# Search for a Range

Medium

## Question

Given a sorted array of n integers, find the starting and ending position of a given target value.

If the target is not found in the array, return [-1, -1].

Example
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

O(logn) time

## Thinking

Use two binary search tree to find the Range

## Solution

### Java

``````class Solution {
public int[] searchRange(int[] A, int target) {
if (A == null || A.length == 0) {
return new int[]{-1, -1};
}
int len = A.length;
if (target < A[0] || target > A[len - 1]) {
return new int[]{-1, -1};
}
int start = 0, end = len - 1, mid = 0, start2 = -1;
// first binary search to find start index
while (start + 1 < end) {
mid =  start + (end - start) / 2;
if (A[mid] == target) {
if (start2 == -1) {
start2 = mid;
}
end = mid;
} else if (A[mid] > target) {
end = mid;
} else {
start = mid;
}
}
if (A[start] == target) {
// noop
} else if (A[end] == target) {
start = end;
} else {
return new int[]{-1, -1};
}
end = len - 1;
// 2nd binary search to search right index
while (start2 + 1 < end) {
mid = start2 + (end - start2) / 2;
if (A[mid] == target) {
start2 = mid;
} else if (A[mid] > target) {
end = mid;
} else {
start2 = mid;
}
}
if (A[end] == target) {
// noop
} else if (A[start2] == target) {
end = start2;
} else {
return new int[]{-1, -1};
}
return new int[] {start, end};
}
}
``````