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++
// 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 Codeint 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
// 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 Codepublic 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(); }} |
Python 3
# Class to make a Nodeclass 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 classif __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#
// 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 Codepublic 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 |
Javascript
// 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 Codelet s = new MyStack;// Function callss.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 |
Output
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++
// 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 codeint 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
// Java program for the above approachimport 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 codeclass 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 |
Python3
# Python program for the above approachclass 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].mins = MinStack()# Function callss.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#
// C# program for the above approachusing 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 codepublic 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 |
Javascript
// JS program for the above approachfunction 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 codelet s = new MinStack();// Function callss.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 |
Output
-4 0 -4 -1
Related Article: An approach that uses O(1) time and O(n) extra space is discussed here.
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