London | 25-SDC-July | Rihanna Poursoltani | Sprint 2 | Improve code with precomputing

[RihannaP]

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Changelist

For optimising I tried to

• Store the list and the compare only with neighbours
• I added a set of all char and then check if a lowercase letter exists

Questions

No question

[cjyuan]

Code looks good. Can you explain using big-O notation why your approach is better?

[RihannaP]

Code looks good. Can you explain using big-O notation why your approach is better?

Using a set makes membership checks O(1) on average. Creating the set takes O(n) and we iterate over at most n unique characters, so the overall complexity is O(n).

[cjyuan]

Code looks good. Can you explain using big-O notation why your approach is better?

Using a set makes membership checks O(1) on average. Creating the set takes O(n) and we iterate over at most n unique characters, so the overall complexity is O(n).

Can you also do the same for common_prefix.py?

[RihannaP]

Can you also do the same for common_prefix.py?

Sorting the strings takes O(n log n). After sorting, we only compare adjacent strings, and each prefix comparison takes O(m) ( m is the length of the shorter string). This results in O(n·m) for the comparisons.

Overall complexity is O(n log n + n·m), which is better than comparing every pair of strings (O(n²·m)).

[cjyuan]

If we are factoring in the length of the strings, m, then the complexity of sorting won't just be O(n log n).
n log n measures only the number of comparisons needed.

[RihannaP]

If we are factoring in the length of the strings, m, then the complexity of sorting won't just be O(n log n). n log n measures only the number of comparisons needed.

You’re right. so sorting and comparison cost is O(m · n log n)
Adjust scan O(n · m) so overall:
O(n log n · m + n · m)= O(n log n · m)

[cjyuan]

Yes. O(m · n log n)

O(n log n · m) may be misinterpreted as O(n log (n · m)).