Skip to content

[sonjh1217] WEEK 02 solutions #1762

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 2 commits into from
Aug 3, 2025
Merged

[sonjh1217] WEEK 02 solutions #1762

Changes from 1 commit
Commits
Show all changes
2 commits
Select commit Hold shift + click to select a range
d859932
valid anagram
jihyun-fastforward Jul 28, 2025
cef1274
product of array except self
sonjh1217 Aug 1, 2025
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
22 changes: 22 additions & 0 deletions valid-anagram/sonjh1217.swift
View file
Open in desktop
Copy link
Contributor

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

시간복잡도 뿐만 아니라 공간복잡도까지 고려하여 코드를 작성하신 부분이 인상깊습니다. 저와 다른 풀이로, 문자의 개수를 세어서 count를 확인하고 있어서 배워갑니다.

Original file line number Diff line number Diff line change
@@ -0,0 +1,22 @@
class Solution {
func isAnagram(_ s: String, _ t: String) -> Bool {
var count = [Character:Int]()
for char in s {
count[char, default: 0] += 1
}

for char in t {
count[char, default: 0] -= 1

if count[char]! < 0 {
return false
}
}

return count.values.allSatisfy {$0 == 0}

//시간 복잡도 O(n)
//공간 복잡도 O(n)
}
}