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

# 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.

# Greedy solution.
class Solution(object):
    def isSubsequence(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if not s:
            return True

        i = 0
        for c in t:
            if c == s[i]:
                i += 1
            if i == len(s):
                break
        return i == len(s)