feat(392-is-subsequence): added py3 soln and readme

0392_is-subsequence/README.md

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+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, ..., sk` 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?

0392_is-subsequence/python3/main.py

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+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        slen = len(s)
+        tlen = len(t)
+        si = 0
+        ti = 0
+        
+        # If s is longer than t, it's obviously not
+        # going to be a subsequence
+        if slen > tlen:
+            return False
+        
+        while ti < tlen and si < slen:
+            found = False
+            
+            while ti < tlen:
+                if s[si] == t[ti]:
+                    found = True
+                    ti += 1
+                    break
+
+                ti += 1
+            
+            if not found:
+                return False
+            
+            si += 1
+        
+        # We reached end of t but we still have
+        # characters remaining in s and hence
+        # it's not a subsequence
+        if si < slen:
+            return False
+            
+        return True
+        
+