Write a recursive function in Python which should take a nested/embedded list and return us the number of string elements present in it (also counting the strings present in inner lists). Sample calls are shown below;                  >>> countstr([10,'abc',20,'hello']) 2    >>> countstr([10,'abc',2,'20']) 2 >>> countstr([10,'abc',20,['20','fahad',100], None, 10.5]) 3

Write a function definition named first_and_second that takes in sequence and returns the first and second value of that sequence as a list. This is in python Also assert, assert first_and_second([1, 2, 3, 4]) == [1, 2] assert first_and_second(["python", "is", "awesome"]) == ["python", "is"] assert first_and_second(["strawberry", "kiwi", "mango", "guava"]) == ["strawberry", "kiwi"]

Python data structures: write a function that takes in a list (L) as input, and returns the number of inversions. An inversion in L is a pair of elements x and y such that x appears before y in L, but x > y. Your function must run in -place and implement the best possible algoritm of time complexity.  def inversions(L: List) -> int:'''finds and returns number of inversions'''

Complete the recursive function remove_last which creates a new list identical to the input list s but with the last element in the sequence that is equal to x removed. Hint: Remember that you can use negative indexing on lists! For example 1st [-1] refers to the last element in a list 1st, lst [-2] refers to the second to last element... def remove_last(x, s): """Create a new list that is identical to s but with the last element from the list that is equal to x removed. >>> remove_last(1, []) [] >>> remove_last(1, [1]) [] >>> [1] >>> [2] remove_last(1, [1,1]) remove_last(1, [2,1]) >>> remove_last(1, [3,1,2]) [3, 2] >>> remove_last(1, [3,1,2,1]) [3, 1, 2] >>> remove_last (5, [3, 5, 2, 5, 11]) [3, 5, 2, 11] IIIII "*** YOUR CODE HERE ***" Illustrated here is a more complete doctest that shows good testing methodology. It is a little cumbersome as documentation, but you'll want to think about it for your projects. Test every condition that might come up. Then you won't be surprised when…

Please do it on c++ don't use stl function must read the instruction

Python

A sequential search of a sorted list can halt when the target is less than a given element in the list. Modify the program to stop when the target becomes less than the current value being compared. In a sorted list, this would indicate that the target is not in the list and searching the remaining values is unnecessary.   -----------------------------------------------------------------------------------   """ File: search.py Project 11.1 """ def sequentialSearch(target, lyst):     """Returns the position of the target item if found,     or -1 otherwise.     The lyst is assumed to be sorted in ascending order."""     position = 0     while position < len(lyst):         if target == lyst[position]:             return position         position += 1     return -1 def main():     """Tests with three lists."""     print(sequentialSearch(3, [0, 1, 2, 3, 4]))     print(sequentialSearch(3, [0, 1, 2]))     # Should stop at second position.     print(sequentialSearch(3, [0, 4, 5, 6])) if…

No explicit loops must be used. In C# Suppose variable nums contains a list of integers.List«Integer> nums = Arrays.asList(1, 1, 2, 3, 4, 1, 3, 2, 4, 3, 3, 1);var nums = new List<int> { 1, 1, 2, 3, 4, 1, 3, 2, 4, 3, 3, 1 }; // C#Using Java Stream API *or* LINQ, write a short code that prints the following.For each section indicate the platform (whether the solution is given in Java orC#  ILINQ).A) Print all numbers.B) Print the sum() of the numbers.C) count how many times number '1' appears in the list.