Solution 1 - Two Pointer

Scan t and advance a pointer in s when characters match.

TimeComplexity: O(|t|)

SpaceComplexity: O(1)

""" 0392.1 - Is Subsequence - Solution 1 - Two Pointer """

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        i = 0
        for ch in t:
            if i < len(s) and s[i] == ch:
                i += 1
        return i == len(s)

if __name__ == "__main__":
    print(Solution().isSubsequence("abc", "ahbgdc"))  # True

fn is_subsequence(s: &str, t: &str) -> bool {
	let sb = s.as_bytes();
	let mut i = 0usize;
	for &ch in t.as_bytes() {
		if i < sb.len() && sb[i] == ch {
			i += 1;
		}
	}
	i == sb.len()
}

fn main() {}

0392 - Is Subsequence

Solution 1 - Two Pointer