London | 25-SDC-NOV | Jesus del Moral | Sprint 1 | Analyse and Refactor Functions#94
London | 25-SDC-NOV | Jesus del Moral | Sprint 1 | Analyse and Refactor Functions#94delmorallopez wants to merge 4 commits intoCodeYourFuture:mainfrom
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| * Optimal Time Complexity: O(n) - We must examine each element at least once to calculate the sum and product, so O(n) is the best we can achieve for this problem. | ||
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| * we can reduce the constant factor by eliminating the redundant second loop | ||
| * Instead of two passes (2n operations), we can do both calculations in a single pass (n operations) |
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What do you mean by "two passes == 2n operations" and "1 pass == n operation"?
What are these operations? Does that mean one pass is guaranteed to take half as much time to complete?
The code still needs to perform sum += num and product *= num whether we use one loop or two loops.
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When I said “2n vs n operations”, I meant loop iterations. However, both implementations still perform n additions and n multiplications. The single-loop version reduces loop overhead and improves cache efficiency, but asymptotically both are O(n). One pass is not guaranteed to take half the time, because Big-O ignores constant factors and the arithmetic work remains the same.
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You have a good understanding and the performance is indeed improved when two loops are reduced to one.
However, the improvement is not really "doubled", so in complexity analysis we don't focus on the constant factor.
| common_items.append(i) | ||
| return common_items | ||
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| return list(set(first_sequence) & set(second_sequence)) |
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