Find First and Last Position of Element in Sorted Array Test

The solution approach of the problem is to initialize two pointers, left and right, which represent the boundaries of the search range. Initially, set left = 0 and right = len(nums) - 1. a. Calculate the mid index as mid = (left + right) // 2. b. If the value at nums[mid] is greater than or equal to the target, update right = mid - 1 to search the left half of the array. c. If the value at nums[mid] is less than the target, update left = mid + 1 to search the right half of the array. Repeat steps b-c until left > right. After the binary search, check if the value at index left is equal to the target. If it is not, return [-1, -1] since the target is not present in the array. If the value at index left is equal to the target, it represents the leftmost occurrence of the target in the array. We can then perform a second binary search to find the rightmost occurrence of the target. Let's call this function findRightPosition(nums, target). Re-initialize left and right to the original values: left = 0 and right = len(nums) - 1. a. Calculate the mid index as mid = (left + right) // 2. b. If the value at nums[mid] is greater than the target, update right = mid - 1 to search the left half of the array. c. If the value at nums[mid] is less than or equal to the target, update left = mid + 1 to search the right half of the array. Repeat steps b-c until left > right. After the binary search, check if the value at index right is equal to the target. If it is not, decrement right by 1 to obtain the index of the rightmost occurrence of the target. Finally, return the results as [left, right], representing the first and last positions of the target in the sorted array.