WatchGuard Technologies Software Developer Intern Interview Questions, Process, and Tips

First round was aptitude + coding. 60 questions in 60 mi , 30 about aptitude (mostly quants) , 30 about coding mcqs related to dsa ,oops. There was negative marking also.

Q1. 3 dsa question and 1 pattern question. Interviewer give me choice to select 3 data structures and asked about them with optimal solution and time complexity for each.
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Q2. Oops questions, oops is very important for this company in cs fundamental subject
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Q1. There were basic questions, but it depends on their mood. Tell they all true and what inside your mind spontaneous. Do not take time to think.
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Round duration - 60 Minutes
Round difficulty - Medium

It was basically to check my problem solving and OOPS concepts. Also , concepts related to development were asked

Q1. Alpha Pattern
Ninja loves playing with characters. So one day, he wants to arrange a few characters in the ‘N’ number of rows. The first row contains 1 character, the second row has two characters, the thi...read more
Ans. Brute force
The key here is to traverse all the lines sequentially, and for each line, we print the required character at the given index.

 
The steps are as follows:
• We define an ‘ans’ matrix to store the final pattern.
• We initialize a variable ‘k’ to 0, which will be the starting value for the pattern.
• We will iterate over all the rows, i.e., i = 0 to i = N - 1:
  • We will run a for loop starting from j = 0 with the condi...read more
Answered by CodingNinjas
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Q2. Covid Vaccination
We are suffering from the Second wave of Covid-19. The Government is trying to increase its vaccination drives. Ninja wants to help the Government to plan an effective method to help incr...read more
Ans. Brute force
The idea is to choose a peak value at the ‘dayNumber’ th index. Then we can create the array like a mountain with the peak of the mountain being at the  ‘dayNumber’ th index. The sum of the elements of this array must be less than or equal to maxVaccines.If we find that the sum is greater, then we have chosen a high peak value, and if it is less, then it means we have chosen a smaller peak value. So we ...read more
Answered by CodingNinjas
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Q3. Treasure Hunt
Alex and Rome are playing treasure hunting. Rome has a treasure box that Alex wants to open. Rome has N keys, each of which has a unique key number. She placed the keys in boxes numbered from...read more
Ans. Brute Force
The idea here is to traverse the entire array ‘key’ linearly and find the box which contains the required key. 
The steps are as follows:
• We will initialize a variable ‘answer’ to -1.
• Iterate from i = 0 to N - 1:
  • If the current box contains the required key, then update ‘answer’ to the current index.
  • We will start from the 1st box and iteratively move to the Nth box.
• If there is no box with the required key, ...read more
Answered by CodingNinjas
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Round duration - 90 Minutes
Round difficulty - Easy

It was an HR cum Technical round taken by a senior member of the technical team

Q1. Party Over
Ninja is coming after a long party to his home, but he faces a monster while returning. Monster puts up a condition to Ninja in order to free him. The monster gives him ‘n’ strings and asks him ...read more
Ans. Brute Force
The idea is to reverse all the strings and then sort them in alphabetical order and return the sorted strings back in their original form. We can use a nested loop and compare each string with all the preceding strings to sort the strings.
 
The steps are as follows:
• Initially, input a string array “arr” containing all the strings needed to be sorted based on the last character.
• Reverse the array containing...read more
Answered by CodingNinjas
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Q2. Mirror Diagonals
Ninja has a 2D array ‘arr’ with ‘N’ rows and each row contains ‘N’ elements. Your task is to check if the left diagonal numbers and right diagonal numbers are equal in the 2D Array.
Inpu...read more
Ans. Greedy Approach
We only need to check whether the diagonal elements are equal or not. So we can only iterate over the main diagonal elements and do the checking.
For the left diagonal, 'i' should be equal to ‘j’ and for the right diagonal, sum of 'i' and ‘j’ must be ‘N’ - 1 where 'i' is the row index and 'j' is the column index. For the left diagonal, we can use ‘i’ in place of ‘j’ as they both are the same and for the r...read more
Answered by CodingNinjas
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