📄 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


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

🔨 My Solution

  • using Two Pointers

💻 My Submission

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        if len(s)==0: return True
        while p1<len(s) and p2<len(t):
            if s[p1]==t[p2]:
        if p1==len(s): return True
        return False


Time Complexity: O(n)
Space Complexity: O(1)

🎈 Follow up:

Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 10910^9, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?

❗ Follow up - 첫 번째 시도: 실패

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        # 순서 고려 안됨...!! 
        t_count=Counter([x for x in t if x in s])
        if s_count==t_count: return True
        return False


  • 개수만 비교할 뿐, 순서가 고려되지 않았다.


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