Imagine that we have a list of names called nameList. There are various algorithms we could run on it. Here's one such algorithm, which compares each name in the list to every other name:

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01-01-2021
Computers and Technology

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Imagine that we have a list of names called nameList. There are various algorithms we could run on it. Here's one such algorithm, which compares each name in the list to every other name:

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IfeanyiEze8899idn IfeanyiEze8899idn · Answer 1 · 2021-01-02T01:52:41-05:00

IfeanyiEze8899idn IfeanyiEze8899idn

02-01-2021

Answer:

The algorithm has a quadratic time complexity of O(n^2)

Explanation:

The algorithm has two nested for-loop statements. The length of the name list is denoted as "n". Every name searched in the name list has a time complexity of n, therefore the time complexity of finding two names is (n x n) = n^2, which is denoted as O(n^2) in Big-O notation.

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