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Susquehanna International Group (SIG)
Software Developer Co-Op was asked...January 14, 2014

Write a function that is passed a string and determines whether or not the string is a palindrome. (They had me write on the whiteboard in front of them.)

4 Answers

I used for-loops and character arrays and got the question right.

I think the double index way is pretty neat. for(int i=0,j=n-1;i

I think the double index way is pretty neat. for(int i=0,j=n-1;i

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A Thinking Ape

Finding the length of the longest substring containing unique characters

3 Answers

static int solve(String s) { Map positions = new HashMap(); int maxLength = 0; int length = 0; for (int index = 0; index length) { length++; if (length > maxLength) { maxLength = length; } } else { length = index - lastPosition; } positions.put(c, index); } return maxLength; } Less

public int longgestUniqueSubstring(String arr){ HashMap map=new HashMap(); int max=0,temp=0; for(int i=0;imax)max=temp; } return max; } Less

public int longgestUniqueSubstring(String arr){ HashMap map=new HashMap(); int max=0,temp=0; for(int i=0;imax)max=temp; } return max; } Less

Communications Security Establishment Canada | Centre de la sécurité des télécommunications

hackerrank easy, basic java and web dev knowledge, sql

3 Answers


I see why you didn't get an offer.

This organization is clearly too serious for you 😂. In case you didn’t realize, the security clearance isn’t an “interview”. It’s just a test to make sure you are not crazy. If you start the clearance, you basically have the job (unless you did drugs or something else really stupid). Less

Susquehanna International Group (SIG)

If you were given two words and needed to determine if the words were anagrams, how would you go about doing this?

3 Answers

sort both words and compare them letter by letter.

Use a hash table, like a unordered_map then loop over one word to record the number of times of each char. Use the table and loop over the second word to see if they have some set of chars with same number of times of occurrence. If the word is ascii then an integer array of length 256 can be used as the hash table. Less

Sorting sucks, it takes nlogn Instead use a frequency array: int f[26]; for(char c:s) f[c-'a']++; for(char c:t) f[c-'a']--; for(int i=26;i--;) if(f[i]) return false; return true; Less


When is the right time to optimize the code?

2 Answers

After you finish all the minimum implementation of the project.

From scratch writing as possible as optimized is good. After completing the logic and everything is ok reoptimizing might be necessary Less

Schneider Electric

Do you have experience with source-control?

2 Answers

Yes, I have experience using git from my previous position and from school.

Yes I am experienced in assurance indestries


How do you swap two integer variables without creating a temporary variable?

2 Answers

int x = 10; int y = 12; x = x+y //x = 22, y = 12 y = x - y //x = 22, y = 10 x = x-y // x= 12, y=10 Less

In python: a, b = b, a


Given two sorted lists and k, find k lowest elements from the two lists.

2 Answers

The key in these questions is to cover the fundamentals, and be ready for the back-and-forth with the interviewer. Might be worth doing a mock interview with one of the HubSpot or ex-HubSpot Software Engineer Co-Op experts on Prepfully? They give real-world practice and guidance, which is pretty helpful. Less

Were you able to answer? what was the process like in the interview?


Given an array, write a function to output the ranking of each value in the array. Equal values should have the same ranking. For example: input of [25, 7, 54, 90, 7] should result in [2, 1, 3, 4, 1]

2 Answers

# without sorting def rankings(arr): ranking=[0]*len(arr) rank=0 maximum=max(arr) tracking=[] while sum(arr)!=(maximum+1)*len(arr): minimum=min(arr) if minimum not in tracking: rank+=1 ranking[arr.index(minimum)]=rank tracking.append(minimum) arr[arr.index(minimum)]=maximum+1 return ranking arr=[25,7,54,90,7] ranking=rankings(arr) print(ranking) Less

def solution(a): sortedArray = sorted(a) resultArray = a d = {} rank = 1 m = sortedArray[0] d[m] = 1 for item in sortedArray[1:]: if item > m: m = item rank += 1 d[item] = rank for index in range(len(a)): resultArray[index] = d[a[index]] return resultArray Less


The coding challenge required getting data from an API, manipulating it, and submitting the answer.

2 Answers

Would you mind what exactly was the coding challenge and how you tackled it ?

What about your onsight question? what was that around ? Thanks

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