Learnings from my experience on both sides of the interview table. Mainly composed of a live walkthrough I code up a solution to an interview question, on the spot. This talk was given at McMaster University to Engineering students who wanted to learn more about technical software interviews. Recording - https://www.youtube.com/watch?v=ylkRLBVbVMk

1. Demystifying
Software Interviews
Learnings from Being on Both Sides of the Table

3. Intro
Me - Michael Viveros:
• Sfwr Eng & Mgmt, 2017
• Software Engineer II, PagerDuty
• Interviewed dozens of candidates
You:
• Applying? Interviewing?

4. Live Walkthrough

5. Time Complexity
arr = [1, 2, 3]
for i in arr: # O(n)
print(i) # O(1)
# O(n * 1) = O(n)
# n = 3, a = [1, 2, 3], 3 iterations
# n = 4, a = [1, 2, 3, 4], 4 iterations
# n = 5, a = [1, 2, 3, 4, 5], 5 iterations

6. Time Complexity
arr = [1, 2, 3]
for i in arr: # O(n)
print(i) # O(1)
# O(n * 1) = O(n)
# n = 3, a = [1, 2, 3], 3 iterations
# n = 4, a = [1, 2, 3, 4], 4 iterations
# n = 5, a = [1, 2, 3, 4, 5], 5 iterations

7. Time Complexity
arr = [1, 2, 3]
for i in arr: # O(n)
print(i) # O(1)
# O(n * 1) = O(n)
# n = 3, a = [1, 2, 3], 3 iterations
# n = 4, a = [1, 2, 3, 4], 4 iterations
# n = 5, a = [1, 2, 3, 4, 5], 5 iterations

8. Time Complexity
arr = [1, 2, 3]
for i in arr: # O(n)
print(i) # O(1)
# O(n * 1) = O(n)
# n = 3, a = [1, 2, 3], 3 iterations
# n = 4, a = [1, 2, 3, 4], 4 iterations
# n = 5, a = [1, 2, 3, 4, 5], 5 iterations

9. Time Complexity
arr = [1, 2, 3]
for i in arr: # O(n)
for j in arr: # O(n)
print(i, j)
# O(n * n) = O(n^2)
# n = 3, a = [1, 2, 3], 9 iterations
# n = 4, a = [1, 2, 3, 4], 16 iterations
# n = 5, a = [1, 2, 3, 4, 5], 25 iterations

10. Time Complexity
arr = [1, 2, 3]
for i in arr: # O(n)
for j in arr: # O(n)
print(i, j)
# O(n * n) = O(n^2)
# n = 3, a = [1, 2, 3], 9 iterations
# n = 4, a = [1, 2, 3, 4], 16 iterations
# n = 5, a = [1, 2, 3, 4, 5], 25 iterations

11. Time Complexity
arr = [1, 2, 3]
for i in arr: # O(n)
for j in arr: # O(n)
print(i, j)
# O(n * n) = O(n^2)
# n = 3, a = [1, 2, 3], 9 iterations
# n = 4, a = [1, 2, 3, 4], 16 iterations
# n = 5, a = [1, 2, 3, 4, 5], 25 iterations

12. Time Complexity
arr = [1, 2, 3]
for i in arr: # O(n)
for j in arr: # O(n)
print(i, j)
# O(n * n) = O(n^2)
# n = 3, a = [1, 2, 3], 9 iterations
# n = 4, a = [1, 2, 3, 4], 16 iterations
# n = 5, a = [1, 2, 3, 4, 5], 25 iterations

13. Time Complexity
arr = [1, 2, 3]
num = 5
for i in x:
if i == num:
return True
# O(n)
dict = {1: 10, 2: 20}
num = 5
return num in dict
# O(1)

14. Time Complexity
arr = [1, 2, 3]
num = 5
for i in x:
if i == num:
return True
# O(n)
dict = {1: 10, 2: 20}
num = 5
return num in dict
# O(1)

15. Space Complexity
arr = [2, 1, 3]
minNum = 100
for num in arr:
if num < minNum:
minNum = num
return minNum
# O(1)
arr = [2, 1, 3]
arrSorted = sorted(arr)
return arrSorted[0]
# O(n)

16. Space Complexity
arr = [2, 1, 3]
minNum = 100
for num in arr:
if num < minNum:
minNum = num
return minNum
# O(1)
arr = [2, 1, 3]
arrSorted = sorted(arr)
return arrSorted[0]
# O(n)

17. Ask Questions
# Design an algorithm to print all pairs of integers
# within an array which sum to a specified value.
#
# Input: a = [1, 1, 3], sum = 4
# Output: [1, 3]
#
# Input: a = [ 2, 3, 4, 7, 8 ], sum = 10
# Output: [3, 7], [2, 8]

18. Start Simple, Then Optimize
# O(n^2)
for numI in arr:
for numJ in arr:
if numI + numJ == x:
# O(n)
for numI in arr:
numJ = x - numI
if numJ in occurrences:

19. Start Simple, Then Optimize
# O(n^2)
for numI in arr:
for numJ in arr:
if numI + numJ == x:
# O(n)
for numI in arr:
numJ = x - numI
if numJ in occurrences:

20. Use Google

21. Don't Freeze, COMMUNICATE!
$ python walkthrough.py
[[1, 3], [1, 3], [3, 1], [3, 1]]
$ python walkthrough.py
...
TypeError: unsupported operand type(s) for -: 'int' and 'NoneType'

22. Don't Freeze, COMMUNICATE!
$ python walkthrough.py
[[1, 3], [1, 3], [3, 1], [3, 1]]
$ python walkthrough.py
...
TypeError: unsupported operand type(s) for -: 'int' and 'NoneType'

23. Add Print Statements
print(arr)
for i in range(len(arr)):
numI = arr[i]
for j in range(len(arr)):
numJ = arr[j]
if numI + numJ == x:
print(f"found pair - i {i}, j {j}, numI {numI}, numJ {numJ}")

24. Analyze Time and Space
Complexity
print(arr)
pairs = []
# time - O(n^2), space - O(n)
for i in range(len(arr)): # O(n)
numI = arr[i]
for j in range(len(arr)): # O(n)
numJ = arr[j]
if numI != None and numJ != None and numI + numJ == x:

25. Incremental Approach
occurrences = {}
for num in arr:
if num not in occurrences:
occurrences[num] = 1
else:
occurrences[num] += 1
# pairs = []
# for numI in arr:

26. Other Tips
• Pick the language you're most comfortable with
• Don't worry if you don't finish
• PRACTICE!

27. Resources

30. Resources
• Mock Interviews
• Wed/Thurs Jan 15/16, evenings
• Hackerrank contest
• Slack mobile app
• Workspace - macinterviews.slack.com

31. Pagerduty
• Hiring summer interns!
• Perks:
• Catered lunch
• Downtown Toronto
• Interns are FULL members of the team
• pagerduty.com/careers