 # Min Stack Leetcode: Efficient Solution

The Min Stack Leetcode problem is an important data structure problem that requires an efficient solution, especially in programming interviews or projects that require handling large amounts of data. The problem involves designing a stack that supports push, pop, top, and retrieving the minimum element in constant time. In this blog post, we will discuss various solutions to this problem and analyze their time and space complexities.

The Min Stack Leetcode problem is an important data structure problem that requires an efficient solution, especially in programming interviews or projects that require handling large amounts of data. The problem involves designing a stack that supports push, pop, top, and retrieving the minimum element in constant time. In this blog post, we will discuss various solutions to this problem and analyze their time and space complexities.

Contents

## What is Min Stack Leetcode

Min Stack Leetcode is a popular problem often asked by companies like Amazon, Microsoft, Yahoo, and Adobe in job interviews. The problem requires designing a special type of stack that supports efficient retrieval of the minimum element in constant time O(1) and space complexity O(n). This means that the minimum element in the stack should be obtained in a single operation without the need to traverse other elements in the stack.

## How to Implement Min Stack Leetcode

One way to implement Min Stack Leetcode problem is to use two stacks, one for the actual data and another for the minimum values so far. The minimum stack keeps track of the minimum value each time a new element is pushed into the stack. When an element is popped from the main stack, the minimum stack also pops the corresponding minimum value. This way, the minimum element can always be retrieved in constant time O(1) by simply accessing the top of the minimum stack.

## Complexity Analysis of Min Stack Leetcode

The space and time complexity of Min Stack Leetcode problem is O(n). The space complexity is O(n) because we need to maintain two stacks, one for the actual data and another for the minimum values. The time complexity is O(1) for all operations, including push, pop, top, and getMin, because they all take constant time.

## Example of Min Stack Leetcode

Suppose we have a Min Stack Leetcode implementation that supports the following operations:

• push(x) – Inserts element x into the stack
• pop() – Removes the element on the top of the stack
• top() – Returns the element on the top of the stack
• getMin() – Returns the minimum element in the stack

Initially, the minimum element minEle in the stack is -1. If we push the numbers -2, 0, -3, and -4 respectively, the stack will look like this:

 Stack Minimum Stack -4 -4 -3 -4 0 -4 -2 -4

Since -3 is less than the minimum element the original number -4 becomes the minimum element. The actual stack will then look like this:

 Stack Minimum Stack -2 -3 0 -3 -3 -4 -4 -4

Now, if we call getMin(), it should return -4, which is the current minimum element in the stack.

Min Stack Leetcode is an important problem that tests one’s ability to design efficient algorithms and data structures.

## Approaches for Solving Min Stack Leetcode Problem

### Rephrased: Using Two Stacks as Method 1

The first approach for solving the Min Stack Leetcode problem involves using two stacks to keep track of the elements and their minimums. One stack represents the actual elements, while the other stack is used to keep track of the minimum element so far. The top element of the minimum stack always contains the minimum element present in the actual stack at any given time.

Here’s the implementation of this approach:

1. Initialize two empty stacks: one for elements and one for minimums.
2. To push an element onto the stack, simply push it onto the element stack.
3. To push a minimum onto the stack, check if the minimum stack is empty or if the element being pushed is less than the current minimum. If so, push the element onto the minimum stack.
4. To pop an element from the stack, pop an element from the element stack. If the popped element is the minimum, pop it from the minimum stack as well.
5. To get the minimum element from the stack, simply return the top element of the minimum stack.

### Method 2: Using a Single Stack

The second approach for solving the Min Stack Leetcode problem involves using a single stack to keep track of the elements and their minimums. Here, we keep track of the minimum element at every step, and store it along with the actual element in the stack.

Here’s how you can implement this approach:

1. Initialize an empty stack to store the elements.
2. When pushing an element onto the stack, check if the stack is empty or if the element being pushed is less than or equal to the current minimum. If so, push the current minimum before pushing the actual element.
3. When popping an element from the stack, if the popped element is the minimum, pop the next element as well, as it represents the minimum at the previous step.
4. To get the minimum element from the stack, simply return the topmost element of the stack.

Both approaches provide solutions to the Min Stack Leetcode problem, but they differ in implementation. Method 1 involves using two stacks to keep track of elements and their minimums, whereas Method 2 involves using a single stack to store the element along with its minimum at every step.

## Code Implementation for Min Stack Leetcode Solution

### C++ Program for Min Stack Leetcode Solution

To implement the Min Stack Leetcode solution in C++, follow the steps given below:

1. First, create two stacks, one for storing actual elements and the other for storing the minimum element.
2. Push the element to the actual element stack.
3. If the element is less than or equal to the top element of the minimum element stack, then push the element to the minimum element stack as well.
4. To pop an element, first check if the top element in the actual element stack is equal to the top element in the minimum element stack. If it is, then pop from both stacks. If not, then simply pop from the actual element stack.
5. To get the top element, return the top element of the actual element stack.
6. To get the minimum element, return the top element of the minimum element stack.

### Java Program for Min Stack Leetcode Solution

To implement the Min Stack Leetcode solution in Java, follow the steps given below:

1. Create a class named MinStack.
2. Define two stacks inside the class, one for storing the actual elements and the other for storing the minimum elements.
3. Implement the push() method to push elements onto both stacks.
4. Implement the pop() method such that if the popped element is equal to the top element of the minimum element stack, pop from both stacks. Else, simply pop from the actual element stack.
5. Implement the top() method to get the top element from the actual element stack.
6. Implement the getMin() method to get the top element from the minimum element stack.

## Complexity Analysis for Min Stack Leetcode Solution

The Min Stack Leetcode problem aims to design a stack that supports push, pop, top, and retrieving the minimum element in the stack. The time and space complexity of the solution should be optimal. The following sections will explain the time and space complexities of the two methods that can be used to solve the problem.

### Method 1: Using Two Stacks

This method uses two stacks, one to hold the actual elements and the other to hold the minimum elements. The minimum stack maintains the minimum element of the actual stack at any given time. The time complexity of this solution is O(1) for all operations including getting the minimum element, which is always at the top of the minimum stack. The space complexity of this method is O(n) where n is the number of elements in the stack. This is because we are maintaining two stacks that hold the same number of elements.

### Method 2: Using One Stack with Pair Values

This method uses a single stack to hold pair values of the elements and the current minimum element. The pair value holds the actual element and the minimum element up to that position in the stack. The time complexity of this method is O(1) for all operations including getting the minimum element, which is stored as part of the pair value. The space complexity of this method is also O(n) where n is the number of elements in the stack since each element requires two spaces in the stack due to the pair values.

It is essential to choose the optimal solution based on the specific requirements and constraints of your use case to ensure high efficiency and performance of your implementation.

## Advantages of Using Min Stack Leetcode

Min Stack Leetcode is a popular problem in the tech industry that requires designing a stack that supports push, pop, and getMin operations in constant time. This algorithmic problem helps programmers optimize code and improve efficiency through the use of the stack data structure. Here are some of the advantages of using Min Stack Leetcode:

### 1. Constant Time Operations

Min Stack Leetcode problem requires that we support all operations in constant time O(1). This means that no matter the size of the stack, the push, pop, and getMin operations all take the same amount of time to execute. This is important for optimizing code and improving efficiency, especially when dealing with larger datasets.

### 2. Easy to Implement

The Min Stack Leetcode problem is easy to implement due to its simple data structure. The stack data structure is a widely used and well-understood concept in computer science, making the problem easy to understand and solve. The problem can be implemented using an array or a linked list, which are basic data structures taught in introductory data structure courses.

### 3. Practical Application

The Min Stack Leetcode problem has practical applications in the real world. For example, the problem can be used in the development of financial software to track daily transactions and calculate running balances. The stack can also be used in web development to implement the “back” functionality in web browsers.

### 4. Enhances Problem-Solving Skills

The Min Stack Leetcode problem is a great way to enhance problem-solving skills. It requires the use of the stack data structure, which is a fundamental concept in computer science. Solving the problem can also improve algorithmic thinking, which is an essential skill for any programmer.

### 5. Competitive Programming

The Min Stack Leetcode problem is a popular problem in competitive programming. It requires a solid understanding of the stack data structure and can be solved using various programming languages. Solving the problem can also lead to job opportunities in the tech industry, as many tech companies use competitive programming as a tool for recruitment.

In conclusion, the Min Stack Leetcode problem is a valuable tool for improving problem-solving skills and optimizing code. With constant time operations, easy implementation, practical applications, enhancement of problem-solving skills, and opportunities in competitive programming, the Min Stack Leetcode problem is an essential concept for any programmer.

## Conclusion

In conclusion, the Min Stack Leetcode solution is a useful algorithm that efficiently retrieves the minimum element in a stack. By creating a integer, minCurrentElement, we are able to store the current minimum element and make it easily accessible. This solution has been utilized by well-known companies like Amazon, Microsoft, Yahoo, and Adobe. With an implementation that utilizes data structures like arrays and linked lists, the solution has proven to be time and space efficient. By utilizing the MinStack class and its methods, we can add and remove elements from the stack while also retrieving the top element and the minimum element. Overall, the Min Stack Leetcode solution is an effective and practical algorithm that is used in a variety of industries and applications. 