INFYTQ Final Round 2020

INFYTQ – The Certification program by Infosys to recognize talent, certifying them, and providing job opportunity. This certification is divided in two rounds this year ie. Qualifier and the final round.
Here, I like to share my experience of the Final round of this certification test.

The final round consists of two parts:-
1. Two Hands-on Coding problem.
2. 20 Multiple choice questions.

Both of these parts need to be completed in 3hr Duration. So let’s Begin with coding problem that i got:-

1st. You are given a string containing at least 1 special character, 1 odd digit, and 1 even digit. We need to count the number of special characters in the string. Based on the parity of count ( ODD/EVEN ) we need to generate the output string.
–> If count is odd, we need to arrange odd and even digits alternatively starting with odd digit first.
–> If count is even, we need to arrange even and odd digits alternatively starting with even digit first.
–> Append the remaining odd/even digits the end

Keep in mind the relative order of odd and even digit should remain the same.

For example – for input string “@2$1347%”, there are 3 special characters, 2 even digits, and 3 odd digit.
so the output string will be – “12347” (Beginning with odd as count is odd).

2nd . You are given a string containing only digits. We need to need to find the special substrings present in input. Special substring is defined as a substring whose integer value can be defined as product of two consecutive integers ie X = n * (n + 1). Note – substring with integer value 0 is not considered as special.

For example – For input string “1242” special substring will be “12”, “2”, “42”, “2”.

So these are the two Coding problems that came.

In the next section, there will 20 MCQ’s based on SQL queries, NoSQL queries, Object-oriented programming, Data structures, and Dry run of given code.

Overall, the test is not very difficult. Just clarity of basics is the requirement.
Now i am waiting for further interviews and other processes.

I hope everyone find this useful. Thanks

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