You are given a 0-indexed circular string array words
and a string target
. A circular array means that the array's end connects to the array's beginning.
- Formally, the next element of
words[i]
iswords[(i + 1) % n]
and the previous element ofwords[i]
iswords[(i - 1 + n) % n]
, wheren
is the length ofwords
.
Starting from startIndex
, you can move to either the next word or the previous word with 1
step at a time.
Return the shortest distance needed to reach the string target
. If the string target
does not exist in words
, return -1
.
Input: words = ["hello","i","am","leetcode","hello"], target = "hello", startIndex = 1 Output: 1 Explanation: We start from index 1 and can reach "hello" by - moving 3 units to the right to reach index 4. - moving 2 units to the left to reach index 4. - moving 4 units to the right to reach index 0. - moving 1 unit to the left to reach index 0. The shortest distance to reach "hello" is 1.
Input: words = ["a","b","leetcode"], target = "leetcode", startIndex = 0 Output: 1 Explanation: We start from index 0 and can reach "leetcode" by - moving 2 units to the right to reach index 3. - moving 1 unit to the left to reach index 3. The shortest distance to reach "leetcode" is 1.
Input: words = ["i","eat","leetcode"], target = "ate", startIndex = 0 Output: -1 Explanation: Since "ate" does not exist in words, we return -1.
1 <= words.length <= 100
1 <= words[i].length <= 100
words[i]
andtarget
consist of only lowercase English letters.0 <= startIndex < words.length
impl Solution {
pub fn closet_target(words: Vec<String>, target: String, start_index: i32) -> i32 {
let start_index = start_index as usize;
let n = words.len();
for i in 0..n {
if words[(start_index + i) % n] == target || words[(start_index + n - i) % n] == target
{
return i as i32;
}
}
-1
}
}