Design a stack that supports getMin() in O(1) time
Last Updated :
26 Apr, 2025
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Design a Data Structure SpecialStack that supports all the stack operations like push(), pop(), peek() and an additional operation getMin() which should return minimum element from the SpecialStack. All these operations of SpecialStack must have a time complexity of O(1).
Example:
Input: queries = [push(2), push(3), peek(), pop(), getMin(), push(1), getMin()]
Output: [3, 2, 1]
Explanation:
push(2): Stack is [2]
push(3): Stack is [2, 3]
peek(): Top element is 3
pop(): Removes 3, stack is [2]
getMin(): Minimum element is 2
push(1): Stack is [2, 1]
getMin(): Minimum element is 1Input: queries = [push(10), getMin(), push(5), getMin(), pop()]
Output: [10, 5]
Explanation:
push(10): Stack is [10]
getMin(): Minimum element is 10
push(5): Stack is [10, 5]
getMin(): Minimum element is 5
pop(): Removes 5, stack is [10]
Table of Content
Using an Auxiliary Stack - O(n) Time and O(n) Space
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)
- push x to the first stack (the stack with actual elements)
- compare x with the top element of the second stack (the auxiliary stack). Let the top element be y.
- If x is smaller than y then push x to the auxiliary stack.
- If x is greater than y then push y to the auxiliary stack.
int Pop()
- pop the top element from the auxiliary stack.
- 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()
- Return the top element of the auxiliary stack.
#include <iostream>
#include <stack>
using namespace std;
class SpecialStack {
stack<int> s;
stack<int> minStack;
public:
void push(int x) {
s.push(x);
// If the minStack is empty or the new element is smaller than
// the top of minStack, push it onto minStack
if (minStack.empty() || x <= minStack.top()) {
minStack.push(x);
} else {
// Otherwise, push the top element of minStack
// again to keep the minimum unchanged
minStack.push(minStack.top());
}
}
// Pop the top element from the stack
int pop() {
if (s.empty()) {
return -1;
}
// Pop from both stacks
int poppedElement = s.top();
s.pop();
minStack.pop();
return poppedElement;
}
// Return the top element of the stack without removing it
int peek() {
if (s.empty()) {
return -1;
}
return s.top();
}
// Check if the stack is empty
bool isEmpty() {
return s.empty();
}
// Get the minimum element in the stack
int getMin() {
if (minStack.empty()) {
return -1;
}
return minStack.top();
}
};
int main() {
SpecialStack stack;
stack.push(18);
stack.push(19);
stack.push(29);
stack.push(15);
stack.push(16);
cout << stack.getMin() << endl;
return 0;
}
import java.util.Stack;
class SpecialStack {
Stack<Integer> s = new Stack<>();
Stack<Integer> minStack = new Stack<>();
public void push(int x) {
s.push(x);
// If the minStack is empty or the new element is smaller than
// the top of minStack, push it onto minStack
if (minStack.isEmpty() || x <= minStack.peek()) {
minStack.push(x);
} else {
// Otherwise, push the top element of minStack
// again to keep the minimum unchanged
minStack.push(minStack.peek());
}
}
// Pop the top element from the stack
public int pop() {
if (s.isEmpty()) {
return -1;
}
// Pop from both stacks
int poppedElement = s.pop();
minStack.pop();
return poppedElement;
}
// Return the top element of the stack without removing it
public int peek() {
if (s.isEmpty()) {
return -1;
}
return s.peek();
}
// Check if the stack is empty
public boolean isEmpty() {
return s.isEmpty();
}
// Get the minimum element in the stack
public int getMin() {
if (minStack.isEmpty()) {
return -1;
}
return minStack.peek();
}
}
public class GfG {
public static void main(String[] args) {
SpecialStack stack = new SpecialStack();
stack.push(18);
stack.push(19);
stack.push(29);
stack.push(15);
stack.push(16);
System.out.println(stack.getMin());
}
}
class specialStack:
def __init__(self):
self.s = []
self.minStack = []
def push(self, x):
self.s.append(x)
# If the minStack is empty or the new element is smaller than
# the top of minStack, push it onto minStack
if not self.minStack or x <= self.minStack[-1]:
self.minStack.append(x)
else:
# Otherwise, push the top element of minStack
# again to keep the minimum unchanged
self.minStack.append(self.minStack[-1])
# Pop the top element from the stack
def pop(self):
if not self.s:
return -1
# Pop from both stacks
poppedElement = self.s.pop()
self.minStack.pop()
return poppedElement
# Return the top element of the stack without removing it
def peek(self):
if not self.s:
return -1
return self.s[-1]
# Check if the stack is empty
def isEmpty(self):
return len(self.s) == 0
# Get the minimum element in the stack
def getMin(self):
if not self.minStack:
return -1
return self.minStack[-1]
if __name__ == '__main__':
stack = specialStack()
stack.push(18)
stack.push(19)
stack.push(29)
stack.push(15)
stack.push(16)
print(stack.getMin())
using System;
using System.Collections.Generic;
class SpecialStack {
Stack<int> s = new Stack<int>();
Stack<int> minStack = new Stack<int>();
public void Push(int x) {
s.Push(x);
// If the minStack is empty or the new element is smaller than
// the top of minStack, push it onto minStack
if (minStack.Count == 0 || x <= minStack.Peek()) {
minStack.Push(x);
} else {
// Otherwise, push the top element of minStack
// again to keep the minimum unchanged
minStack.Push(minStack.Peek());
}
}
// Pop the top element from the stack
public int Pop() {
if (s.Count == 0) {
return -1;
}
// Pop from both stacks
int poppedElement = s.Pop();
minStack.Pop();
return poppedElement;
}
// Return the top element of the stack without removing it
public int Peek() {
if (s.Count == 0) {
return -1;
}
return s.Peek();
}
// Check if the stack is empty
public bool IsEmpty() {
return s.Count == 0;
}
// Get the minimum element in the stack
public int GetMin() {
if (minStack.Count == 0) {
return -1;
}
return minStack.Peek();
}
}
class GfG {
static void Main() {
SpecialStack stack = new SpecialStack();
stack.Push(18);
stack.Push(19);
stack.Push(29);
stack.Push(15);
stack.Push(16);
Console.WriteLine(stack.GetMin());
}
}
class specialStack {
constructor() {
this.s = [];
this.minStack = [];
}
push(x) {
this.s.push(x);
// If the minStack is empty or the new element is smaller than
// the top of minStack, push it onto minStack
if (this.minStack.length === 0 || x <= this.minStack[this.minStack.length - 1]) {
this.minStack.push(x);
} else {
// Otherwise, push the top element of minStack
// again to keep the minimum unchanged
this.minStack.push(this.minStack[this.minStack.length - 1]);
}
}
// Pop the top element from the stack
pop() {
if (this.s.length === 0) {
return -1;
}
// Pop from both stacks
const poppedElement = this.s.pop();
this.minStack.pop();
return poppedElement;
}
// Return the top element of the stack without removing it
peek() {
if (this.s.length === 0) {
return -1;
}
return this.s[this.s.length - 1];
}
// Check if the stack is empty
isEmpty() {
return this.s.length === 0;
}
// Get the minimum element in the stack
getMin() {
if (this.minStack.length === 0) {
return -1;
}
return this.minStack[this.minStack.length - 1];
}
}
const stack = new specialStack();
stack.push(18);
stack.push(19);
stack.push(29);
stack.push(15);
stack.push(16);
console.log(stack.getMin());
Output
15
Time Complexity:
- For insert operation: O(1) (As insertion 'push' in a stack takes constant time)
- For delete operation: O(1) (As deletion 'pop' in a stack takes constant time)
- 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)
Using a Pair in Stack - O(1) Time and O(n) Space
This approach uses a stack where each element is stored as a pair: the element itself and the minimum value up to that point. When an element is pushed, the minimum is updated. The getMin() function directly accesses the minimum value from the top of the stack in constant time, ensuring that both push(), pop(), and getMin() operations are O(1). This approach efficiently tracks the minimum value without needing to traverse the stack.
// C++ program to implement a stack that supports
// all operations in O(1) time and O(n) extra space.
#include <iostream>
#include <stack>
using namespace std;
// A user defined stack that supports getMin() in
// addition to push(), pop() and peek()
class SpecialStack {
private:
stack<pair<int, int> > s;
public:
SpecialStack() {
}
// Add an element to the top of Stack
void push(int x) {
int newMin = s.empty() ? x : min(x, s.top().second);
// we push the pair of given element and newMin into stack
s.push({ x, newMin });
}
// Remove the top element from the Stack
void pop() {
if (!s.empty()) {
s.pop();
}
}
// Returns top element of the Stack
int peek() {
if (s.empty()) {
return -1;
}
int top = s.top().first;
return top;
}
// Finds minimum element of Stack
int getMin() {
if (s.empty()) {
return -1;
}
int mn = s.top().second;
return mn;
}
};
int main() {
SpecialStack ss;
// Function calls
ss.push(2);
ss.push(3);
cout << ss.peek() << " ";
ss.pop();
cout << ss.getMin() << " ";
ss.push(1);
cout << ss.getMin() << " ";
}
// Java program to implement a stack that supports
// all operations in O(1) time and O(n) extra space.
import java.util.Stack;
class SpecialStack {
private Stack<int[]> s;
public SpecialStack() {
s = new Stack<>();
}
// Add an element to the top of Stack
public void push(int x) {
int newMin = s.isEmpty() ? x : Math.min(x, s.peek()[1]);
s.push(new int[]{x, newMin});
}
// Remove the top element from the Stack
public void pop() {
if (!s.isEmpty()) {
s.pop();
}
}
// Returns top element of the Stack
public int peek() {
return s.isEmpty() ? -1 : s.peek()[0];
}
// Finds minimum element of Stack
public int getMin() {
return s.isEmpty() ? -1 : s.peek()[1];
}
public static void main(String[] args) {
SpecialStack ss = new SpecialStack();
// Function calls
ss.push(2);
ss.push(3);
System.out.print(ss.peek() + " ");
ss.pop();
System.out.print(ss.getMin() + " ");
ss.push(1);
System.out.print(ss.getMin() + " ");
}
}
# Python program to implement a stack that supports
# all operations in O(1) time and O(n) extra space.
class SpecialStack:
def __init__(self):
self.s = []
# Add an element to the top of Stack
def push(self, x):
newMin = x if not self.s else min(x, self.s[-1][1])
self.s.append((x, newMin))
# Remove the top element from the Stack
def pop(self):
if self.s:
self.s.pop()
# Returns top element of the Stack
def peek(self):
return -1 if not self.s else self.s[-1][0]
# Finds minimum element of Stack
def getMin(self):
return -1 if not self.s else self.s[-1][1]
if __name__ == "__main__":
ss = SpecialStack()
# Function calls
ss.push(2)
ss.push(3)
print(ss.peek(), end=" ")
ss.pop()
print(ss.getMin(), end=" ")
ss.push(1)
print(ss.getMin(), end=" ")
// C# program to implement a stack that supports
// all operations in O(1) time and O(n) extra space.
using System;
using System.Collections.Generic;
class SpecialStack {
private Stack<(int, int)> s;
public SpecialStack() {
s = new Stack<(int, int)>();
}
// Add an element to the top of Stack
public void Push(int x) {
int newMin = s.Count == 0 ? x : Math.Min(x, s.Peek().Item2);
s.Push((x, newMin));
}
// Remove the top element from the Stack
public void Pop() {
if (s.Count > 0) {
s.Pop();
}
}
// Returns top element of the Stack
public int Peek() {
return s.Count == 0 ? -1 : s.Peek().Item1;
}
// Finds minimum element of Stack
public int GetMin() {
return s.Count == 0 ? -1 : s.Peek().Item2;
}
public static void Main() {
SpecialStack ss = new SpecialStack();
// Function calls
ss.Push(2);
ss.Push(3);
Console.Write(ss.Peek() + " ");
ss.Pop();
Console.Write(ss.GetMin() + " ");
ss.Push(1);
Console.Write(ss.GetMin() + " ");
}
}
// JavaScript program to implement a stack that supports
// all operations in O(1) time and O(n) extra space.
class SpecialStack {
constructor() {
this.s = [];
}
// Add an element to the top of Stack
push(x) {
let newMin = this.s.length === 0 ? x :
Math.min(x, this.s[this.s.length - 1][1]);
this.s.push([x, newMin]);
}
// Remove the top element from the Stack
pop() {
if (this.s.length > 0) {
this.s.pop();
}
}
// Returns top element of the Stack
peek() {
return this.s.length === 0 ? -1 : this.s[this.s.length - 1][0];
}
// Finds minimum element of Stack
getMin() {
return this.s.length === 0 ? -1 : this.s[this.s.length - 1][1];
}
}
// Driver Code
const ss = new SpecialStack();
ss.push(2);
ss.push(3);
console.log(ss.peek() + " ");
ss.pop();
console.log(ss.getMin() + " ");
ss.push(1);
console.log(ss.getMin() + " ");
Output
-4 0 -4 -1
Without Extra Space- O(1) Time and O(1) Space
The idea is to use a variable minEle to track the minimum element in the stack. Instead of storing the actual value of minEle in the stack, we store a modified value when pushing an element smaller than minEle. This allows retrieving the previous minimum in O(1) time and space.
Operations
- Push(x)
- If the stack is empty, push x and set minEle = x.
- If x >= minEle, push x normally.
- If x < minEle, push 2*x - minEle and update minEle = x (this encodes the previous min).
- Pop()
- Remove the top element.
- If the removed element is >= minEle, no change in minEle.
- If the removed element is < minEle, update minEle = 2*minEle - top (decoding the previous min).
- Peek()
- Returns minEle if the top is modified (encoded) or top otherwise.
- getMin()
- Returns minEle, the current minimum in O(1) time.
// C++ program to implement a stack that supports
// all operations in O(1) time and O(1) extra space.
#include <iostream>
#include <stack>
using namespace std;
// A user defined stack that supports getMin() in
// addition to push(), pop() and peek()
class SpecialStack {
private:
stack<int> s;
int minEle;
public:
SpecialStack() {
minEle = -1;
}
// Add an element to the top of Stack
void push(int x) {
if (s.empty()) {
minEle = x;
s.push(x);
}
// If new number is less than minEle
else if (x < minEle) {
s.push(2 * x - minEle);
minEle = x;
}
else {
s.push(x);
}
}
// Remove the top element from the Stack
void pop() {
if (s.empty()) {
return ;
}
int top = s.top();
s.pop();
// Minimum will change, if the minimum element
// of the stack is being removed.
if (top < minEle) {
minEle = 2 * minEle - top;
}
}
// Returns top element of the Stack
int peek() {
if (s.empty()) {
return -1;
}
int top = s.top();
// If minEle > top means minEle stores value of top.
return (minEle > top) ? minEle : top;
}
// Finds minimum element of Stack
int getMin() {
if (s.empty())
return -1;
// variable minEle stores the minimum element
// in the stack.
return minEle;
}
};
int main() {
SpecialStack ss;
// Function calls
ss.push(2);
ss.push(3);
cout << ss.peek() << " ";
ss.pop();
cout << ss.getMin() << " ";
ss.push(1);
cout << ss.getMin() << " ";
}
// Java program to implement a stack that supports
// all operations in O(1) time and O(1) extra space.
import java.util.Stack;
class SpecialStack {
private Stack<Integer> s;
private int minEle;
public SpecialStack() {
s = new Stack<>();
minEle = -1;
}
// Add an element to the top of Stack
public void push(int x) {
if (s.isEmpty()) {
minEle = x;
s.push(x);
}
// If new number is less than minEle
else if (x < minEle) {
s.push(2 * x - minEle);
minEle = x;
} else {
s.push(x);
}
}
// Remove the top element from the Stack
public void pop() {
if (s.isEmpty()) {
return;
}
int top = s.pop();
// Minimum will change, if the minimum element
// of the stack is being removed.
if (top < minEle) {
minEle = 2 * minEle - top;
}
}
// Returns top element of the Stack
public int peek() {
if (s.isEmpty()) {
return -1;
}
int top = s.peek();
// If minEle > top means minEle stores value of top.
return (minEle > top) ? minEle : top;
}
// Finds minimum element of Stack
public int getMin() {
if (s.isEmpty()) {
return -1;
}
// variable minEle stores the minimum element
// in the stack.
return minEle;
}
public static void main(String[] args) {
SpecialStack ss = new SpecialStack();
// Function calls
ss.push(2);
ss.push(3);
System.out.print(ss.peek() + " ");
ss.pop();
System.out.print(ss.getMin() + " ");
ss.push(1);
System.out.print(ss.getMin() + " ");
}
}
# Python program to implement a stack that supports
# all operations in O(1) time and O(1) extra space.
class SpecialStack:
def __init__(self):
self.s = []
self.minEle = -1
# Add an element to the top of Stack
def push(self, x):
if not self.s:
self.minEle = x
self.s.append(x)
# If new number is less than minEle
elif x < self.minEle:
self.s.append(2 * x - self.minEle)
self.minEle = x
else:
self.s.append(x)
# Remove the top element from the Stack
def pop(self):
if not self.s:
return
top = self.s.pop()
# Minimum will change, if the minimum element
# of the stack is being removed.
if top < self.minEle:
self.minEle = 2 * self.minEle - top
# Returns top element of Stack
def peek(self):
if not self.s:
return -1
top = self.s[-1]
# If minEle > top means minEle stores value of top.
return self.minEle if self.minEle > top else top
# Finds minimum element of Stack
def getMin(self):
if not self.s:
return -1
# variable minEle stores the minimum element
# in the stack.
return self.minEle
if __name__ == '__main__':
ss = SpecialStack()
# Function calls
ss.push(2)
ss.push(3)
print(ss.peek(), end=" ")
ss.pop()
print(ss.getMin(), end=" ")
ss.push(1)
print(ss.getMin(), end=" ")
// C# program to implement a stack that supports
// all operations in O(1) time and O(1) extra space.
using System;
using System.Collections.Generic;
class SpecialStack {
private Stack<int> s;
private int minEle;
public SpecialStack() {
s = new Stack<int>();
minEle = -1;
}
// Add an element to the top of Stack
public void Push(int x) {
if (s.Count == 0) {
minEle = x;
s.Push(x);
}
// If new number is less than minEle
else if (x < minEle) {
s.Push(2 * x - minEle);
minEle = x;
} else {
s.Push(x);
}
}
// Remove the top element from the Stack
public void Pop() {
if (s.Count == 0) {
return;
}
int top = s.Pop();
// Minimum will change, if the minimum element
// of the stack is being removed.
if (top < minEle) {
minEle = 2 * minEle - top;
}
}
// Returns top element of Stack
public int Peek() {
if (s.Count == 0) {
return -1;
}
int top = s.Peek();
// If minEle > top means minEle stores value of top.
return (minEle > top) ? minEle : top;
}
// Finds minimum element of Stack
public int GetMin() {
if (s.Count == 0) {
return -1;
}
// variable minEle stores the minimum element
// in the stack.
return minEle;
}
static void Main() {
SpecialStack ss = new SpecialStack();
// Function calls
ss.Push(2);
ss.Push(3);
Console.Write(ss.Peek() + " ");
ss.Pop();
Console.Write(ss.GetMin() + " ");
ss.Push(1);
Console.Write(ss.GetMin() + " ");
}
}
// JavaScript program to implement a stack that supports
// all operations in O(1) time and O(1) extra space.
class SpecialStack {
constructor() {
this.s = [];
this.minEle = -1;
}
// Add an element to the top of Stack
push(x) {
if (this.s.length === 0) {
this.minEle = x;
this.s.push(x);
}
// If new number is less than minEle
else if (x < this.minEle) {
this.s.push(2 * x - this.minEle);
this.minEle = x;
} else {
this.s.push(x);
}
}
// Remove the top element from the Stack
pop() {
if (this.s.length === 0) {
return;
}
let top = this.s.pop();
// Minimum will change, if the minimum element
// of the stack is being removed.
if (top < this.minEle) {
this.minEle = 2 * this.minEle - top;
}
}
// Returns top element of Stack
peek() {
if (this.s.length === 0) {
return -1;
}
let top = this.s[this.s.length - 1];
// If minEle > top means minEle stores value of top.
return this.minEle > top ? this.minEle : top;
}
// Finds minimum element of Stack
getMin() {
if (this.s.length === 0) {
return -1;
}
// variable minEle stores the minimum element
// in the stack.
return this.minEle;
}
}
// Driver Code
let ss = new SpecialStack();
ss.push(2);
ss.push(3);
console.log(ss.peek(), " ");
ss.pop();
console.log(ss.getMin(), " ");
ss.push(1);
console.log(ss.getMin(), " ");
Output
3 2 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 = xnew minEle = 2 * minEle - y
= 2*x - (2*x - prevMinEle)
= prevMinEle // This is what we wanted
