392. Is Subsequence

Description

Given two strings s and t, return true if s is a subsequence of t, or false otherwise.

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:

• Input: s = “abc”, t = “ahbgdc”
• Output: true

Example 2:

• Input: s = “axc”, t = “ahbgdc”
• Output: false

Constraints:

• 0 <= s.length <= 100
• 0 <= t.length <= 10⁴
• s and t consist only of lowercase English letters.

Follow up: Suppose there are lots of incoming s, say s₁, s₂, …, s_k where k >= 10⁹, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?

Submitted Code

class Solution(object):
    def isSubsequence(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if s == t or not s:     # s와 t가 같거나, s가 빈 문자열이면 무조건 True
            return True
        
        sc = 0                  # s 포인터

        for tc in t:            # t 포인터(ch)
            if tc == s[sc]:
                sc += 1

            if sc > len(s) - 1:     # sc가 s의 마지막 인덱스보다 크다면 모든 문자가 t에 있는 것
                return True
        
        return False

Runtime: ** ms | Beats **100.00%
Memory: 12.33 MB | Beats 94.21%

두 개의 포인터를 사용한 방법이다.

Other Solutions

1st

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        sp = tp = 0

        while sp < len(s) and tp < len(t):
            if s[sp] == t[tp]:
                sp += 1
            tp += 1
        
        return sp == len(s)

time complexity: 𝑂(𝑛) ← s나 t 중 더 긴 길이
space complexity: 𝑂(1)

좀 더 깔끔하게 작성된 코드를 참고했다.