes top element from min stack also. */
int SpecialStack::pop()
{
    int x = Stack::pop();
    int y = min.pop();
 
    /* Push the popped element y back
       only if it is not equal to x */
    if (y != x)
        min.push(y);
 
    return x;
}

Java




/* SpecialStack's member method to
insert an element to it. This method
makes sure that the min stack is
also updated with appropriate minimum
values */
 
void push(int x)
{
    if (isEmpty() == true) {
        super.push(x);
        min.push(x);
    }
    else {
        super.push(x);
        int y = min.pop();
        min.push(y);
 
        /* push only when the incoming
           element of main stack is smaller
        than or equal to top of auxiliary stack */
        if (x <= y)
            min.push(x);
    }
}
 
/* SpecialStack's member method to
   remove an element from it. This method
   removes top element from min stack also. */
public Integer pop()
{
    int x = super.pop();
    int y = min.pop();
 
    /* Push the popped element y back
       only if it is not equal to x */
    if (y != x)
        min.push(y);
    return x;
}
 
// This code is contributed by Sumit Ghosh

Complexity Analysis:  

  • Time Complexity: 
    1. For Insert operation: O(1) (As insertion ‘push’ in a stack takes constant time)
    2. For Delete operation: O(1) (As deletion ‘pop’ in a stack takes constant time)
    3. For ‘Get Min’ operation: O(1) (As we have used an auxiliary which has it’s top as the minimum element)
  • Auxiliary Space: O(n). 
    The complexity in the worst case is the same as above but in other cases, it will take slightly less space than the above approach as repetition is neglected.

Further optimized O(1) time complexity and O(1) space complexity solution :
The above approach can be optimized further and the solution can be made to work in O(1) time complexity and O(1) space complexity. The idea is to store min element found till current insertion) along with all the elements as a reminder of a DUMMY_VALUE, and the actual element as a multiple of the DUMMY_VALUE.
For example, while pushing an element ‘e’ into the stack, store it as (e * DUMMY_VALUE + minFoundSoFar), this way we know what was the minimum value present in the stack at the time ‘e’ was being inserted.

To pop the actual value just return e/DUMMY_VALUE and set the new minimum as (minFoundSoFar % DUMMY_VALUE).

Note: Following method will fail if we try to insert DUMMY_VALUE in the stack, so we have to make our selection of DUMMY_VALUE carefully.
Let’s say the following elements are being inserted in the stack – 3 2 6 1 8 5



d is dummy value.

s is wrapper stack

top is top element of the stack

min is the minimum value at that instant when the elements were inserted/removed

The following steps shows the current state of the above variables at any instant – 

  1. s.push(3);
    min=3 //updated min as stack here is empty 
    s = {3*d + 3}
    top = (3*d + 3)/d = 3 
     
  2. s.push(2);
    min = 2 //updated min as min > current element
    s = {3*d + 3-> 2*d + 2}
    top = (2*d + 2)/d = 2
     
  3. s.push(6);
    min = 2
    s = {3*d + 3-> 2*d + 2-> 6*d + 2}
    top = (6*d + 2)/d = 6
     
  4. s.push(1);
    min = 1 //updated min as min > current element
    s = {3*d + 3-> 2*d + 2-> 6*d + 2 -> 1*d + 1}
    top = (1*d + 1)/d = 1
     
  5. s.push(8);
    min = 1
    s = {3*d + 3-> 2*d + 2-> 6*d + 2 -> 1*d + 1 -> 8*d + 1}
    top = (8*d + 1)/d = 8
     
  6. s.push(5);
    min = 1
    s = {3*d + 3-> 2*d + 2-> 6*d + 2 -> 1*d + 1 -> 8*d + 1 -> 5*d + 1}
    top = (5*d + 1)/d = 5
     
  7. s.pop();
    s = {3*d + 3 -> 2*d + 2 -> 6*d + 2 -> 1*d + 1 -> 8*d + 1 -> 5*d + 1}
    top = (5*d + 1)/d = 5
    min = (8*d + 1)%d = 1 // min is always remainder of the second top element in stack.
     
  8. s.pop();
    s = {3*d + 3 -> 2*d + 2-> 6*d + 2 -> 1*d + 1 -> 8*d + 1}
    top = (8*d + 1)/d = 8
    min = (1*d + 1)%d = 1
     
  9. s.pop()
    s = {3*d + 3 -> 2*d + 2-> 6*d + 2 -> 1*d + 1} 
    top = (1*d + 1)/d = 1
    min = (6*d + 2)%d = 2
     
  10. s.pop()
    s = {3*d + 3-> 2*d + 2-> 6*d + 2}
    top = (6*d + 2)/d = 6
    min = (2*d + 2)%d = 2
     
  11. s.pop()
    s = {3*d + 3-> 2*d + 2}
    top = (2*d + 2)/d = 2
    min = (3*d + 3)%d = 3
     
  12. s.pop()
    s = {3*d + 3}
    top = (3*d + 3)/d = 3
    min  = -1 // since stack is now empty

C++




#include <iostream>
#include <stack>
#include <vector>
using namespace std;
 
/* A special stack having peek() pop() and
 * push() along with additional getMin() that
 * returns minimum value in a stack without
 * using extra space and all operations in O(1)
 * time.. ???? */
class SpecialStack
{
     
    // Sentinel value for min
    int min = -1; 
     
    // DEMO_VALUE
    static const int demoVal = 9999;
    stack<int> st;
 
public:
 
    void getMin()
    {
        cout << "min is: " << min << endl;
    }
 
    void push(int val)
    {
         
        // If stack is empty OR current element
        // is less than min, update min.
        if (st.empty() || val < min)
        {
            min = val;
        }
         
        // Encode the current value with
        // demoVal, combine with min and
        // insert into stack
        st.push(val * demoVal + min);
        cout << "pushed: " << val << endl;
    }
 
    int pop()
    {   
        // if stack is empty return -1;
        if ( st.empty() ) {
          cout << "stack underflow" << endl ;
          return -1;
        }
       
        int val = st.top();
        st.pop();
         
        // If stack is empty, there would
        // be no min value present, so
        // make min as -1
        if (!st.empty())
            min = st.top() % demoVal;
        else
            min = -1;
             
        cout << "popped: " << val / demoVal << endl;
         
        // Decode actual value from
        // encoded value
        return val / demoVal;
    }
 
    int peek()
    {
         
        // Decode actual value
        // from encoded value
        return st.top() / demoVal;
    }
};
 
// Driver Code
int main()
{
    SpecialStack s;
 
    vector<int> arr = { 3, 2, 6, 1, 8, 5, 5, 5, 5 };
 
    for(int i = 0; i < arr.size(); i++)
    {
        s.push(arr[i]);
        s.getMin();
    }
     
    cout << endl;
    for(int i = 0; i < arr.size(); i++)
    {
        s.pop();
        s.getMin();
    }
    return 0;
}
 
// This code is contributed by scisaif

Java




import java.util.Stack;
 
/* A special stack having peek() pop() and push() along with
 * additional getMin() that returns minimum value in a stack
 * without using extra space and all operations in O(1)
 * time.. 🙂
 * */
public class SpecialStack {
 
    int min = -1; // sentinel value for min
    static int demoVal = 9999; // DEMO_VALUE
    Stack<Integer> st = new Stack<Integer>();
 
    void getMin() { System.out.println("min is: " + min); }
 
    void push(int val)
    {
        // if stack is empty OR current element is less than
        // min, update min..
        if (st.isEmpty() || val < min) {
            min = val;
        }
 
        st.push(val * demoVal
                + min); // encode the current value with
                        // demoVal, combine with min and
                        // insert into stack
        System.out.println("pushed: " + val);
    }
 
    int pop()
    {   
        // if stack is empty return -1;
        if (st.isEmpty() ) {
             System.out.println("stack underflow");
               return -1;
          }
       
        int val = st.pop();
 
        if (!st.isEmpty()) // if stack is empty, there would
                           // be no min value present, so
                           // make min as -1
            min = st.peek() % demoVal;
        else
            min = -1;
        System.out.println("popped: " + val / demoVal);
        return val / demoVal; // decode actual value from
                              // encoded value
    }
 
    int peek()
    {
        return st.peek() / demoVal; // decode actual value
                                    // from encoded value
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        SpecialStack s = new SpecialStack();
 
        int[] arr = { 3, 2, 6, 1, 8, 5, 5, 5, 5 };
 
        for (int i = 0; i < arr.length; i++) {
            s.push(arr[i]);
            s.getMin();
        }
        System.out.println();
        for (int i = 0; i < arr.length; i++) {
            s.pop();
            s.getMin();
        }
    }
}
Output
pushed: 3
min is: 3
pushed: 2
min is: 2
pushed: 6
min is: 2
pushed: 1
min is: 1
pushed: 8
min is: 1
pushed: 5
min is: 1
pushed: 5
min is: 1
pushed: 5
min is: 1
pushed: 5
min is: 1

popped: 5
min is: 1
popped: 5
min is: 1
popped: 5
min is: 1
popped: 5
min is: 1
popped: 8
min is: 1
popped: 1
min is: 2
popped: 6
min is: 2
popped: 2
min is: 3
popped: 3
min is: -1

Complexity Analysis:  

For push() operation: O(1) (As insertion ‘push’ in a stack takes constant time)
For pop() operation: O(1) (As pop operation in a stack takes constant time)

For ‘Get Min’ operation: O(1) (As we have maintained min variable throughout the code)

Auxiliary Space: O(1). No extra space is used.

https://youtu.be/1Ld7gbW1oHc
Design a stack that supports getMin() in O(1) time and O(1) extra space
Thanks to @Venki, @swarup, and @Jing Huang for their inputs.
Please write comments if you find the above code incorrect, or find other ways to solve the same problem.

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