# Leetcode: Search in a Sorted Array of Unknown Size

Search in a Sorted Array of Unknown Size

Similar Problems:

Given an integer array sorted in ascending order, write a function to search target in nums. If target exists, then return its index, otherwise return -1. However, the array size is unknown to you. You may only access the array using an ArrayReader interface, where ArrayReader.get(k) returns the element of the array at index k (0-indexed).

You may assume all integers in the array are less than 10000, and if you access the array out of bounds, ArrayReader.get will return 2147483647.

Example 1:

```Input: array = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4
```

Example 2:

```Input: array = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1
```

Note:

1. You may assume that all elements in the array are unique.
2. The value of each element in the array will be in the range [-9999, 9999].

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution: search from beginning to the end
```## Blog link: https://code.dennyzhang.com/search-in-a-sorted-array-of-unknown-size
## Basic Ideas:
##  The array has no more than 10000 items.
##  So we simply search from begining to the end.
## Complexity: Time O(1), Space O(1)
class Solution:
"""
:type target: int
:rtype: int
"""
i = 0
while True:
if v == 2147483647: return -1
if v == target: return i
i += 1
return -1
```

• Solution: binary search
```## Blog link: https://code.dennyzhang.com/search-in-a-sorted-array-of-unknown-size
## Basic Ideas: binary search
##
##  Find the right boundry
##
## Complexity: Time O(1), Space O(1)
class Solution:
"""
:type target: int
:rtype: int
"""
hi = 1
while reader.get(hi) < target: hi *= 2
lo = int(hi/2)

# find the first element which equals the target
while lo<=hi:
mid = lo + int((hi-lo)/2)