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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 } 
  • 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 } }

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