ACCIONA Interview Questions and Answers

Round duration - 60 Minutes
Round difficulty - Medium

This round was purely based on Data Structures and Algorithms . One has to be fairly comfortable in solving Algorithmic problems to pass this round . Both the questions asked were quite common and luckily I had already prepared them from CodeStudio and LeetCode.

Q1. 
Binary Tree Traversals Problem Statement
Given a Binary Tree with 'N' nodes, where each node holds an integer value, your task is to compute the In-Order, Pre-Order, and Post-Order traversals of the binar...read more
Ans. 
Compute the In-Order, Pre-Order, and Post-Order traversals of a Binary Tree given in level-order format.
• Implement functions to perform In-Order, Pre-Order, and Post-Order traversals of a Binary Tree.

• Use level-order input to construct the Binary Tree.

• Traverse the Binary Tree recursively to generate the required traversals.

• Ensure proper handling of null nodes represented by -1 in the input.

• Return the three traversals as
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Q2. 
Reverse Linked List Problem Statement
Given a Singly Linked List of integers, your task is to reverse the Linked List by altering the links between the nodes.
Input:
The first line of input is an intege...read more
Ans. 
Reverse a singly linked list by altering the links between nodes.
• Iterate through the linked list and reverse the links between nodes

• Use three pointers to keep track of the current, previous, and next nodes

• Update the links between nodes to reverse the list

• Return the head of the reversed linked list
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Round duration - 45 Minutes
Round difficulty - Medium

This round basically tested some concepts from Data Structures and File Manipulation .

Q1. 
Intersection of Two Arrays Problem Statement
Given two arrays A and B with sizes N and M respectively, both sorted in non-decreasing order, determine their intersection.
The intersection of two arrays in...read more
Ans. 
The problem involves finding the intersection of two sorted arrays efficiently.
• Use two pointers to iterate through both arrays simultaneously.

• Compare elements at the pointers and move the pointers accordingly.

• Handle cases where elements are equal and update the intersection array.

• Return the intersection array as the result.
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