def ALGORITHM (text): processor := list containing '0' i = 1 while i < length (text): if text[i] == "s": push "s" onto processor else if text[i] == "d": push "d" onto processor else if text[i] == "p": push "p" onto processor else if text[i] == "q": push "q" onto processor else: v = pop from processor v := v * 10 v := v + numeric value of text[i] digit push v onto processor i := i + 1 loop while length (processor) > 1 t2 = processor.pop() c = processor.pop() t1 = processor.pop() if c == "s": push t1 + t2 onto processor else if c == "d": push tl else if c == push tl else if c == "q": t2 onto processor "p": t2 onto processor push t1 / t2 onto processor loop return = pop from processor

Develop Pseudo-code (English-like) for:    In the Find Largest pseudo-code algorithm of Figure 2.14, listed below, if the numbers in our list were not unique and therefore the largest number could occur more than once. Modify the algorithm below to find all the occurrences of the largest number and their position in the list.

Python

c++

ORIGINAL Q: Instead of reading in an entire list N1,N2,⋯N1​,N2,⋯​ all at once, some algorithms (depending on the task to be done) can read in only one element at a time and process that single element completely before inputting the next one. This can be a useful technique when the list is very big (e.g., billions of elements) and there might not be enough memory in the computer to store it in its entirety. Write an algorithm that reads in a sequence of values v≥0, one at a time, and computes the average of all the numbers. You should stop the computation when you input a value of V = −1. Do not include this negative value in your computations; it is not a piece of data but only a marker to identify the end of the list. PLEASE MODIFY THE FOLLOWING CODE SO THE USER INPUTS THE NUMBERS ONE BY ONE: ORIGINAL Q: Instead of reading in an entire list N1,N2,⋯N1 ,N2,⋯ all at once, some algorithms (depending on the task to be done) can read in only one element at a time and process that single…

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…

Simple Python Coding Exercise   Code your program according to the following algorithm 1. Prompt the user for a string of words on one line separated by spaces.2. Convert the input string a list of words called input_list. (Hint: use split())3. Create a new empty list called output_list (i.e., output_list=[])4. Scan input_list to find the smallest word.    When the smallest word is found, append that word to output_list and remove it from input_list.5. If input_list still has words in it, go to step 4.6. When input_list is empty, convert output_list into a string of words separated by spaces. (Hint: use join())7. Print the string.

Exercises Input a list from user. User should enter the length first and follow that by the elements (element by element). Find the location of an input value inside a list. > Count how many times an input value appears in a list. > Sort the elements of a list in descending order. * Use bubble sorting