carloscn
commented
11 months ago Analysis
pub fn check_distances(s: &str, distance: Vec<i32>) -> bool
{
let mut ret:bool = false;
if s.len() < 1 || distance.len() < 1 {
return ret;
}
let dup_s:Vec<char> = s.chars().collect();
let mut dup_v:Vec<i32> = vec![];
for i in 0..dup_s.len() {
let target = dup_s[i];
for j in (i + 1)..dup_s.len() {
if dup_s[j] == target {
dup_v.push((j - i - 1) as i32);
}
}
}
for i in 0..distance.len() {
if (i < dup_v.len() && dup_v[i] != distance[i]) {
return false;
}
}
ret = true;
return ret;
}
Description
You are given a 0-indexed string s consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26.
Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0, 'b' -> 1, 'c' -> 2, ... , 'z' -> 25).
In a well-spaced string, the number of letters between the two occurrences of the ith letter is distance[i]. If the ith letter does not appear in s, then distance[i] can be ignored.
Return true if s is a well-spaced string, otherwise return false.
Example 1:
Input: s = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0] Output: true Explanation:
Example 2:
Input: s = "aa", distance = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0] Output: false Explanation:
Constraints:
2 <= s.length <= 52 s consists only of lowercase English letters. Each letter appears in s exactly twice. distance.length == 26 0 <= distance[i] <= 50