O(n) time and O(1) space using extra variable

Author: 149ps

392. Is Subsequence/392. Is Subsequence.py

@@ -0,0 +1,36 @@
+"""
+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 Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        if not s and t:
+            return True
+        if s and not t:
+            return False
+        if not s and not t:
+            return True
+        list_t = list(t)
+        count = 0
+        for i,char in enumerate(t):
+            if char == s[count]:
+                count += 1
+                if count >= len(s):
+                    return True
+        return count == len(s)