Description
https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/
This is an interactive problem.
You have a sorted array of unique elements and an unknown size. You do not have an access to the array but you can use the ArrayReader
interface to access it. You can call ArrayReader.get(i)
that:
- returns the value at the
ith
index (0-indexed) of the secret array (i.e.,secret[i]
), or - returns
231 - 1
if thei
is out of the boundary of the array.
You are also given an integer target
.
Return the index k
of the hidden array where secret[k] == target
or return -1
otherwise.
You must write an algorithm with O(log n)
runtime complexity.
Example 1:
Input: secret = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in secret and its index is 4.
Example 2:
Input: secret = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in secret so return -1.
Constraints:
1 <= secret.length <= 104
-104 <= secret[i], target <= 104
secret
is sorted in a strictly increasing order.
Python Solution
First to find an index where its value is no less than the target. Then we can use a binary search to find the index of the target.
# """
# This is ArrayReader's API interface.
# You should not implement it, or speculate about its implementation
# """
#class ArrayReader:
# def get(self, index: int) -> int:
class Solution:
def search(self, reader, target):
"""
:type reader: ArrayReader
:type target: int
:rtype: int
"""
end = 1
while reader.get(end) < target:
end *= 2
start = 0
while start + 1 < end:
mid = start + (end - start) // 2
if reader.get(mid) == target:
return mid
elif reader.get(mid) < target:
start = mid
else:
end = mid
if reader.get(start) == target:
return start
if reader.get(end) == target:
return end
return -1
- Time Complexity: O(log(T)), where T is the index of the target.
- Space Complexity: O(1).
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