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]
iswords[(i + 1) % n]
and the previous element ofwords[i]
iswords[(i - 1 + n) % n]
, wheren
is the length ofwords
.
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]
andtarget
consist of only lowercase English letters.0 <= startIndex < words.length
Solution
shortest-distance-to-target-string-in-a-circular-array.py
class Solution:
def closetTarget(self, words: List[str], target: str, startIndex: int) -> int:
N = len(words)
res = inf
for i, x in enumerate(words):
if x == target:
left = abs(startIndex - i)
right = N - left
# print(words, i, left, right)
res = min(res, min(left, right))
return -1 if res == inf else res