CHALLENGE 2.7.2: Alphabetic detect. ACTIVITY Write code to print the location of any alphabetic character in the 2-character string passCode. Each alphabetic character detected should print a separate statement followed by a newline. Ex: If passCode is "9a", output is: Alphabetic at 1 Hint: Use two if statements to check each of the two characters in the string, using Character.isLetter(). 422784.2543518.qx3zqy7 1 import java.util.Scanner; 2 3 public class FindAlpha { 4 5 6 7 8 9 10 11 12 13} Run public static void main (String [] args) { Scanner scnr = new Scanner(System.in); String passCode; passCode = scnr.next(); if | } Doll 1 test passed All tests passed

CHALLENGE 2.7.2: Alphabetic detect. ACTIVITY Write code to print the location of any alphabetic character in the 2-character string passCode. Each alphabetic character detected should print a separate statement followed by a newline. Ex: If passCode is "9a", output is: Alphabetic at 1 Hint: Use two if statements to check each of the two characters in the string, using Character.isLetter(). 422784.2543518.qx3zqy7 1 import java.util.Scanner; 2 3 public class FindAlpha { 4 5 6 7 8 9 10 11 12 13} Run public static void main (String [] args) { Scanner scnr = new Scanner(System.in); String passCode; passCode = scnr.next(); if | } Doll 1 test passed All tests passed

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

Q4: Colouring Book IHaving spent the last couple of hours grading assignments, I decide it's time for a break. I take out my favourite colouring book, turn to a random page I haven't coloured in yet, and lay it on my desk. I then take out all my n crayons and line them up on the desk (it's a very long desk). The colour of the " erayon is a string c11 (eg. "alue"). Many of the crayons have the same colour. In fact, no matter how many crayons I have, there are at most 30 distinct colours amongst them. To start colouring, I always take a sublist (see Qi for a definition) of the crayons laid on the desk and put away the rest (too many options can be overwhelming and can lead to indecision). I take a look at the line art in front of me and wonder, "Ilow many different colours do I need to make this lonk great? One? Two? Mayhe three?". Now you understand my dilemma and are fully aware of my indecision. You kindly decide to help me out by telling me for every number k, if I were to use k…
Write an application that counts by five from 5 through 500 inclusive and that starts a new line after every multiple of 50 (50,100,150 and so on).Save the file as CountByAnything.java
modifying following code and write comments for each line of codes. #include <stdio.h>#include <stdlib.h>#include <stdbool.h>#include <string.h> #define MAX_PASSWORD_LENGTH 128#define MIN_PASSWORD_LENGTH 10#define ALLOW_PASSPHRASE true#define MIN_PHRASE_LENGTH 20#define OPTIONAL_TESTS_REQUIRED true#define MIN_OPTIONAL_TESTS_TO_PASS 4 bool isStrongPassword(char *password);void printTestResults(char *password, bool isPassphrase, int optionalTestsPassed); int main() { char passwords[][128] = { "password", "mypassword", "thisismypassword", "passssword", "This is my password phrase1", "Tinypw1", "Ireallydontlikehavingtomakeupnewpasswordsallthetime1", "Iloveyouxxxooo1", "Boom**********!", "IHATEPWORDS1!", "ihatepwords1!", "IHatePwords!", "IHatePwords", "my pass phrase does not need to pass tests", "short pass phrase", "x", "x1",…
"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;
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+…
Complete the rotate_text() function that takes 2 parameters, a string data and an integer n. If n is positive, then the function will shift all the characters in data forward by n positions, with characters at the end of the string being moved to the start of the string. If n is 0 then the text remains the same. For example: rotate_text('abcde', rotate_text('abcde', rotate_text('abcde', 1) would return the string 'eabcd' 3) would return the string 'cdeab' 5) would return the string 'abcde' rotate_text('abcde', 6) would return the string 'eabcd' ... and so on. If n is negative, then the function will shift the characters in data backward by n positions, with characters at the start of the string being moved to the end of the string. For example: rotate text('abcde', -1) would return the string 'bcdea'
DO NOT COPY FROM OTHER WEBSITES Correct and detailed answer will be Upvoted else downvoted. Thank you!
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.