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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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