README_EN.md

392. Is Subsequence

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Description

Given two strings s and t, check if s is a subsequence of t.

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: If 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?

Solutions

Python3

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        m, n = len(s), len(t)
        i = j = 0
        while i < m and j < n:
            if s[i] == t[j]:
                i += 1
            j += 1
        return i == m

Java

class Solution {
    public boolean isSubsequence(String s, String t) {
        int m = s.length(), n = t.length();
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (s.charAt(i) == t.charAt(j)) {
                ++i;
            }
            ++j;
        }
        return i == m;
    }
}

TypeScript

function isSubsequence(s: string, t: string): boolean {
    let m = s.length, n = t.length;
    let i = 0;
    for (let j = 0; j < n && i < m; ++j) {
        if (s.charAt(i) == t.charAt(j)) {
            ++i;
        }
    }
    return i == m;
};

C++

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int m = s.size(), n = t.size();
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (s[i] == t[j]) {
                ++i;
            }
            ++j;
        }
        return i == m;
    }
};

Go

func isSubsequence(s string, t string) bool {
	i, j, m, n := 0, 0, len(s), len(t)
	for i < m && j < n {
		if s[i] == t[j] {
			i++
		}
		j++
	}
	return i == m
}

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