Editorial

Solution walkthrough

Review the explanation, then compare the reference implementation.

Video explanation

Solution

Intuition

Between two indices in a circular array, the two directional distances sum to the array length. If the direct absolute difference is d, the shortest route is min(d, n - d). Compare this for every target occurrence.

Approach

Scan the words and, whenever a word equals target, update the minimum with both its direct and wraparound distance from startIndex. Return the minimum, or -1 if no matching word was found. A match at the starting index yields zero, and repeated target words are all considered because the nearest one may lie across the array boundary.

class Solution:
    def closestTarget(self, words: List[str], target: str, startIndex: int) -> int:
        n = len(words)
        res = inf

        for i in range(n):
            if words[i] == target:
                d = abs(i - startIndex)
                res = min(res, d, n - d)

        return res if res != inf else -1

Complexity analysis

Let n = words.length. Time: O(n) for the circular-distance scan. Auxiliary space: O(1).