is_subsequence 

 1# @leet start
 2class Solution:
 3    def isSubsequence(self, s: str, t: str) -> bool:
 4        """
 5        This question asks us if `s` is a subsequence of `t`. To find that out,
 6        we can use two pointers. We start out at the beginning of `s` and `t`.
 7        If the current char of `s`, $s_i$ is equal to the current char of `t`,
 8        $t_j$, we increment both `i` and `j` since we've found a match. Otherwise,
 9        this is a miss, so we only increment `j`, since we want to continue on
10        for `t`. Finally, we make sure we've reached the end of `s`, so we make
11        sure that `i` is equal to the length of `s`.
12        """
13        i, j, m, n = 0, 0, len(s), len(t)
14
15        while i < m and j < n:
16            l, r = s[i], t[j]
17            if l == r:
18                i += 1
19            j += 1
20        return i == m
21
22
23# @leet end
24
25
26def test():
27    assert 2 + 2 == 4
class Solution: 
 3class Solution:
 4    def isSubsequence(self, s: str, t: str) -> bool:
 5        """
 6        This question asks us if `s` is a subsequence of `t`. To find that out,
 7        we can use two pointers. We start out at the beginning of `s` and `t`.
 8        If the current char of `s`, $s_i$ is equal to the current char of `t`,
 9        $t_j$, we increment both `i` and `j` since we've found a match. Otherwise,
10        this is a miss, so we only increment `j`, since we want to continue on
11        for `t`. Finally, we make sure we've reached the end of `s`, so we make
12        sure that `i` is equal to the length of `s`.
13        """
14        i, j, m, n = 0, 0, len(s), len(t)
15
16        while i < m and j < n:
17            l, r = s[i], t[j]
18            if l == r:
19                i += 1
20            j += 1
21        return i == m
def isSubsequence(self, s: str, t: str) -> bool: 
 4    def isSubsequence(self, s: str, t: str) -> bool:
 5        """
 6        This question asks us if `s` is a subsequence of `t`. To find that out,
 7        we can use two pointers. We start out at the beginning of `s` and `t`.
 8        If the current char of `s`, $s_i$ is equal to the current char of `t`,
 9        $t_j$, we increment both `i` and `j` since we've found a match. Otherwise,
10        this is a miss, so we only increment `j`, since we want to continue on
11        for `t`. Finally, we make sure we've reached the end of `s`, so we make
12        sure that `i` is equal to the length of `s`.
13        """
14        i, j, m, n = 0, 0, len(s), len(t)
15
16        while i < m and j < n:
17            l, r = s[i], t[j]
18            if l == r:
19                i += 1
20            j += 1
21        return i == m

This question asks us if s is a subsequence of t. To find that out, we can use two pointers. We start out at the beginning of s and t. If the current char of s, $s_i$ is equal to the current char of t, $t_j$, we increment both i and j since we've found a match. Otherwise, this is a miss, so we only increment j, since we want to continue on for t. Finally, we make sure we've reached the end of s, so we make sure that i is equal to the length of s.

def test(): 
27def test():
28    assert 2 + 2 == 4