Author: akshay

Welcome file Mobile app development in 2025 is no longer just about building a great app — it’s about delivering continuous improvements instantly, securely, and without waiting for App Store reviews. Over‑the‑air (OTA) updates have become the backbone of modern mobile release cycles, especially for teams working with React Native, Android, and iOS apps. Below is a complete deep‑dive explaining why OTA updates are no longer optional and why every serious product team must adopt them in 2025. Introduction: The New Era of Mobile App Development The mobile ecosystem is evolving rapidly. User expectations are higher than ever — they…

Welcome file The mobile app industry has entered a new phase where traditional update cycles are no longer keeping up with user expectations. Users want instant fixes, instant improvements, and instant updates without waiting for the Play Store or App Store to approve every small change. As a result, traditional store driven updates are slowly losing relevance. A new approach known as Over The Air updates has started replacing them for modern apps across categories like e commerce, fintech, health tech, entertainment and social. This shift is not a minor change. It is a complete transformation in how teams build,…

Welcome file Fortune 500 companies are making a strategic shift in how they deploy mobile applications. Traditional app store release cycles—once considered standard practice—are now seen as competitive liabilities. Enterprise leaders are rapidly adopting over-the-air (OTA) update solutions to stay agile, reduce costs, and deliver superior user experiences. Major financial institutions, retailers, healthcare providers, and tech giants have quietly transformed their mobile deployment strategies. They’re shipping updates 10-20 times faster, reducing critical bug fix times from weeks to hours, and saving millions in operational costs annually. This shift isn’t just about speed—it’s about survival in markets where mobile apps drive…

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes p and q. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in the tree that has both p and q as descendants (where we allow a node to be a descendant of itself).” Examples Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Input: root = [2,1], p = 2, q = 1…

Given the root of a Binary Search Tree (BST) and an integer k, return the kth smallest value (1-indexed) among all the values of the nodes in the tree. Examples Input: root = [3,1,4,null,2], k = 1 Output: 1 Input: root = [5,3,6,2,4,null,null,1], k = 3 Output: 3 Approach Since it’s a BST, an in-order traversal gives values in ascending order. Perform in-order traversal and decrement k until it reaches 0. Return the node’s value where k == 0. Time & Space Complexity Time Complexity: O(n) in the worst case (if k is near n) Space Complexity: O(n) due to…

You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST. There may be multiple valid insertion ways as long as the tree remains a valid BST. Examples Input: root = [4,2,7,1,3], val = 5 Output: [4,2,7,1,3,5] Input: root = [40,20,60,10,30,50,70], val = 25 Output: [40,20,60,10,30,50,70,null,null,25] Approach Use recursion to find the correct insertion point. If the node is null, create a new TreeNode and return it.…

You are given the root of a binary search tree (BST) and an integer val. Find the node in the BST such that the node’s value equals val, and return the subtree rooted at that node. If such a node does not exist, return null. Input / Output Input: root = [4,2,7,1,3], val = 2 Output: [2,1,3] Input: root = [4,2,7,1,3], val = 5 Output: [] Approach This is a classic BST property search problem. At each node: If the node is null, return null. If node.val === val, return node. If val < node.val, search the left subtree. If…

Given the root of a binary tree, determine if it is a valid binary search tree (BST). Definition of BST The left subtree of a node contains only nodes with keys strictly less than the node’s key. The right subtree of a node contains only nodes with keys strictly greater than the node’s key. Both left and right subtrees must also be binary search trees. Input / Output Input: root = [2,1,3] Output: true Input: root = [5,1,4,null,null,3,6] Output: false Explanation: The root node’s value is 5 but its right child’s value is 4, which violates BST property. Approach Use…

A Binary Search Tree (BST) is a binary tree where each node follows the following properties: The left child contains only nodes with values less than the parent node. The right child contains only nodes with values greater than the parent node. Both the left and right subtrees must also be binary search trees. Why is it Called a Search Tree? It is called a “Search Tree” because it allows efficient search operations using the tree structure. If the tree is balanced, searching has a time complexity of: O(log n) Inorder Traversal If we perform an inorder traversal (left →…

React Blind 75 This guide is a carefully curated collection of 75 React interview questions designed to help learners build a deep understanding of React and confidently crack technical interviews at product-based companies, startups, and top-tier organizations. The questions are structured progressively, starting from the fundamentals and moving toward advanced concepts, including state management, hooks, performance optimization, server-side rendering, error handling, and more. Each question is accompanied by: The full question text. The topic it belongs to. Difficulty level (Easy / Intermediate / Advanced). A detailed explanation with code examples where applicable. Common interview scenarios where the question might arise.…