London| 26 March SDC | Jamal Laqdiem | Sprint 1 | Refactor Functions

[jamallaqdiem]

Self checklist

• I have titled my PR with Region | Cohort | FirstName LastName | Sprint | Assignment Title
• My changes meet the requirements of the task
• I have tested my changes
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Changelist

Refactored the scripts js and py to optimise the time and space complexity.

[cjyuan]

Code is correct.

• Could you also include the time complexity of the original implementation?

  Note: I think "Time Complexity" in the files refers to the time complexity of the original implementation, and "Optimal Time Complexity" refers to the time complexity of the refactored version.

[cjyuan]

Could you explain how you derived the time complexity of $O(1)$ and the space complexity of $O(n+m)$?

[jamallaqdiem]

Thanks for reviewing, is the the Optimal time complexity that should be O(n+m), we have to spend O(m) time to build the set and O(n) time to filter firstArray.
the space Com should be O(u) to store the matching items.

[cjyuan]

What is $u$? Normally we express all complexity based on the size of the input, which are $n$ and $m$.

[cjyuan]

• Why use $m$ in the complexity of your algorithm but not in the complexity of the original algorithm?

• How does the algorithm use $O(min(n, m))$ space?

[jamallaqdiem]

You are completely right about the inconsistency between old and new Time complex:

1. let's define n as len(first_sequence) and m as len(second_sequence), if the input sizes differ, m should be used in both.
   Original Algorithm: as explained earlier the outer loop runs n times, the inner loop runs m times, and the not in list takes up to O(n) time. Therefore, time complexity of the original code is O(n^2 \cdot m). If we assume n = m, it simplifies to O(n^3).
   New Algorithm: Converting the sequences to sets takes O(n) + O(m) time. The Python set intersection '&' average-case time complexity is O(\min(n, m)) as documented in this link :(https://wiki.python.org/moin/TimeComplexity). the overall optimal time complexity of O(n + m).

2. Space Complexity Proof for O(\min(n, m)):
   the original algorithm allocates memory for the common_items list, an item is only appended to common_items if it exists in both first_sequence and second_sequence
   Mathematically, the intersection of two sets cannot contain more elements than the smaller of the two sets

[cjyuan]

How much space needed by first_set and second_set?