L= (0'12 I i = j or j = k} and generate leftmost derivation for the string 01122. %3D
Q: L= (0'12 I i = j orj = k} and generate leftmost derivation for the string 01122. %3D
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Construct the CFG for the following languages
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- An string should taken from user and check if it contains uppercase charcters or not if yes print those else print no such elementComputer Science This is a codingbat problem. Please answer in Java and do not change the starter code. You have been asked to write the game of tic tac toe. In the first version of the game, we will only be comparing horizontally for a win. A win is determine by all of the elements in a row either being an X or an O. You need to return a String with the winner or TIE if there is no winner. ticTacToe(["X", "O", "O"], ["X", "O", "O"], ["X", "X", "X"]) → "X"ticTacToe(["X", "O", "X"], ["X", "X", "O"], ["O", "O", "O"]) → "O"ticTacToe(["X", "O", "X"], ["X", "X", "O"], ["O", "X", "O"]) → "TIE" public String ticTacToe(String[] top, String[] middle, String[] bottom) { }public static String pancakeScramble(String text) This nifty little problem is taken from the excellent Wolfram Challenges problem site where you can also see examples of what the result should be for various arguments. Given a text string, construct a new string by reversing its first two characters, then reversing the first three characters of that, and so on, until the last round where you reverse your entire current string.This problem is an exercise in Java string manipulation. For some mysterious reason, the Java String type does not come with a reverse method. The canonical way to reverse a Java string str is to first convert it to mutable StringBuilder, reverse its contents, and convert the result back to an immutable string, that is,str = new StringBuilder(str).reverse().toString(); Here's the tester it must pass: @Test public void testPancakeScramble() throws IOException {// Explicit test casesassertEquals("", P2J3.pancakeScramble(""));assertEquals("alu",…
- String spell = "Redikulus"; How would you find the index of the 'u' just before the last 's' using a single line of code? Multiple answers may be correct. Given the string String spell = "Redikulus"; How would you find the index of the 'u' just before the last 's' using a single line of code? Multiple answers may be correct. A. spell.substring(5, spell.length()).indexOf('u'); B. spell.indexOf('u', 5); C. spell.lastIndexOf('u'); D. spell.contains('u'); E. spell.indexOf('u', spell.length() - 2); F. spell.indexOf('u', spell.indexOf('l'));Alert dont submit AI generated answer. The dfa M in Jexercise2.6 accepts the string w = a. Modify M to create a new automaton M0, so that L(M0) = L(M) − {a}For any element in keysList with a value smaller than 40, print the corresponding value in itemsList, followed by a comma (no spaces). Ex: If keysList = {32, 105, 101, 35} and itemsList = {10, 20, 30, 40}, print: 10,40, #include <iostream>#include <string.h>using namespace std; int main() { const int SIZE_LIST = 4; int keysList[SIZE_LIST]; int itemsList[SIZE_LIST]; int i; cin >> keysList[0]; cin >> keysList[1]; cin >> keysList[2]; cin >> keysList[3]; cin >> itemsList[0]; cin >> itemsList[1]; cin >> itemsList[2]; cin >> itemsList[3]; /* Your code goes here */ cout << endl; return 0;}
- Please help me with this task. Thank you! public class CustomString { String myString; boolean isSet; public CustomString() { this.myString = null; this.isSet = false; } /** * Returns a new string version of the current string where the alphabetical characters specified in the given arg, are removed. * * The alphabetical characters to be removed are case insensitive. * All non-alphabetical characters are unaffected. * If the current string is null, empty, or has not been set to a value, this method should return an empty string. * * Example(s): * - For a current string "my lucky numbers are 6, 8, and 19.", calling remove("ra6") would return "my lucky numbes e 6, 8, nd 19." * - For a current string "my lucky numbers are 6, 8, and 19.", calling remove("6,.") would return "my lucky numbers are 6, 8, and 19." * - For a current string "my lucky numbers are 6, 8, and 19.", calling remove("") would return "my…Create an iterator that returns numbers, starting with 1, and each sequence will increase by one (returning 1,2,3,4,5 etc.):Not allowed to add variables in brackets next to originIndex * @return index of the point that is closest to the origin, which is (0, 0) In * case of a tie, return the lowest index */ public int closestToOriginIndex() {
- PART E: ROTATE LEFTGiven a string str, print out a rotated left 2 version where the first 2 chars are moved to the end. Assume thestring length will be at least 2.Methods to use: substring(), length(), concat()Testing cases: Hello => lloHe java => vaja Hi => Hi code => deco cat => tca 12345 => 34512 Chocolate => ocolateCh bricks => icksbFind errors in this code also: internalstatdchkPalindrome(string str) { for(int i = 0, j = str.Length - 1; i < str.Length / 2; i++, j--) { if(str[i] != str[j]) { flag = false; break; } } if(flag) { ome"); } else Console.ine("Not Palindrome"); }For any element in keysList with a value smaller than 40, print the corresponding value in itemsList, followed by a space. Ex: If keysList = {32, 105, 101, 35} and itemsList = {10, 20, 30, 40}, print: 10 40 #include <iostream>#include <string.h>using namespace std; int main() { const int SIZE_LIST = 4; int keysList[SIZE_LIST]; int itemsList[SIZE_LIST]; int i; cin >> keysList[0]; cin >> keysList[1]; cin >> keysList[2]; cin >> keysList[3]; cin >> itemsList[0]; cin >> itemsList[1]; cin >> itemsList[2]; cin >> itemsList[3]; /* Your code goes here */ cout << endl; return 0;}