London | July SDC | Ali Qassab | Sprint 4 | Implement laptop allocation #46
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Good start on this task - I've left some comments for places where you could think more about the efficiency of the implementation.
prep-exercises/laptop_allocation.py
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| sadness = calculate_sadness(person, laptop) | ||
| print(f"{person.name} -> {laptop.manufacturer} {laptop.model} ({laptop.operating_system.value}) - Sadness: {sadness}") | ||
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| print(f"\nTotal Sadness: {calculate_total_sadness(allocation)}") |
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Thinking about efficiency again, you just looped over everyone to get their individual sadness. The function you're calling here uses a loop as well. Is there any other way of achieving this without needing to loop twice here?
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Yes, with these changes:
- Add total_sadness = 0 before the loop
- Accumulate total_sadness += sadness inside the loop
- Replace the calculate_total_sadness(allocation) call with total_sadness
prep-exercises/laptop_allocation.py
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| # Create a priority queue of (person, laptop, sadness) tuples | ||
| # Sort by sadness to allocate best matches first | ||
| preferences = [] | ||
| for person in people: |
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In this approach you are looping over all people and laptops,then afterwards sorting and then doing the assignement.
Is there a way you could achieve this with fewer loops, or not going fully around the loops you have so much?
Imagine if you were doing this for a class full of hundreds of laptops and people - how many times would you end up going around the loops?
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For 100 people and 100 laptops:
Before: 10,000 pairs created + sorted
After: At most 10,000 sadness calculations, but typically much fewer due to early exits and shrinking sets
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