TCS iON Software Developer Intern Interview Questions, Process, and Tips

Technical MCQ questions on core computer science subjects were asked.

Q1. How to find whether the given linked list has loop in it?
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Q2. Explain osi model with example of browser.
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There was apti round with good apti questions and also code debugging

Q1. About polymorphism ,code polymorphism and inheritance
Ans. Coded it in cpp in notepad
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Q2. 3 apti question and also discussion on resume
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Q3. Angle between hands of clock on 10:25? Trains traveling towards eachother give speed where will they collide?
Ans. 
Angle between hands of clock at 10:25 is 147.5 degrees. Trains will collide at midpoint of their initial positions.
• To calculate angle between hands of clock, use formula: |(30*H)-(11/2)*M|

• For 10:25, H=10 and M=25, so angle = |(30*10)-(11/2)*25| = 147.5 degrees

• When two trains are traveling towards each other, their relative speed is added to get the collision speed.

• The collision point is the midpoint of their initial po...read more
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Q1. General resume based
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Q2. Strength and weakness
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Round duration - 30 Minutes
Round difficulty - Easy

Timing of test start from 11 AM and everything goes well here .

Q1. What is segmentation in the context of operating systems?
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Q2. How many levels of normalization are there in database management systems?
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Round duration - 60 Minutes
Round difficulty - Easy

It happen nicely and no problem occur.

Q1. 
Trapping Rain Water Problem Statement
You are given a long type array/list ARR of size N, representing an elevation map. The value ARR[i] denotes the elevation of the ith bar. Your task is to determine th...read more
Ans. Brute Force Approach
The idea here is to travel over every elevation on the map and calculate the units of water the elevation can store.
 
Here is the algorithm :
 
1. Iterate over every elevation or element and find the maximum elevation on to the left and right of it. Say, the maximum elevation on to the left of the current elevation or element that we are looking at is ‘maxLeftHeight’ and the maximum elevation on...read more
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Q2. 
Sort 0 1 2 Problem Statement
Given an integer array arr of size 'N' containing only 0s, 1s, and 2s, write an algorithm to sort the array.
Input:
The first line contains an integer 'T' representing the n...read more
Ans. Sorting
Use any good sorting algorithm like Merge Sort, Quick Sort or inbuilt sorting function of different languages.
• Sort the Array and just return.
Space Complexity: O(1)Explanation:
O(1), As we are using constant space.
Time Complexity: O(nlogn)Explanation:
O(N*log(N)), where ‘N’ is the size of the array.
We are using inbuilt sort algorithm which has Overall Time Complexity O(N*log(N))
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Round duration - 90 mintues
Round difficulty - Hard

Environment is very bad the intervier video call is lagging.

Q1. What are the sub-parts or phases of the analysis part in compiler design?
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Q2. 
Left Rotations of an Array
Given an array of size N and Q queries, each query requires left rotating the original array by a specified number of elements. Return the modified array for each query.
Input:
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Ans. Rotate One Element At A Time
The idea is to create a function which would rotate the array one element at a time. This can be done by shifting the array towards left by one element and copying the first element to the end of the array. For every query repeatedly call the above function, until the desired rotation is obtained.
 
If the number of rotations required, say ‘K’, is greater than the number of elements, ‘N’,...read more
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Round duration - 60 Minutes
Round difficulty - Medium

It was a Coding round with 2 programming questions which need to solved in 60 minutes in any of the languages (C/C++/Java/Python)

Q1. 
Ninja and the Maze Problem Statement
Ninja is stuck in a maze represented as a 2D grid. He can move in four directions (Up, Down, Left, Right) until he hits a wall ('1'). Once stopped, he can choose a new...read more
Ans. Depth First Search
• Ninja can move in at most four directions i.e left, right, up, and down.
• We will take a 2-D array ‘directions’ of size 4 x 2  in which the first column denotes the shift in the horizontal direction and the second column denotes a shift in the vertical direction.
    We will store all the four directions { (0,1) (0,-1) ( 1,0) ( -1,0 ) )} in 2-d array.
• In each direction, Ninja continues to m...read more
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Q2. 
Kth Largest Element Problem
Given an array containing N distinct positive integers and a number K, determine the Kth largest element in the array.
Example:
Input:
N = 6, K = 3, array = [2, 1, 5, 6, 3, ...read more
Ans. Brute Force
• The most obvious brute force approach would be to sort the array in descending order and return the ‘K’th element from the beginning of the array.
• Sort the array in descending order, for sorting most of the languages have their inbuilt sort methods which are usually very fast.
• After sorting, return the element arr['K'-1](i.e. element at index ‘K’-1, considering 0-based indexing).
Space Complexity: O(1)Explanati...read more
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Round duration - 60 minutes
Round difficulty - Easy

It was in the morning at around 10 a.m conducted on Skype.

Q1. 
Median of Two Sorted Arrays
Given two sorted arrays A and B of sizes N and M, find the median of the merged array formed by combining arrays A and B. If the total number of elements, N + M, is even, the m...read more
Ans. Sorting
The first approach is by joining the array and again sorting is to get median.
 
The algorithm is as follows:
• Create a new array of size of total number of elements.
• Insert elements of both the arrays in the array.
• Sort the array and find the median.
Space Complexity: OtherExplanation:
O(N + M) per test case, where ‘N’ and ‘M’ are the sizes of the arrays.
 
In the worst case, we will be creating a single array...read more
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Q2. 
Distance Between Two Nodes in a Binary Tree
Given a binary tree and the values of two distinct nodes, determine the distance between these two nodes in the tree. The distance is defined as the minimum num...read more
Ans. Naive Approach
In any rooted tree, the distance between two nodes 'U' and 'V' can be found by finding the lowest common ancestor (LCA), ‘x’ of two nodes. The lowest common ancestor (LCA) between nodes 'U' and 'V' is defined as the lowest node in the tree that has both 'U' and 'V' as descendants, where we define each node to be a descendant of itself (so if 'U' has a direct connection from 'V', 'V' is the lowest common a...read more
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Round duration - 20 minutes
Round difficulty - Easy

A HR Round in which the interviewer discussed about me and some things about my resume (Achievements, Projects etc.)