34. Search for a Range

https://leetcode.com/problems/search-for-a-range/description/

Given an array of integers sorted in ascending order, 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].

Sol:

class Solution {
    public int[] searchRange(int[] nums, int target) {
        
        // ordered sequence. Binary search
        //Time O(logn) Space O(1)
        // implement upper and lower bound function
        
        
        // lower and upper are index elements
        int lower = lower_bound(nums, 0, nums.length, target);
        int upper = upper_bound(nums, 0, nums.length, target);
        
        if (lower == nums.length || nums[lower] != target)
            return new int[]{-1, -1};
        else
            return new int[]{lower, upper - 1};
        
    }
    
    int lower_bound (int[] A, int first, int last, int target){
        while(first != last){
            int mid = first + (last - first) / 2;
            if(target > A[mid]) first = ++ mid;
            else
                last = mid;
        }
        
        return first;
    }
    
    int upper_bound (int[] A, int first, int last, int target){
        while (first != last){
            int mid = first + (last - first) / 2;
            if (target >= A[mid]) first = ++ mid;
            else
                last = mid;
        }
        
        return first;
    }
    
}