Easy
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.
words[i]
is words[(i + 1) % n]
and the previous element of words[i]
is words[(i - 1 + n) % n]
, where n
is the length of words
.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
.
Example 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.
Example 2:
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.
Example 3:
Input: words = [“i”,”eat”,”leetcode”], target = “ate”, startIndex = 0
Output: -1
Explanation: Since “ate” does not exist in words
, we return -1.
Constraints:
1 <= words.length <= 100
1 <= words[i].length <= 100
words[i]
and target
consist of only lowercase English letters.0 <= startIndex < words.length
class Solution {
fun closetTarget(words: Array<String>, target: String, startIndex: Int): Int {
val n = words.size
if (words[startIndex] == target) {
return 0
}
var ld = -1
var rd: Int
var ans = Int.MAX_VALUE
var i = (startIndex + 1) % n
while (i != startIndex) {
if (words[i] == target) {
ld = if (i > startIndex) startIndex + (n - i) else startIndex - i
rd = if (i > startIndex) i - startIndex else n - startIndex + i
ans = Math.min(ans, Math.min(ld, rd))
}
i = (i + 1) % n
}
return if (ld == -1) {
-1
} else ans
}
}