# coding=utf8
'''
392. Is Subsequence
Given a string s and a string t, check if s is subsequence of t.

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).

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

Example 1:
s = "abc", t = "ahbgdc"

Return true.

Example 2:
s = "axc", t = "ahbgdc"

Return false.

Follow up:
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?
'''

class Solution1:
    def isSubsequence(self, s: str, t: str) -> bool:
        res = False
        s_curr, t_curr = 0, 0
        if len(s)==0:
            res = True
            return res
        if not set(s).issubset(set(t)):
            return res
        while t_curr < len(t):
            if s[s_curr] == t[t_curr]:
                s_curr += 1
                if s_curr == len(s):
                    res = True
                    break
            t_curr += 1
        return res

class Solution2:
    def isSubsequence(self, s: str, t: str) -> bool:
        t = iter(t)
        return all(c in t for c in s)

# expansion for solution2
class Solution3:
    def isSubsequence(self, s: str, t: str) -> bool:
        for c in s:
            i = t.find(c)
            if i == -1:
                return False
            else:
                t = t[i+1:]
        return True
    
if __name__ =='__main__':
    s = "abc"
    t = "ahbgdc"
    res = Solution3().isSubsequence(s, t)
    print(res)
