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2515. Shortest Distance to Target String in a Circular Array

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 stringtarget. 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 (Rust)

1. Solution

implSolution{pubfncloset_target(words:Vec<String>,target:String,start_index:i32) -> i32{let start_index = start_index asusize;let n = words.len();for i in0..n {if words[(start_index + i) % n] == target || words[(start_index + n - i) % n] == target {return i asi32;}} -1}}