Min Stack LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Min Stack in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
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  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
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  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemMin Stack– LeetCode Problem

Min Stack– LeetCode Problem

Problem:

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the MinStack class:

  • MinStack() initializes the stack object.
  • void push(int val) pushes the element val onto the stack.
  • void pop() removes the element on the top of the stack.
  • int top() gets the top element of the stack.
  • int getMin() retrieves the minimum element in the stack.

Example 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]
Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

Constraints:

  • -231 <= val <= 231 - 1
  • Methods poptop and getMin operations will always be called on non-empty stacks.
  • At most 3 * 104 calls will be made to pushpoptop, and getMin.
Min Stack– LeetCode Solutions
Min Stack Solution in C++:
class MinStack {
 public:
  void push(int x) {
    if (stack.empty())
      stack.emplace(x, x);
    else
      stack.emplace(x, min(x, stack.top().second));
  }

  void pop() {
    stack.pop();
  }

  int top() {
    return stack.top().first;
  }

  int getMin() {
    return stack.top().second;
  }

 private:
  stack<pair<int, int>> stack;  // {x, min}
};
Min Stack Solution in Java:
class MinStack {
  public void push(int x) {
    if (stack.isEmpty())
      stack.push(new int[] {x, x});
    else
      stack.push(new int[] {x, Math.min(x, stack.peek()[1])});
  }

  public void pop() {
    stack.pop();
  }

  public int top() {
    return stack.peek()[0];
  }

  public int getMin() {
    return stack.peek()[1];
  }

  private Stack<int[]> stack = new Stack<>(); // {x, min}
}
Min Stack Solution in Python:
class MinStack:
  def __init__(self):
    self.stack = []

  def push(self, x: int) -> None:
    mini = x if not self.stack else min(self.stack[-1][1], x)
    self.stack.append([x, mini])

  def pop(self) -> None:
    self.stack.pop()

  def top(self) -> int:
    return self.stack[-1][0]

  def getMin(self) -> int:
    return self.stack[-1][1]

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