The Hidden Cost Of A Line of Code
"Given an integer array nums , return true if any value appears at least twice in the array, and return false if every element is distinct." My first approach was to use a list class Solution : def containsDuplicate ( self , nums : List [ int ]) -> bool : container = [] for elem in nums : if elem in container : return True container . append ( elem ) return False I was hoping to optimise for time…
The cost of a single line of code can be hidden and overlooked until it becomes a significant problem. In the context of determining if an integer array contains duplicate values, a variety of solutions were explored, each with varying degrees of efficiency. The initial approach involved using a list to store unique elements, but this resulted in a time complexity of O(n^2), as each lookup required a full traversal of the list.
The author then discovered the superior performance of using a set, which reduces the time complexity to O(n). However, the set-based solution still had hidden costs due to the need for repeated lookups within the set. The author ultimately found the most efficient solution by comparing the lengths of the original list and the set created from it.
This single line of code, `return len(set(nums)) != len(nums)`, provides an optimal O(n) solution while also improving clarity and simplification.
Written by urgent.news from Dev.to's reporting — not their text. Machine-written — may contain errors; check the original before relying on it.
