Jio Insurance Broking Interview Questions and Answers

Attended the codevita competition in final year of college.

Q1. About College Project
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Q2. Some very basics of java
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Q3. Experience of college project as a team
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Round duration - 25 Minutes
Round difficulty - Medium

Very friendly interviewer. Although waiting time was very high

Q1. 
Pair Sum Problem Statement
You are given an integer array 'ARR' of size 'N' and an integer 'S'. Your task is to find and return a list of all pairs of elements where each sum of a pair equals 'S'.
Note:
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Ans. 
Find pairs of elements in an array that sum up to a given value, sorted in a specific order.
• Iterate through the array and for each element, check if the complement (S - current element) exists in a hash set.

• If the complement exists, add the pair to the result list.

• Sort the result list based on the criteria mentioned in the question.

• Return the sorted list of pairs.
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Round duration - 90 minutes
Round difficulty - Easy

This is a written round on paper for everyone. Three coding questions were given. Two out of three must be correct covering every single edge case to qualify for the next round. Only the most optimal solution was to be considered. 

Q1. 
Maximum Subarray Sum Problem Statement
Given an array arr of length N consisting of integers, find the sum of the subarray (including empty subarray) with the maximum sum among all subarrays.
Explanation...read more
Ans. 
Find the sum of the subarray with the maximum sum among all subarrays in an array of integers.
• Iterate through the array and keep track of the maximum sum subarray encountered so far.

• Use Kadane's algorithm to efficiently find the maximum subarray sum.

• Consider the sum of an empty subarray as 0.

• Example: For input arr = [-2, 1, -3, 4, -1], the maximum subarray sum is 4.
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Q2. 
Connecting Ropes with Minimum Cost
You are given 'N' ropes, each of varying lengths. The task is to connect all ropes into one single rope. The cost of connecting two ropes is the sum of their lengths. Yo...read more
Ans. 
Connect ropes with minimum cost by merging two smallest ropes at a time.
• Sort the array of rope lengths in ascending order.

• Merge the two smallest ropes at a time and update the cost.

• Repeat the process until all ropes are merged.

• Return the total cost as the minimum cost to connect all ropes.
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Q3. 
Left View of a Binary Tree Problem Statement
Given a binary tree, your task is to print the left view of the tree.
Example:
Input:
The input will be in level order form, with node values separated by a...read more
Ans. 
Print the left view of a binary tree given in level order form.
• Traverse the tree level by level and print the first node encountered at each level

• Use a queue to perform level order traversal

• Keep track of the level while traversing the tree
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Round duration - 50 minutes
Round difficulty - Easy

This was face to face interview round.

Q1. 
Finding Triplets in a Binary Tree Problem Statement
You are given a Binary Tree of integers and an integer 'X'. Your task is to find all the triplets in the tree whose sum is strictly greater than 'X'. Th...read more
Ans. 
Find all triplets in a binary tree whose sum is greater than a given integer X, with a grandparent-parent-child relationship.
• Traverse the binary tree to find all possible triplets with the required relationship.

• Keep track of the sum of each triplet and compare it with the given integer X.

• Return the triplets that satisfy the condition in any order.

• Ensure each triplet follows the format (grand-parent, parent, child).
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Q2. 
Loot Houses Problem Statement
A thief is planning to steal from several houses along a street. Each house has a certain amount of money stashed. However, the thief cannot loot two adjacent houses. Determi...read more
Ans. 
Determine the maximum amount of money a thief can steal from houses without looting two consecutive houses.
• Use dynamic programming to keep track of the maximum amount of money that can be stolen up to each house.

• At each house, the thief can either choose to steal from the current house or skip it and steal from the previous house.

• The maximum amount of money that can be stolen at the current house is the maximum of the ...read more
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Round duration - 60 minutes
Round difficulty - Easy

This was face to face interview round.