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392. Is Subsequence

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

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:

s = "abc", t = "ahbgdc"

Return true.

Example 2:

s = "axc", t = "ahbgdc"

Return false.

Follow up:

If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

Credits:

Special thanks to @pbrother for adding this problem and creating all test cases.

Solutions (Rust)

1. Two Pointers

impl Solution {
    pub fn is_subsequence(s: String, t: String) -> bool {
        let mut s = s.chars();
        let mut t = t.chars();
        while let Some(ch_s) = s.next() {
            loop {
                match t.next() {
                    Some(ch_t) => {
                        if ch_t == ch_s {
                            break;
                        }
                    },
                    None => return false,
                };
            }
        }
        true
    }
}

2. Binary Search

use std::collections::HashMap;

impl Solution {
    pub fn is_subsequence(s: String, t: String) -> bool {
        let mut char_index = HashMap::new();
        for (i, c) in t.chars().enumerate() {
            char_index.entry(c).or_insert(Vec::new()).push(i);
        }

        let mut prev = 0;
        for c in s.chars() {
            if let Some(v) = char_index.get(&c) {
                match v.binary_search(&prev) {
                    Ok(i) => prev += 1,
                    Err(i) => {
                        if i == v.len() {
                            return false;
                        }
                        prev = v[i] + 1;
                    },
                };
            } else {
                return false;
            }
        }
        true
    }
}