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August 2, 2020 03:41
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| Operation Example Big O Time Complexity | |
| Length: len(s) O(1) | |
| Add: s.add(2) O(1) | |
| Contains: x in/not in s O(1) | |
| Remove: s.remove(2) O(1) | |
| Discard: s.discard(2) O(1) | |
| Pop: s.pop() O(1) | |
| Clear: s.clear() O(1) | |
| Construct: set(iterable) O(length of iterable) | |
| Iterate: for v in s: O(n) | |
| Copy: s.copy() O(n) | |
| Equal/Not Equal: s1 == s2 O(min(len(s1), len(s2))) | |
| Comparison: s1 <= s2 O(len(s1)) | |
| Bitwise or: s1 | s2 O(len(s1)+len(s2)) | |
| Bitwise and: s1 & s2 O(min(len(s), len(t))) | |
| Difference: s1 - s2 O(len(s2)) | |
| Bitwise xor: s1 ^ s2 O(len(s)) | |
| Iteration: for v in s: O(n) | |
| Copy: s.copy() O(n) |
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