JAVA DEVELOPER

• There is an integer array nums sorted in ascending order (with distinct values). Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], …, nums[n-1], nums[0], nums[1], …, nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

• Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums. You must write an algorithm with O(log n) runtime complexity.

  Example 1:
  Input: nums = [4,5,6,7,0,1,2], target = 0
  Output: 4

  Example 2:
  Input: nums = [4,5,6,7,0,1,2], target = 3
  Output: -1

  Example 3:
  Input: nums = [1], target = 0
  Output: -1

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

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

Example 1:
Input: nums = [1,3,5,6], target = 5
Output: 2

Example 2:
Input: nums = [1,3,5,6], target = 2
Output: 1

Example 3:
Input: nums = [1,3,5,6], target = 7
Output: 4

After applying the sorting algorithm Arrays.sort(numbers);, what will be the content of the numbers array?

Now, consider the following MathOperations class that implements the Calculator interface:

Now, how could you use the MathOperations interface
Options: Calculator and class

to calculate the multiplication of 3.5 and 2.0?

• 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]