This function takes a string, text, and returns number of spaces in that string. Identify the loop exit condition, the loop post-condition, and the loop invariant at the end of each iteration of the loop, which shows the algorithm's correctness. def count_spaces (text

"Simon Says" is a memory game where "Simon" outputs a sequence of 10 characters (R, G, B, Y) and the user must repeat the sequence. Create a for loop that compares the two strings starting from index 0. For each match, add one point to userScore. Upon a mismatch, exit the loop using a break statement. Assume simonPattern and userPattern are always the same length. Ex: The following patterns yield a userScore of 4:simonPattern: RRGBRYYBGY userPattern: RRGBBRYBGY #include <iostream>#include <string>using namespace std; int main() {   string simonPattern;   string userPattern;   int userScore;   int i;    userScore = 0;    cin >> simonPattern;   cin >> userPattern;    /* Your solution goes here  */    cout << "userScore: " << userScore << endl;    return 0;

Loops: Based on the following code segment: int val = 10; String s = String.format("%d", val - 5); while (val < 25) { s +=String.format(",%3d", val); val += 5; } System.out.println(s); Select the code using a for loop that produces identical output: String s = String.format("%d", val); for (int val = 10; val < 25; val + 5) { s += String.format(",%3d", val); } System.out.println(s); Strings = String.format("%d", 5); for (int val = 10; val <= 25; val += 5) { s +=String.format(",%3d", val); } System.out.println(s); String s = String.format("%d", 5); for (int val s += String.format(",%3d", val); } == 10; val < 25; val + 5) { System.out.println(s); None of these will produce the same output. String s = String.format("%d", 5); for (int val = 10, val < 25, val += 5) { s += String.format(",%3d", val); } System.out.println(s); Strings = String.format("%d", 5); for (int val = 10; val < 25; val += 5) { s +=String.format(",%3d", val); } System.out.println(s); String s = String.format("%d", 5); for…

Lab Goal : This lab was designed to teach you how to use a while loop.   Lab Description (JAVA PROGRAMMING): Remove all occurrences of the removal string from the original string. Each time you take out the removal string you must also remove the letter that preceded the removal string. After you remove a string and its leading character, you must check to see if any more occurrences of the removal string exist. Sample Data : xR-MxR-MHelloxR-M R-M sxsssxssxsxssexssxsesss xs fuxqwexqwertyxqwexqwe     Sample Output : xR-MxR-MHelloxR-M - String to remove R-M Hello sxsssxssxsxssexssxsesss - String to remove xs sesss fuxqwexqwertyxqwexqwertyrtyxqwertyrtyn - String to remove qwerty fun dogdogcatddodogdogdoggog - String to remove dog catgog   algorithm help use indexOf to look for more removals while you have more removals { take out the current removal using substring use indexOf to look for more removals }           I need help with assignment: "String Cleaner"   StringRemover.js: //(c) A+…

Enter a list of at least three elements for y such that the value of x is 10 when the loop completes. x = 0 for value in y: if len(value > x: x = len(value)

exercise5.py 1 Your code goes here... Guide Collapse Strings -> Coding Exercises - Strings Exercise 5 Exercise 5 Problem Write a program that captures input from the user. Then, swap the letters two at a time in the string. The first two characters change places, the third and fourth characters change places, etc. Assume that the user will only input strings with an even number of characters. Important, do not put a prompt when asking for user input. Just use input(). Adding a prompt will cause your program to not pass the tests. Expected Output • If the user inputs home, then the output is ohem . If the user inputs cars then the output is acer Code Visualizer TRY IT Reminder, do not put a prompt when collecting user input. Just use input() Check It! (1 left) TTT

Java Task Write a for loop that prints the characters stored in a String object backward. The String is entered by the user.  ***Use ch = phrase.charAt(i); to get the character that is in index i of the String phrase.**** For example The user enters “Good morning!” The printout should be “!gninrom dooG”.

True or False: When asking user for data, we can always turn an algorithm that uses a user-confirmation loop into one that uses a sentinel loop. Select one: a. False b. True

PART F: MIDDLE TWOGiven a string str, print out a string made of the middle two chars. Assume the string length is at least 2 and iseven.Methods to use: substring(), length()Testing cases: string => ri code => od Practice => ct ab => ab 0123456789 => 45

Aim: Read a string, S and print its integer value. If S cannot be converted to   an integer, print Bad String.       '''       # getting the input   S = input().strip()   try:   # if it's possible to convert the entered string into an integer then this block will execute   print(int(S))   except:   # if it's not possible, then this block will be executed   print('Bad String')       '''       COMPLEXITY:       Time Complexity -> O(1)   Space Complexity -> O(1)       Sample Input 1:   3   Sample Output 1:   3   Sample Input 2:   SB   Sample Output 2:   Bad String       Explaination:   '3' as a string can be converted into the integer 3, whereas SB can't be   converted to an integer hence the 'except' block is executed and 'Bad String'   is printed.