Description:
Solution:
- countT, window will be what we "have" & what we "need"
- We update our window[c].
- Does window[c] satisfiy 'window[c] == countT'?
Time Complexity: O(n)
'LeetCode ๐๏ธ > String' ์นดํ ๊ณ ๋ฆฌ์ ๋ค๋ฅธ ๊ธ
| 20. Valid Parentheses (0) | 2023.06.18 |
|---|---|
| 49. Group Anagrams (0) | 2023.06.17 |
| 242. Valid Anagram (0) | 2023.06.16 |
| 424. Longest Repeating Character Replacement (0) | 2023.06.14 |
| 3. Longest Substring Without Repeating Characters (0) | 2023.06.13 |
