Completed all 3 solution

[avcode3]

No description provided.

[super30admin]

Your solution for Problem 1 is correct and efficient. You have correctly implemented two binary search functions to find the first and last occurrences. The code is clean and readable. Well done!

One suggestion for improvement: you can optimize the search for the last occurrence by starting from the first occurrence found (if it exists) instead of starting from the beginning. For example, if you have found the first occurrence at index first, then you know the last occurrence must be at first or to the right. So you can call rightBinary with low = first instead of low = 0. This might reduce the number of comparisons in some cases.

Also, note that in the leftBinary function, the condition if mid == low or nums[mid-1] != nums[mid] is correct. However, using mid == 0 (as in the reference solution) is equally clear and might be slightly more straightforward. Similarly for rightBinary, you can use mid == len(nums)-1 but your condition mid == high is also correct because the high passed in is the last index of the current segment. Since the initial call passes the entire array, it is correct.

Overall, your solution is excellent.