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 have a time and space complexity of O(1).
Note: To implement SpecialStack, you should only use standard Stack data structure and no other data structure like arrays, lists, etc
Example:
Input: Consider the following SpecialStack
16 –> TOP
15
29
19
18When 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
18When getMin() is called, it should return 18
which is the minimum in the current stack.
Approach: To solve the problem follow the below idea:
We define a variable minEle that stores the current minimum element in the stack. Now the interesting part is, how to handle the case when the minimum element is removed. To handle this, we push “2x – minEle” into the stack instead of x so that the previous minimum element can be retrieved using the current minEle and its value stored in the stack
Follow the given steps to implement the stack operations:
Push(x): Insert x at the top of the stack
- If the stack is empty, insert x into the stack and make minEle equal to x.
- If the stack is not empty, compare x with minEle. Two cases arise:
- If x is greater than or equal to minEle, simply insert x.
- If x is less than minEle, insert (2*x – minEle) into the stack and make minEle equal to x.
For example, let the previous minEle be 3. Now we want to insert 2. We update minEle as 2 and insert 2*2 – 3 = 1 into the stack
Pop(): Removes an element from the top of the stack
- Remove the element from the top. Let the removed element be y. Two cases arise:
- If y is greater than or equal to minEle, the minimum element in the stack is still minEle.
- If y is less than minEle, the minimum element now becomes (2*minEle – y), so update (minEle = 2*minEle – y). This is where we retrieve the previous minimum from the current minimum and its value in the stack.
For example, let the element to be removed be 1 and minEle be 2. We remove 1 and update minEle as 2*2 – 1 = 3
Important Points:
- Stack doesn’t hold the actual value of an element if it is minimum so far.
- The actual minimum element is always stored in the minEle variable
Below is the illustration of the above approach:
Push(x)
- Number to be Inserted: 3, Stack is empty, so insert 3 into stack and minEle = 3.
- Number to be Inserted: 5, Stack is not empty, 5> minEle, insert 5 into stack and minEle = 3.
- Number to be Inserted: 2, Stack is not empty, 2< minEle, insert (2*2-3 = 1) into stack and minEle = 2.
- Number to be Inserted: 1, Stack is not empty, 1< minEle, insert (2*1-2 = 0) into stack and minEle = 1.
- Number to be Inserted: 1, Stack is not empty, 1 = minEle, insert 1 into stack and minEle = 1.
- Number to be Inserted: -1, Stack is not empty, -1 < minEle, insert (2*-1 – 1 = -3) into stack and minEle = -1.
Pop()
- Initially the minimum element minEle in the stack is -1.
- Number removed: -3, Since -3 is less than the minimum element the original number being removed is minEle which is -1, and the new minEle = 2*-1 – (-3) = 1
- Number removed: 1, 1 == minEle, so number removed is 1 and minEle is still equal to 1.
- Number removed: 0, 0< minEle, original number is minEle which is 1 and new minEle = 2*1 – 0 = 2.
- Number removed: 1, 1< minEle, original number is minEle which is 2 and new minEle = 2*2 – 1 = 3.
- Number removed: 5, 5> minEle, original number is 5 and minEle is still 3
Below is the implementation of the above approach:
// C++ program to implement a stack that supports // getMinimum() in O(1) time and O(1) extra space. #include <bits/stdc++.h> using namespace std;
// A user defined stack that supports getMin() in // addition to push() and pop() struct MyStack {
stack< int > s;
int minEle;
// Prints minimum element of MyStack
void getMin()
{
if (s.empty())
cout << "Stack is empty\n" ;
// variable minEle stores the minimum element
// in the stack.
else
cout << "Minimum Element in the stack is: "
<< minEle << "\n" ;
}
// Prints top element of MyStack
void peek()
{
if (s.empty()) {
cout << "Stack is empty " ;
return ;
}
int t = s.top(); // Top element.
cout << "Top Most Element is: " ;
// If t < minEle means minEle stores
// value of t.
(t < minEle) ? cout << minEle : cout << t;
}
// Remove the top element from MyStack
void pop()
{
if (s.empty()) {
cout << "Stack is empty\n" ;
return ;
}
cout << "Top Most Element Removed: " ;
int t = s.top();
s.pop();
// Minimum will change as the minimum element
// of the stack is being removed.
if (t < minEle) {
cout << minEle << "\n" ;
minEle = 2 * minEle - t;
}
else
cout << t << "\n" ;
}
// Removes top element from MyStack
void push( int x)
{
// Insert new number into the stack
if (s.empty()) {
minEle = x;
s.push(x);
cout << "Number Inserted: " << x << "\n" ;
return ;
}
// If new number is less than minEle
else if (x < minEle) {
s.push(2 * x - minEle);
minEle = x;
}
else
s.push(x);
cout << "Number Inserted: " << x << "\n" ;
}
}; // Driver Code int main()
{ MyStack s;
// Function calls
s.push(3);
s.push(5);
s.getMin();
s.push(2);
s.push(1);
s.getMin();
s.pop();
s.getMin();
s.pop();
s.peek();
return 0;
} |
// Java program to implement a stack that supports // getMinimum() in O(1) time and O(1) extra space. import java.util.*;
// A user defined stack that supports getMin() in // addition to push() and pop() class MyStack {
Stack<Integer> s;
Integer minEle;
// Constructor
MyStack() { s = new Stack<Integer>(); }
// Prints minimum element of MyStack
void getMin()
{
// Get the minimum number in the entire stack
if (s.isEmpty())
System.out.println( "Stack is empty" );
// variable minEle stores the minimum element
// in the stack.
else
System.out.println( "Minimum Element in the "
+ " stack is: " + minEle);
}
// prints top element of MyStack
void peek()
{
if (s.isEmpty()) {
System.out.println( "Stack is empty " );
return ;
}
Integer t = s.peek(); // Top element.
System.out.print( "Top Most Element is: " );
// If t < minEle means minEle stores
// value of t.
if (t < minEle)
System.out.println(minEle);
else
System.out.println(t);
}
// Removes the top element from MyStack
void pop()
{
if (s.isEmpty()) {
System.out.println( "Stack is empty" );
return ;
}
System.out.print( "Top Most Element Removed: " );
Integer t = s.pop();
// Minimum will change as the minimum element
// of the stack is being removed.
if (t < minEle) {
System.out.println(minEle);
minEle = 2 * minEle - t;
}
else
System.out.println(t);
}
// Insert new number into MyStack
void push(Integer x)
{
if (s.isEmpty()) {
minEle = x;
s.push(x);
System.out.println( "Number Inserted: " + x);
return ;
}
// If new number is less than original minEle
if (x < minEle) {
s.push( 2 * x - minEle);
minEle = x;
}
else
s.push(x);
System.out.println( "Number Inserted: " + x);
}
}; // Driver Code public class Main {
public static void main(String[] args)
{
MyStack s = new MyStack();
// Function calls
s.push( 3 );
s.push( 5 );
s.getMin();
s.push( 2 );
s.push( 1 );
s.getMin();
s.pop();
s.getMin();
s.pop();
s.peek();
}
} |
# Class to make a Node class Node:
# Constructor which assign argument to nade's value
def __init__( self , value):
self .value = value
self . next = None
# This method returns the string representation of the object.
def __str__( self ):
return "Node({})" . format ( self .value)
# __repr__ is same as __str__
__repr__ = __str__
class Stack:
# Stack Constructor initialise top of stack and counter.
def __init__( self ):
self .top = None
self .count = 0
self .minimum = None
# This method returns the string representation of the object (stack).
def __str__( self ):
temp = self .top
out = []
while temp:
out.append( str (temp.value))
temp = temp. next
out = '\n' .join(out)
return ( 'Top {} \n\nStack :\n{}' . format ( self .top, out))
# __repr__ is same as __str__
__repr__ = __str__
# This method is used to get minimum element of stack
def getMin( self ):
if self .top is None :
return "Stack is empty"
else :
print ( "Minimum Element in the stack is: {}" . format ( self .minimum))
# Method to check if Stack is Empty or not
def isEmpty( self ):
# If top equals to None then stack is empty
if self .top = = None :
return True
else :
# If top not equal to None then stack is empty
return False
# This method returns length of stack
def __len__( self ):
self .count = 0
tempNode = self .top
while tempNode:
tempNode = tempNode. next
self .count + = 1
return self .count
# This method returns top of stack
def peek( self ):
if self .top is None :
print ( "Stack is empty" )
else :
if self .top.value < self .minimum:
print ( "Top Most Element is: {}" . format ( self .minimum))
else :
print ( "Top Most Element is: {}" . format ( self .top.value))
# This method is used to add node to stack
def push( self , value):
if self .top is None :
self .top = Node(value)
self .minimum = value
elif value < self .minimum:
temp = ( 2 * value) - self .minimum
new_node = Node(temp)
new_node. next = self .top
self .top = new_node
self .minimum = value
else :
new_node = Node(value)
new_node. next = self .top
self .top = new_node
print ( "Number Inserted: {}" . format (value))
# This method is used to pop top of stack
def pop( self ):
if self .top is None :
print ( "Stack is empty" )
else :
removedNode = self .top.value
self .top = self .top. next
if removedNode < self .minimum:
print ( "Top Most Element Removed :{} " . format ( self .minimum))
self .minimum = (( 2 * self .minimum) - removedNode)
else :
print ( "Top Most Element Removed : {}" . format (removedNode))
# Driver program to test above class if __name__ = = '__main__' :
stack = Stack()
# Function calls
stack.push( 3 )
stack.push( 5 )
stack.getMin()
stack.push( 2 )
stack.push( 1 )
stack.getMin()
stack.pop()
stack.getMin()
stack.pop()
stack.peek()
# This code is contributed by Blinkii |
// C# program to implement a stack // that supports getMinimum() in O(1) // time and O(1) extra space. using System;
using System.Collections;
// A user defined stack that supports // getMin() in addition to Push() and Pop() public class MyStack {
public Stack s;
public int minEle;
// Constructor
public MyStack() { s = new Stack(); }
// Prints minimum element of MyStack
public void getMin()
{
// Get the minimum number
// in the entire stack
if (s.Count == 0)
Console.WriteLine( "Stack is empty" );
// variable minEle stores the minimum
// element in the stack.
else
Console.WriteLine( "Minimum Element in the "
+ " stack is: " + minEle);
}
// prints top element of MyStack
public void Peek()
{
if (s.Count == 0) {
Console.WriteLine( "Stack is empty " );
return ;
}
int t = ( int )s.Peek(); // Top element.
Console.Write( "Top Most Element is: " );
// If t < minEle means minEle stores
// value of t.
if (t < minEle)
Console.WriteLine(minEle);
else
Console.WriteLine(t);
}
// Removes the top element from MyStack
public void Pop()
{
if (s.Count == 0) {
Console.WriteLine( "Stack is empty" );
return ;
}
Console.Write( "Top Most Element Removed: " );
int t = ( int )s.Pop();
// Minimum will change as the minimum element
// of the stack is being removed.
if (t < minEle) {
Console.WriteLine(minEle);
minEle = 2 * minEle - t;
}
else
Console.WriteLine(t);
}
// Insert new number into MyStack
public void Push( int x)
{
if (s.Count == 0) {
minEle = x;
s.Push(x);
Console.WriteLine( "Number Inserted: " + x);
return ;
}
// If new number is less than original minEle
if (x < minEle) {
s.Push(2 * x - minEle);
minEle = x;
}
else
s.Push(x);
Console.WriteLine( "Number Inserted: " + x);
}
} // Driver Code public class main {
public static void Main(String[] args)
{
MyStack s = new MyStack();
// Function calls
s.Push(3);
s.Push(5);
s.getMin();
s.Push(2);
s.Push(1);
s.getMin();
s.Pop();
s.getMin();
s.Pop();
s.Peek();
}
} // This code is contributed by Arnab Kundu |
// JS program to implement a stack that supports // getMinimum() in O(1) time and O(1) extra space. // A user defined stack that supports getMin() in // addition to push() and pop() class MyStack { constructor() {
this .s = [];
this .minEle;
}
// Prints minimum element of MyStack
getMin() {
if ( this .s.length == 0)
console.log( "Stack is empty" );
// variable minEle stores the minimum element
// in the stack.
else
console.log( "Minimum Element in the stack is: " , this .minEle);
}
// Prints top element of MyStack
peek() {
if ( this .s.length == 0) {
console.log( "Stack is empty " );
return ;
}
let t = this .s[0]; // Top element.
console.log( "Top Most Element is: " );
// If t < minEle means minEle stores
// value of t.
(t < this .minEle) ? console.log( this .minEle) : console.log(t);
}
// Remove the top element from MyStack
pop() {
if ( this .s.length == 0) {
console.log( "Stack is empty " );
return ;
}
console.log( "Top Most Element Removed: " );
let t = this .s[0]; // Top element.
this .s.shift();
// Minimum will change as the minimum element
// of the stack is being removed.
if (t < this .minEle) {
console.log( this .minEle);
this .minEle = (2 * this .minEle) - t;
}
else
console.log(t);
}
// Removes top element from MyStack
push(x) {
// Insert new number into the stack
if ( this .s.length == 0) {
this .minEle = x;
this .s.unshift(x);
console.log( "Number Inserted: " , x);
return ;
}
// If new number is less than minEle
else if (x < this .minEle) {
this .s.unshift(2 * x - this .minEle);
this .minEle = x;
}
else
this .s.unshift(x);
console.log( "Number Inserted: " , x);
}
}; // Driver Code let s = new MyStack;
// Function calls s.push(3); s.push(5); s.getMin(); s.push(2); s.push(1); s.getMin(); s.pop(); s.getMin(); s.pop(); s.peek(); // This code is contributed by adityamaharshi21 |
Number Inserted: 3 Number Inserted: 5 Minimum Element in the stack is: 3 Number Inserted: 2 Number Inserted: 1 Minimum Element in the stack is: 1 Top Most Element Removed: 1 Minimum Element in the stack is: 2 Top Most Element Removed: 2 Top Most Element is: 5
Time Complexity: O(1)
Auxiliary Space: O(1)
How does this approach work?
When the element to be inserted is less than minEle, we insert “2x – minEle”. The important thing to note is, that 2x – minEle will always be less than x (proved below), i.e., new minEle and while popping out this element we will see that something unusual has happened as the popped element is less than the minEle. So we will be updating minEle.
How 2*x – minEle is less than x in push()?
x < minEle which means x – minEle < 0
// Adding x on both sides
x – minEle + x < 0 + x
2*x – minEle < x
We can conclude 2*x – minEle < new minEle
While popping out, if we find the element(y) less than the current minEle, we find the new minEle = 2*minEle – y
How previous minimum element, prevMinEle is, 2*minEle – y
in pop() is y the popped element?// We pushed y as 2x – prevMinEle. Here
// prevMinEle is minEle before y was insertedy = 2*x – prevMinEle
// Value of minEle was made equal to x
minEle = x .new minEle = 2 * minEle – y
= 2*x – (2*x – prevMinEle)
= prevMinEle // This is what we wanted
Design a stack that supports getMin() in O(1) time and O(1) extra space by creating a MinStack class:
To solve the problem follow the below idea:
Create a class node that has two variables Val and min. Val will store the actual value that we are going to insert in the stack, whereas min will store the min value so far seen up to that node
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
int mini( int a, int b) { return a > b ? b : a; }
class MinStack {
public :
stack<pair< int , int > > s;
void push( int element)
{
/* new max will be given no. if stack is empty else
we compare given no. to max at current top of
stack*/
int new_min = s.empty()
? element
: mini(element, s.top().second);
// we push the pair of given_element,new_min in s
s.push({ element, new_min });
}
int pop()
{
int popped;
if (!s.empty()) {
// popped has popped number
popped = s.top().first;
s.pop();
}
else {
// print a message or throw exception etc
}
return popped;
}
int minimum()
{
int min_elem = s.top().second;
return min_elem;
}
}; // Driver code int main()
{ MinStack s;
// Function calls
s.push(-1);
s.push(10);
s.push(-4);
s.push(0);
cout << s.minimum() << endl;
cout << s.pop() << endl;
cout << s.pop() << endl;
cout << s.minimum();
return 0;
} // this code is contributed by Apoorv Shrivastava - VITB |
// Java program for the above approach import java.io.*;
import java.util.*;
class MinStack {
Stack<Node> s;
class Node {
int val;
int min;
public Node( int val, int min)
{
this .val = val;
this .min = min;
}
}
/** initialize your data structure here. */
public MinStack() { this .s = new Stack<Node>(); }
public void push( int x)
{
if (s.isEmpty()) {
this .s.push( new Node(x, x));
}
else {
int min = Math.min( this .s.peek().min, x);
this .s.push( new Node(x, min));
}
}
public int pop() { return this .s.pop().val; }
public int top() { return this .s.peek().val; }
public int getMin() { return this .s.peek().min; }
} // Driver code class GFG {
public static void main(String[] args)
{
MinStack s = new MinStack();
// Function calls
s.push(- 1 );
s.push( 10 );
s.push(- 4 );
s.push( 0 );
System.out.println(s.getMin());
System.out.println(s.pop());
System.out.println(s.pop());
System.out.println(s.getMin());
}
} // this code is contributed by gireeshgudaparthi |
# Python program for the above approach class MinStack:
# initialize your data structure here.
def __init__( self ):
self .s = []
class Node:
def __init__( self , val, Min ):
self .val = val
self . min = Min
def push( self , x):
if not self .s:
self .s.append( self .Node(x, x))
else :
Min = min ( self .s[ - 1 ]. min , x)
self .s.append( self .Node(x, Min ))
def pop( self ):
return self .s.pop().val
def top( self ):
return self .s[ - 1 ].val
def getMin( self ):
return self .s[ - 1 ]. min
s = MinStack()
# Function calls s.push( - 1 )
s.push( 10 )
s.push( - 4 )
s.push( 0 )
print (s.getMin())
print (s.pop())
print (s.pop())
print (s.getMin())
# This code is contributed by lokesh |
// C# program for the above approach using System;
using System.Collections.Generic;
public class MinStack {
Stack<Node> s;
public class Node {
public int val;
public int min;
public Node( int val, int min)
{
this .val = val;
this .min = min;
}
}
/** initialize your data structure here. */
public MinStack() { this .s = new Stack<Node>(); }
public void push( int x)
{
if (s.Count == 0) {
this .s.Push( new Node(x, x));
}
else {
int min = Math.Min( this .s.Peek().min, x);
this .s.Push( new Node(x, min));
}
}
public int pop() { return this .s.Pop().val; }
public int top() { return this .s.Peek().val; }
public int getMin() { return this .s.Peek().min; }
} // Driver code public class GFG {
public static void Main(String[] args)
{
MinStack s = new MinStack();
// Function call
s.push(-1);
s.push(10);
s.push(-4);
s.push(0);
Console.WriteLine(s.getMin());
Console.WriteLine(s.pop());
Console.WriteLine(s.pop());
Console.WriteLine(s.getMin());
}
} // This code contributed by gauravrajput1 |
// JS program for the above approach function mini(a,b) { return a > b ? b : a; }
class MinStack { constructor()
{
this .s = new Array();
}
pushE(element)
{
/* new max will be given no. if stack is empty else
we compare given no. to max at current top of
stack*/
let new_min = 0;
if ( this .s.length==0)
new_min = element;
else
new_min = mini(element, this .s[ this .s.length-1].second);
// we push the pair of given_element,new_min in s
this .s.push({
first:element,
second:new_min
});
}
popE()
{
let popped;
if ( this .s.length>0) {
// popped has popped number
popped = this .s[ this .s.length-1].first;
this .s.pop();
}
else {
// print a message or throw exception etc
}
return popped;
}
minimum()
{
let min_elem = this .s[ this .s.length-1].second;
return min_elem;
}
}; // Driver code let s = new MinStack();
// Function calls s.pushE(-1); s.pushE(10); s.pushE(-4); s.pushE(0); console.log(s.minimum()); console.log(s.popE()); console.log(s.popE()); console.log(s.minimum()); // This code is contributed by adityamaharshi21 |
-4 0 -4 -1
Related Article: An approach that uses O(1) time and O(n) extra space is discussed here.