Search in a Sorted Array of Unknown Size

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

You may assume all integers in the array are less than 10000, and if you access the array out of bounds,ArrayReader.getwill return2147483647.

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].

Analysis

int hi = 1;
while (reader.get(hi) < target) {
hi = hi * 2;
}
int low = hi / 2;

Solution

Binary Search - Template #3 (end = Integer.MAX_VALUE) - (5 ms, faster than 17.24%)

class Solution {
public int search(ArrayReader reader, int target) {
int start = 0, end = Integer.MAX_VALUE;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (target == reader.get(mid)) {
return mid;
} else if (target < reader.get(mid)) {
end = mid - 1;
} else {
start = mid + 1;
}
}
if (reader.get(start) == target) {
return start;
} else if (reader.get(end) == target) {
return end;
}
return -1;
}
}

Binary Search - Template #1 (hi = hi * 2 while val < target)

class Solution {
public int search(ArrayReader reader, int target) {
int hi = 1;
while (reader.get(hi) < target) {
hi = hi * 2;
}
int low = hi / 2;
while (low <= hi) {
int mid = low + (hi - low) / 2;
int val = reader.get(mid);
if (val == target) {
return mid;
} else if ( val > target) {
hi = mid - 1;
} else {
low = mid + 1;
}
}
return -1;
}
}