main.cpp

// Source: https://leetcode.com/problems/shortest-distance-to-target-string-in-a-circular-array
// Title: Shortest Distance to Target String in a Circular Array
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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 2.
// - moving 1 unit to the left to reach index 2.
// 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`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <string>
#include <vector>

using namespace std;

// Single Pass
class Solution {
 public:
  int closestTarget(const vector<string>& words, const string& target, const int startIndex) {
    const int n = words.size();

    int minDist = n;
    for (int i = 0; i < n; ++i) {
      if (words[i] != target) continue;

      int dist = abs(startIndex - i);
      minDist = min({minDist, dist, n - dist});
    }

    return minDist == n ? -1 : minDist;
  }
};