Leetcode: Search for a Range

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

Your algorithm's runtime complexity must be in the order of O(log n).

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

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

Understand the problem:
The problem gives a sorted array of integers, find the starting and ending position of a given target value. It requires the time complexity as O(logn), which indicates to use binary search. If not found, return [-1, -1]. 

Solution:
Since the problem asks for O(logn) solution, we can solve this problem by using the binary search for the first and last index of the target.

Code (Java):

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public class Solution {     public int[] searchRange(int[] A, int target) {         if (A == null || A.length == 0) {             int[] result = {-1, -1};             return result;         }         int[] result = new int[2];         // Find the first index of the target value         result[0] = binarySearchFirstIndex(A, target);                   if (result[0] == -1) {             result[1] = -1;             return result;         }                   // Find the last index of the target value         result[1] = binarySearchLastIndex(A, target);                   return result;     }           private int binarySearchFirstIndex(int[] A, int target) {         int lo = 0;         int hi = A.length - 1;                   while (lo + 1 < hi) {             int mid = lo + (hi - lo) / 2;             if (A[mid] == target) {                 hi = mid;             } else if (A[mid] > target) {                 hi = mid;             } else {                 lo = mid;             }         }                   if (A[lo] == target) {             return lo;         }                   if (A[hi] == target) {             return hi;         }                   return -1;     }           private int binarySearchLastIndex(int[] A, int target) {         int lo = 0;         int hi = A.length - 1;                   while (lo + 1 < hi) {             int mid = lo + (hi - lo) / 2;             if (A[mid] == target) {                 lo = mid;             } else if (A[mid] > target) {                 hi = mid;             } else {                 lo = mid;             }         }                   if (A[hi] == target) {             return hi;         }                   if (A[lo] == target) {             return lo;         }                   return -1;     } }