# 2515. Shortest Distance to Target String in a Circular Array
"""
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. 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.
"""
class Solution(object):
def closestTarget(self, words, target, startIndex):
"""
:type words: List[str]
:type target: str
:type startIndex: int
:rtype: int
"""
ans, n = len(words), len(words)
for i, j in enumerate(words):
if j == target:
ans = min(ans, abs(i - startIndex), n - abs(i - startIndex))
return ans if ans < n else -1
if __name__ == "__main__":
obj = Solution()
print(
obj.closestTarget(
words=["hello", "i", "am", "leetcode", "hello"],
target="hello",
startIndex=1,
)
)
print(
obj.closestTarget(words=["a", "b", "leetcode"], target="leetcode", startIndex=0)
)
print(obj.closestTarget(words=["i", "eat", "leetcode"], target="ate", startIndex=0))
