[LeetCode-String] Split a String in Balanced Strings

CHOI YUN HOยท2021๋…„ 11์›” 8์ผ
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๐Ÿ“ƒ ๋ฌธ์ œ ์„ค๋ช…

Split a String in Balanced Strings

[๋ฌธ์ œ ์ถœ์ฒ˜ : LeetCode]

๐Ÿ‘จโ€๐Ÿ’ป ํ•ด๊ฒฐ ๋ฐฉ๋ฒ•

์ฃผ์–ด์ง„ ๋ฌธ์ž์—ด s๋Š” ๊ฐ™์€ ๊ฐœ์ˆ˜์˜ 'L'๊ณผ 'R'๋กœ ์ด๋ฃจ์–ด์ง„ Balanced String์ด๋‹ค.
์ด s๋ฅผ ์ตœ๋Œ€ ๋ช‡ ๊ฐœ์˜ Balanced String์œผ๋กœ ๋‚˜๋ˆŒ ์ˆ˜ ์žˆ๋Š”์ง€ ๊ตฌํ•˜๋Š” ๋ฌธ์ œ.

๊ท ํ˜•์ด ๋งž์œผ๋ ค๋ฉด ์šฐ์„  ๊ธธ์ด๊ฐ€ ์ง์ˆ˜์—ฌ์•ผํ•˜๋‹ˆ 2๋‹จ์œ„๋กœ ๋‚˜๋ˆˆ๋‹ค.
์•ž์—์„œ๋ถ€ํ„ฐ ๋‘ ๊ธ€์ž์”ฉ ๋Š˜๋ ค๊ฐ€๋ฉฐ L๊ณผ R์˜ ๊ฐœ์ˆ˜๋ฅผ ๋น„๊ตํ•˜์—ฌ ๊ฐ™์œผ๋ฉด ๊ทธ ๋ถ€๋ถ„์„ ๊ธฐ์ค€์œผ๋กœ ๋‚˜๋ˆ„๊ณ  ๊ทธ ์ง€์ ์—์„œ๋ถ€ํ„ฐ ๋‹ค์‹œ ๋‘ ๊ธ€์ž์”ฉ ๋Š˜๋ ค๊ฐ€๋ฉฐ ๊ฒ€์‚ฌํ•œ๋‹ค.

๐Ÿ‘จโ€๐Ÿ’ป ์†Œ์Šค ์ฝ”๋“œ

class Solution:
    def balancedStringSplit(self, s: str) -> int:
        i = res = 0
        for j in range(2, len(s) + 1, 2):
            if s[i:j].count('L') == s[i:j].count('R'):
                res += 1
                i = j

        return res









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