**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

Given two strings `s`

and `t`

, return `true`

* if *`s`

* is a subsequence of *

`t`

`false`

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 `"`

while __a__b__c__d__e__"`"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 <= 10`

^{4}`s`

and`t`

consist only of lowercase English letters.

**Follow up:** Suppose there are lots of incoming `s`

, say `s`

where _{1}, s_{2}, ..., s_{k}`k >= 10`

, and you want to check one by one to see if ^{9}`t`

has its subsequence. In this scenario, how would you change your code?

```
class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s) > len(t):return False
if len(s) == 0:return True
subsequence=0
for i in range(0,len(t)):
if subsequence <= len(s) -1:
print(s[subsequence])
if s[subsequence]==t[i]:
subsequence+=1
return subsequence == len(s)
```

```
class Solution {
public:
bool isSubsequence(string s, string t) {
int n = s.length(),m=t.length();
int j = 0;
// For index of s (or subsequence
// Traverse s and t, and
// compare current character
// of s with first unmatched char
// of t, if matched
// then move ahead in s
for (int i = 0; i < m and j < n; i++)
if (s[j] == t[i])
j++;
// If all characters of s were found in t
return (j == n);
}
};
```

```
public class Solution {
public boolean isSubsequence(String s, String t) {
if (s.length() == 0) return true;
int indexS = 0, indexT = 0;
while (indexT < t.length()) {
if (t.charAt(indexT) == s.charAt(indexS)) {
indexS++;
if (indexS == s.length()) return true;
}
indexT++;
}
return false;
}
}
```

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