384. Shortest Distance to Target String in a Circular Array

Walk the array once. At every index i whose word equals target, compute the circular distance from startIndex — the smaller of going forward |i - startIndex| or going backward n - |i - startIndex|. Track the minimum.

Why circular distance is min(|i - s|, n - |i - s|)

On a circle of n positions, you can travel from s to i two ways: clockwise ((i - s) mod n steps) or counter-clockwise ((s - i) mod n steps). They sum to n, so the shorter is min(d, n - d) where d = |i - s|.

Steps

1. Initialise best = -1 (sentinel for "no occurrence found yet").
2. Walk i from 0 to n - 1:
   • If words[i] != target, skip.
   • Otherwise compute d = abs(i - startIndex), then d = min(d, n - d).
   • If best == -1 or d < best, set best = d.
3. Return best.

Walk-through

words = ["hello", "i", "am", "coding", "hello"], target = "hello", startIndex = 1. n = 5.

Return 1. ✓

Another, target = "coding", startIndex = 0:

• Only i = 3 matches. d = 3, min(3, 2) = 2. Return 2.

Target not in words: loop never updates best; return -1.

Edge cases

• Target appears at startIndex itself: d = 0, min(0, n) = 0 — return 0 immediately on that iteration.
• Single occurrence: the answer is just that one circular distance.
• No occurrence: return -1 via the sentinel.

Complexity

• Time: O(n) — single pass.
• Space: O(1).

import java.util.*;

class Solution {
    public int closetTarget(String[] words, String target, int startIndex) {
        int n = words.length;
        int best = -1;
        for (int i = 0; i < n; i++) {
            if (!words[i].equals(target)) continue;
            int d = Math.min(Math.abs(i - startIndex), n - Math.abs(i - startIndex));
            if (best == -1 || d < best) best = d;
        }
        return best;
    }
}