Wipro pvt interview experience Real time questions & tips from candidates to crack your interview
Application Engineer
Wipro pvt
2 rounds | 7 Coding
problems
Rejected
Interview preparation journey
Preparation
Duration: 2.5 months
Topics: Data Structures, Pointers, OOPS, System Design, Algorithms, Dynamic Programming
Tip 1 : Practice Atleast 250 Questions
Tip 2 :Take time to study DSA
Application process
Where: Campus
Eligibility: 6.5 CGPA
Tip 1 : Have some projects on resume.
Tip 2 : Do not put false things on resume.
Interview rounds
01
Round
Medium
Online Coding Interview
Duration50 minutes
Interview date10 Nov 2021
Coding problem2
2 coding problems and Some mCQS
1. Missing Numbers
Easy
28m average time
85% success
0/40
Asked in companies
You are given an array 'ARR' of distinct positive integers. You need to find all numbers that are in the range of the elements of the array, but not in the array. The missing elements should be printed in sorted order.
Example:
If the given array is [4, 2, 9] then you should print "3 5 6 7 8". As all these elements lie in the range but not present in the array.
Problem approach
Create a temp array temp[] of size n + 1 with all initial values as 0.
Traverse the input array arr[], and do following for each arr[i]
if(temp[arr[i]] == 0) temp[arr[i]] = 1
Traverse temp[] and output the array element having value as 0 (This is the missing element).
2. Middle Of Linked List
Easy
20m average time
80% success
0/40
Asked in companies
Given a singly linked list of 'N' nodes. The objective is to determine the middle node of a singly linked list. However, if the list has an even number of nodes, we return the second middle node.
Note:
1. If the list is empty, the function immediately returns None because there is no middle node to find.
2. If the list has only one node, then the only node in the list is trivially the middle node, and the function returns that node.
Problem approach
Traverse linked list using two-pointers. Move one pointer by one and the other pointers by two. When the fast pointer reaches the end, the slow pointer will reach the middle of the linked list.
02
Round
Easy
Video Call
Duration30 minutes
Interview date20 Nov 2021
Coding problem5
Basic HR types questions and questions related to projects and DS algo
1. Armstrong Number
Easy
15m average time
85% success
0/40
Asked in companies
You are given an integer ‘NUM’ . Your task is to find out whether this number is an Armstrong number or not.
A k-digit number ‘NUM’ is an Armstrong number if and only if the k-th power of each digit sums to ‘NUM’.
Example
153 = 1^3 + 5^3 + 3^3.
Therefore 153 is an Armstrong number.
Problem approach
The approach implemented below is simple. We traverse through all numbers in given range. For every number, we first count number of digits in it. Let the number of digits in current number be n. Them we find sum of n-th power of all digits. If sum is equal to i, we print the number
2. Project Based Question
Tell me about your projects
3. Operating System Questions
what is RTOS
4. Maximum Depth Of A Binary Tree
Easy
15m average time
85% success
0/40
Asked in companies
You are given the root node of a binary tree with N nodes, whose nodes have integer values. Your task is to find the maximum depth of the given Binary tree.
Depth of a binary tree is the same as its height. In simpler terms, you have to find the total number of nodes encountered while moving from the root node to the farthest leaf node, along the longest path of the binary tree.
Example:-
If we are given the above binary tree as input then moving from root node(5) to the farthest leaf node(50), the path formed will be [ 5->10->25->35->40->45->50 ]. The total number of nodes encountered is 7, therefore the maximum depth of the binary tree is 7.
Problem approach
maxDepth()
1. If tree is empty then return -1
2. Else
(a) Get the max depth of left subtree recursively i.e.,
call maxDepth( tree->left-subtree)
(a) Get the max depth of right subtree recursively i.e.,
call maxDepth( tree->right-subtree)
(c) Get the max of max depths of left and right
subtrees and add 1 to it for the current node.
max_depth = max(max dept of left subtree,
max depth of right subtree)
+ 1
(d) Return max_depth
5. Operating System Question
What is thread in OS?
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Skill covered: Programming
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