2515

2515. Shortest Distance to Target String in a Circular Array (Easy)

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

Companies: Bloomberg

Related Topics:
Array, String

Similar Questions:

• Defuse the Bomb (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/shortest-distance-to-target-string-in-a-circular-array
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int closetTarget(vector<string>& A, string target, int startIndex) {
        int N = A.size(), ans = INT_MAX;
        for (int i = 0; i < N; ++i) {
            if (A[i] != target) continue;
            int d = abs(i - startIndex);
            ans = min({ans, d, N - d});
        }
        return ans == INT_MAX ? -1 : ans;
    }
};