Problem: Find First and Last Position of Element in Sorted Array

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

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

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4] Example 2:

Input: nums = [5,7,7,8,8,10], target = 6 Output: [-1,-1] Example 3:

Input: nums = [], target = 0 Output: [-1,-1]

Constraints:

0 <= nums.length <= 105 -109 <= nums[i] <= 109 nums is a non-decreasing array. -109 <= target <= 109

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int len = nums.length-1;
        int[] res = {-1, -1};

        int index = binarySearch(0, len, target, nums);
        if(index == -1) return res;

        int pos = index;

        while(pos != -1){
            res[0] = pos;
            pos = binarySearch(0, pos-1, target, nums);
        }

        pos = index;

        while(pos != -1){
            res[1] = pos;
            pos = binarySearch(pos+1, len, target, nums);
        }

        return res;
    }

    public int binarySearch(int low, int high, int target, int[] nums){
        while(low <= high){
            int mid = low + (high - low)/2;
            if(nums[mid] == target)
                return mid;
            else if(nums[mid] < target)
                low = mid+1;
            else
                high = mid-1;
        }
        return -1;
    }
}