README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0392.Is%20Subsequence/README_EN.md	Two Pointers String Dynamic Programming

392. Is Subsequence

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Description

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

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: 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?

Solutions

Solution 1: Two Pointers

We define two pointers $i$ and $j$ to point to the initial position of the string $s$ and $t$ respectively. Each time we compare the two characters pointed to by the two pointers, if they are the same, both pointers move right at the same time; if they are not the same, only $j$ moves right. When the pointer $i$ moves to the end of the string $s$, it means that $s$ is the subsequence of $t$.

The time complexity is $O(m + n)$, where $m$ and $n$ are the lengths of the strings $s$ and $t$ respectively. The space complexity is $O(1)$.

Python3

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

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;
    }
}

C++

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

TypeScript

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

Rust

impl Solution {
    pub fn is_subsequence(s: String, t: String) -> bool {
        let (s, t) = (s.as_bytes(), t.as_bytes());
        let n = t.len();
        let mut i = 0;
        for &c in s.iter() {
            while i < n && t[i] != c {
                i += 1;
            }
            if i == n {
                return false;
            }
            i += 1;
        }
        true
    }
}

C#

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

C

bool isSubsequence(char* s, char* t) {
    int m = strlen(s);
    int n = strlen(t);
    int i = 0;
    for (int j = 0; i < m && j < n; ++j) {
        if (s[i] == t[j]) {
            ++i;
        }
    }
    return i == m;
}