10405090xyzabc Chief Executive Officer Interview Questions and Answers

Design a stack that supports getmin in O(1) time and O(1) space complexities.

• Use two stacks, one for storing the actual values and the other for storing the minimum values.

• When pushing a new value, check if it is smaller than the current minimum value and push it to the minimum stack if it is.

• When popping a value, check if it is the current minimum value and pop it from the minimum stack if it is.

• To get the minimum val...read more

Binary tree traversal is the process of visiting each node in a binary tree exactly once in a specific order.

• There are three main types of binary tree traversal: inorder, preorder, and postorder.

• Inorder traversal visits the left subtree, then the root, then the right subtree.

• Preorder traversal visits the root, then the left subtree, then the right subtree.

• Postorder traversal visits the left subtree, then the right subt...read more

LRU cache is a data structure that stores the most recently used items and discards the least recently used items.

• LRU stands for Least Recently Used

• It has a fixed size and when the cache is full, the least recently used item is removed to make space for a new item

• It uses a combination of a doubly linked list and a hash map to achieve O(1) time complexity for both insertion and deletion

• Example: A web browser cache that ...read more

Yes, cloud computing refers to the delivery of computing services over the internet.

• Cloud computing allows users to access data and applications from anywhere with an internet connection.

• It offers scalability, flexibility, and cost-effectiveness compared to traditional on-premises computing.

• Examples of cloud computing services include Amazon Web Services, Microsoft Azure, and Google Cloud Platform.

Merge Sort is a divide and conquer algorithm that sorts an array by dividing it into two halves, sorting them separately, and then merging the sorted halves.

• Divide the array into two halves

• Recursively sort the two halves

• Merge the sorted halves

Find pairs of integers in a BST whose sum is equal to a given number.

• Traverse the BST and store the values in a hash set.

• For each node, check if (X - node.value) exists in the hash set.

• If yes, add the pair (node.value, X - node.value) to the result.

• Continue traversal until all nodes are processed.

Merge overlapping time intervals into mutually exclusive intervals.

• Sort the intervals based on their start time.

• Iterate through the intervals and merge overlapping intervals.

• Output the mutually exclusive intervals.

• Example: [(1,3), (2,6), (8,10), (15,18)] -> [(1,6), (8,10), (15,18)]

Different types of hashing and alternative for Linear Chaining

• Different types of hashing include division, multiplication, and universal hashing

• Alternative for Linear Chaining is Open Addressing

• Open Addressing includes Linear Probing, Quadratic Probing, and Double Hashing

An AVL tree is a self-balancing binary search tree where the heights of the left and right subtrees differ by at most one.

• AVL tree is a binary search tree with additional balance factor for each node.

• The balance factor is the difference between the heights of the left and right subtrees.

• Insertion and deletion operations in AVL tree maintain the balance factor to ensure the tree remains balanced.

• Rotations are performed ...read more

Find the minimum number of squares whose sum equals to a given number n.

• Use dynamic programming to solve the problem efficiently.

• Start with finding the square root of n and check if it is a perfect square.

• If not, then try to find the minimum number of squares required for the remaining number.

• Repeat the process until the remaining number becomes 0.

• Return the minimum number of squares required for the given number n.

Insertion sort for a singly linked list.

• Traverse the list and compare each node with the previous nodes

• If the current node is smaller, swap it with the previous node

• Repeat until the end of the list is reached

• Time complexity is O(n^2)