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Formatted question description: https://leetcode.ca/all/2515.html

# 2515. Shortest Distance to Target String in a Circular Array

## Description

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] 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

## Solutions

• class Solution {
public int closetTarget(String[] words, String target, int startIndex) {
int n = words.length;
int ans = n;
for (int i = 0; i < n; ++i) {
String w = words[i];
if (w.equals(target)) {
int t = Math.abs(i - startIndex);
ans = Math.min(ans, Math.min(t, n - t));
}
}
return ans == n ? -1 : ans;
}
}

• class Solution {
public:
int closetTarget(vector<string>& words, string target, int startIndex) {
int n = words.size();
int ans = n;
for (int i = 0; i < n; ++i) {
auto w = words[i];
if (w == target) {
int t = abs(i - startIndex);
ans = min(ans, min(t, n - t));
}
}
return ans == n ? -1 : ans;
}
};

• class Solution:
def closetTarget(self, words: List[str], target: str, startIndex: int) -> int:
n = len(words)
ans = n
for i, w in enumerate(words):
if w == target:
t = abs(i - startIndex)
ans = min(ans, t, n - t)
return -1 if ans == n else ans


• func closetTarget(words []string, target string, startIndex int) int {
n := len(words)
ans := n
for i, w := range words {
if w == target {
t := abs(i - startIndex)
ans = min(ans, min(t, n-t))
}
}
if ans == n {
return -1
}
return ans
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

• function closetTarget(
words: string[],
target: string,
startIndex: number,
): number {
const n = words.length;
for (let i = 0; i <= n >> 1; i++) {
if (
words[(startIndex - i + n) % n] === target ||
words[(startIndex + i) % n] === target
) {
return i;
}
}
return -1;
}


• 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 >> 1 {
if words[(start_index - i + n) % n] == target || words[(start_index + i) % n] == target
{
return i as i32;
}
}
-1
}
}