class Solution {
private:
    int binarySearch(vector<int>& arr, int l, int r, int target)
    {
        if (r >= l) {
            int mid = l + (r - l) / 2;

            // If the element is present at the middle
            // itself
            if (arr[mid] == target)
                return mid;

            // If element is smaller than mid, then
            // it can only be present in left subarray
            if (arr[mid] > target)
                return binarySearch(arr, l, mid - 1, target);

            // Else the element can only be present
            // in right subarray
            return binarySearch(arr, mid + 1, r, target);
        }

        // We reach here when element is not
        // present in array
        return -1;
    }

public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> returnResult;
        int resultindex = binarySearch(nums, 0, nums.size()-1, target);
        if(resultindex == -1) {
            returnResult.push_back(-1);
            returnResult.push_back(-1);
        } else {
            //now that we have an index of 1 item, search start and end.
            int start=resultindex;
            int end=resultindex;

            while(nums[start]==target && start > 0) {
                start--;
            }
            if(nums[start] != target) {
                start++;
            }

            while(nums[end]==target && end < nums.size()) {
                end++;
            }
            if(nums[end] != target) {
                end--;
            }

            returnResult.push_back(start);
            returnResult.push_back(end);
        }

        return returnResult;
    }
};