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).
s = "abc"
, t = "ahbgdc"
Return true
.
s = "axc"
, t = "ahbgdc"
Return false
.
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?
Special thanks to @pbrother for adding this problem and creating all test cases.
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
}
}
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
}
}