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Write a program to reverse a stack using recursion. You are not allowed to use loop constructs like while, for..etc, and you can only use the following ADT functions on Stack S:&nbsp; isEmpty(S)&nbsp; push(S)&nbsp; pop(S)The idea of the solution is to hold all values in Function Call Stack until the stack becomes empty. When the stack becomes empty, insert all held items one by one at the bottom of the stack.&nbsp; For example, let the input stack be &nbsp;Reverse a stack using recursion: <a href="https://www.geeksforgeeks.org/reverse-a-stack-using-recursion/">https://www.geeksforgeeks.org/reverse-a-stack-using-recursion/</a>

Reverse a stack using recursion

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 be O(1).&nbsp;To implement SpecialStack, you should only use standard Stack data structure and no other data structure like arrays, list,Example:&nbsp;Consider the following SpecialStack
16 &nbsp;--&gt; TOP
15
29
19
18

When 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 &nbsp;--&gt; TOP
19
18

When getMin() is called, it should 
return 18 which is the minimum in 
the current stack.Design and Implement Special Stack Data Structure : <a href="https://www.geeksforgeeks.org/design-and-implement-special-stack-data-structure/">https://www.geeksforgeeks.org/design-and-implement-special-stack-data-structure/</a>

Design and Implement Special Stack Data Structure

Given a Binary Tree, write an iterative function to print the Preorder traversal of the given binary tree. Refer to this for recursive preorder traversal of Binary Tree. To convert an inherently recursive procedure to iterative, we need an explicit stack. Following is a simple stack based iterative process to print Preorder traversal.&nbsp; 1) Create an empty stack nodeStack and push root node to stack.&nbsp; 2) Do the following while nodeStack is not empty.&nbsp; ….a) Pop an item from the stack and print it.&nbsp; ….b) Push right child of a popped item to stack&nbsp; ….c) Push left child of a popped item to stackIterative Preorder Traversal : <a href="https://www.geeksforgeeks.org/iterative-preorder-traversal/">https://www.geeksforgeeks.org/iterative-preorder-traversal/</a>

Iterative Preorder Traversal

The idea is to move down to leftmost node using left pointer. While moving down, push root and root’s right child to stack. Once we reach leftmost node, print it if it doesn’t have a right child. If it has a right child, then change root so that the right child is processed before.1.1 Create an empty stack2.1 Do following while root is not NULL &nbsp; &nbsp;a) Push root's right child and then root to stack. &nbsp; &nbsp;b) Set root as root's left child.Iterative Postorder Traversal : <a href="https://www.geeksforgeeks.org/iterative-postorder-traversal-using-stack/">https://www.geeksforgeeks.org/iterative-postorder-traversal-using-stack/</a>

Iterative Postorder Traversal

The stock span problem is a financial problem where we have a series of n daily price quotes for a stock and we need to calculate the span of the stock’s price for all n days. The span Si of the stock’s price on a given day i is defined as the maximum number of consecutive days just before the given day, for which the price of the stock on the current day is less than its price on the given day.Example:Input: N = 7, price[] = [100 80 60 70 60 75 85] Output: 1 1 1 2 1 4 6 Explanation: Traversing the given input span for 100 will be 1, 80 is smaller than 100 so the span is 1, 60 is smaller than 80 so the span is 1, 70 is greater than 60 so the span is 2 and so on. Hence the output will be 1 1 1 2 1 4 6.The Stock Span Problem : <a href="https://www.geeksforgeeks.org/the-stock-span-problem/">https://www.geeksforgeeks.org/the-stock-span-problem/</a>

The Stock Span Problem

Given an array of numbers, return true if given array can represent preorder traversal of a Binary Search Tree, else return false. Expected time complexity is O(n).Examples:Input: &nbsp;pre[] = {2, 4, 3} Output: true Given array can represent preorder traversal of below tree &nbsp; &nbsp;2 &nbsp; &nbsp; \ &nbsp; &nbsp; &nbsp;4 &nbsp; &nbsp; / &nbsp; &nbsp;3Check if a given array can represent Preorder Traversal of Binary Search Tree: <a href="https://www.geeksforgeeks.org/check-if-a-given-array-can-represent-preorder-traversal-of-binary-search-tree/">https://www.geeksforgeeks.org/check-if-a-given-array-can-represent-preorder-traversal-of-binary-search-tree/</a>

Check if a given array can represent Preorder Traversal of Binary Search Tree

The Postfix notation is used to represent algebraic expressions. The expressions written in postfix form are evaluated faster compared to infix notation as parenthesis are not required in postfix. We have discussed infix to postfix conversion. In this post, evaluation of postfix expressions is discussed.

Following is an algorithm for evaluation postfix expressions. 
1) Create a stack to store operands (or values). 
2) Scan the given expression and do the following for every scanned element. 


Stack _ Set 4 (Evaluation of Postfix Expression): https://www.geeksforgeeks.org/stack-set-4-evaluation-postfix-expression/

Stack | Set 4 (Evaluation of Postfix Expression)

Welcome to the daily solving of our PROBLEM OF THE DAY with Devashish Khare We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Stack but also build up problem-solving skills. Given an expression string x. Examine whether the pairs and the orders of {,},(,),[,] are correct in exp. For example, the function should return 'true' for exp = [()]{}{[()()]()} and 'false' for exp = [(]).Note: The driver code prints "balanced" if function return true, otherwise it prints "not balanced".Examples :Input: {([])}
Output: true
Explanation: { ( [ ] ) }.&nbsp;Same colored brackets can form balanced pairs, with 0 number of unbalanced bracket.Give the problem a try before going through the video. All the best!!! Problem Link: <a href="https://practice.geeksforgeeks.org/problems/parenthesis-checker2744/1">https://practice.geeksforgeeks.org/problems/parenthesis-checker2744/1</a>

PROBLEM OF THE DAY : 18/09/2024 | Parenthesis Checker

<h2>Largest Rectangular Area in a Histogram Using Stack | Step-by-Step Guide</h2>In this video, we’ll explore how to find the largest rectangular area in a histogram using a stack-based approach. This problem is a classic computational geometry problem where we need to determine the maximum area that can be formed between the bars of a histogram. Using a stack helps to solve the problem efficiently, reducing the time complexity compared to a brute-force approach. By the end of this tutorial, you will understand the stack-based method for solving this problem and how to apply it in real-world scenarios.<h2>What is the Largest Rectangular Area in a Histogram?</h2>In a histogram, we are given the heights of multiple bars, each having a uniform width of 1 unit. The goal is to find the largest rectangular area that can be enclosed by these bars. The rectangle must be formed by contiguous bars, and the height of the rectangle is limited by the shortest bar in the selected range.For example, if you have the following histogram with heights:Height array: [6, 2, 5, 4, 5, 1, 6]
The largest rectangular area that can be formed here is 12, formed by bars of heights 5, 4, and 5.<h3>Key Points Covered:</h3>Brute-Force Approach and Limitations: A brute-force method would involve checking every possible rectangle formed by different sets of bars, calculating their areas, and keeping track of the maximum area. However, this approach has a time complexity of O(n2)O(n^2)O(n2), which is inefficient for large histograms.Optimized Stack-Based Approach: The stack-based approach significantly improves the efficiency of the algorithm by reducing the time complexity to O(n)O(n)O(n). The stack helps in maintaining a list of bar indices in a way that allows us to efficiently calculate the maximum rectangular area.<h3>Steps in the Stack-Based Approach:</h3>Iterating Over Histogram Bars:<ul><li>Traverse through each bar in the histogram one by one.</li><li>If the current bar is taller than or equal to the bar at the top of the stack, push the index of the current bar onto the stack.</li><li>If the current bar is shorter, we start popping from the stack to calculate the area of the rectangle that can be formed with the bar at the top of the stack as the shortest bar.</li></ul>Calculating Area When Bars Are Popped:<ul><li>For each bar popped from the stack, calculate the width of the rectangle using the difference between the current index and the index of the bar now at the top of the stack.</li><li>Multiply this width by the height of the bar being popped to calculate the area.</li></ul>Finalizing the Calculations:<ul><li>After processing all bars, we continue popping any remaining bars in the stack and calculate the corresponding areas until the stack is empty.</li></ul><h3>Example Walkthrough:</h3>Consider the histogram with heights [6, 2, 5, 4, 5, 1, 6].<ul><li>We push the first index (0) since the first bar (6) is taller than the stack (which is empty initially).</li><li>We continue pushing indices as long as the current bar is taller than the one at the top of the stack.</li><li>When we encounter a shorter bar, we pop from the stack and calculate the area of the rectangle formed by the popped bar. This process continues until we find the largest possible rectangular area.</li></ul><h3>Edge Cases:</h3><ul><li>Single Bar: If the histogram contains only one bar, the largest rectangle is simply the height of that bar.</li><li>All Bars of Equal Height: If all bars are of the same height, the largest rectangle spans the entire width of the histogram.</li><li>Decreasing Heights: If the bars are arranged in strictly decreasing order, each smaller bar defines the width of a potential rectangle.</li></ul><h2>Why Use a Stack for This Problem?</h2>The stack helps to efficiently manage the heights and indices of the bars, ensuring that for each bar, we calculate the largest possible rectangle that can be formed. By pushing and popping indices in a controlled manner, we can compute the areas in linear time, which is much faster than the brute-force approach.Topics Included:Understanding the Problem: Learn how to find the largest rectangular area in a histogram and why the brute-force approach is inefficient.Stack-Based Solution: Detailed explanation of the stack-based approach to solving the problem, including how to calculate areas efficiently.Example Walkthrough: Step-by-step example demonstrating the stack-based approach to find the largest rectangular area.Edge Cases and Optimizations: Handling edge cases like single bars, bars of equal height, and bars in decreasing order, along with optimizations for larger datasets.For a detailed guide and further explanation, check out the full article on GeeksforGeeks: <a href="https://www.geeksforgeeks.org/largest-rectangular-area-in-a-histogram-using-stack/">https://www.geeksforgeeks.org/largest-rectangular-area-in-a-histogram-using-stack/</a>.

Largest Rectangular Area in Histogram

Welcome to the daily solving of our PROBLEM OF THE DAY with Nitin Kaplas We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Tree but also build up problem-solving skills. Given a string str consisting of opening and closing parenthesis '(' and ')'. Find length of the longest valid parenthesis substring.A parenthesis string is valid if:<ul><li>For every opening parenthesis, there is a closing parenthesis.</li><li>Opening parenthesis must be closed in the correct order.</li></ul>Examples :Input: str = ((()
Output: 2
Explaination: The longest valid parenthesis substring is "()".Give the problem a try before going through the video. All the best!!! Problem Link: <a href="https://practice.geeksforgeeks.org/problems/longest-valid-parentheses5657/1">https://practice.geeksforgeeks.org/problems/longest-valid-parentheses5657/1</a>

PROBLEM OF THE DAY : 16/09/2024 | Longest Valid Parentheses

This video is part of the Stack and Queue section under GFG SDE Sheet. Design a data structure that works like a LRU Cache. Here cap&nbsp;denotes&nbsp;the capacity of the cache and Q denotes the number of queries. Query can be of two types:<ol><li>SET x y: sets the value of the key x with value y</li><li>GET x: gets the key of x if present else returns -1.</li></ol> The LRUCache class has two methods get() and set() which are defined as follows.<ol><li>get(key): returns the value of the key if it&nbsp;already exists in the cache otherwise returns -1.</li><li>set(key, value): if the key is already present, update its value. If not present, add the key-value pair to the cache.&nbsp;If the cache reaches its capacity it should invalidate&nbsp;the least recently used item before inserting the new item.</li><li>In the constructor of the class the capacity of the cache should be initialized.</li></ol>&nbsp;Example 1:Input:
cap = 2
Q = 2
Queries = SET 1 2 GET 1
Output: 2
Explanation: 
Cache Size = 2

SET 1 2 GET 1
SET 1 2 : 1 -&gt; 2

GET 1 : Print the value corresponding
to Key 1, ie 2.Do check out:- Problem: <a href="https://www.geeksforgeeks.org/problems/lru-cache/1">https://www.geeksforgeeks.org/problems/lru-cache/1</a> SDE Sheet Link: <a href="https://www.geeksforgeeks.org/sde-sheet-a-complete-guide-for-sde-preparation/">https://www.geeksforgeeks.org/sde-sheet-a-complete-guide-for-sde-preparation/</a> Article Link:<a href=" https://www.geeksforgeeks.org/lru-cache-implementation/"> https://www.geeksforgeeks.org/lru-cache-implementation/</a>

SDE Sheet - LRU Cache

Welcome to the daily solving of our PROBLEM OF THE DAY with Yash Dwivedi We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Arrays but also build up problem-solving skills. A celebrity is a person who is known to all but does not know anyone at a party. A party is being organized by some people.&nbsp; A square matrix mat is used to represent people at the party such that if an element of row i and column j is set to 1 it means ith person knows jth person.&nbsp;You need to return the index of the celebrity in the party, if the celebrity does not exist, return&nbsp;-1.Note: Follow 0-based indexing.Examples:Input: mat[][] = [[0 1 0],
 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[0 0 0], 
 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[0 1 0]]
 Output: 1
 Explanation: 0th and 2nd person both know 1. Therefore, 1 is the celebrity.&nbsp; &nbsp;Give the problem a try before going through the video. All the best!!! Problem Link: <a href="https://practice.geeksforgeeks.org/problems/the-celebrity-problem/1">https://practice.geeksforgeeks.org/problems/the-celebrity-problem/1</a>

PROBLEM OF THE DAY : 03/08/2024 | The Celebrity Problem

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