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):

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;
    }
}