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# Is Subsequence LeetCode Solution

## Problem – Is Subsequence

Given two strings `s` and `t`, return `true` if `s` is a subsequence of `t`, or `false` otherwise.

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?

### Is Subsequence LeetCode Solution in Python

``````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)``````

### Is Subsequence LeetCode Solution in C++

``````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);
}
};
``````

### Is Subsequence LeetCode Solution in Java

``````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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