Given a string, write a method to find the first non-repeating vowel (a, e, i, o, u) in it and return its index. If it doesn't exist, return -1. Assume that all the characters of the String is lowercase. Do not use String class functions other than charAt() and length() You can also try to solve with Map data structure like using a Map variable.
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- Write the value_equality method.Given the two expressions ? and ?, first compute the set of variables ? that appear in either ? or ?. Then, the idea consists in performing num_sample times the following test for equality: First, produce a variable assignment (a dictionary) mapping each variable in ? to a random value. Choose these random values from the gaussian distribution centered around 0 and with standard deviation 10 (for instance; any continuous distribution with infinite domain would work). You can obtain such numbers using random.gauss(0, 10). Then, compute the values of ? and ? with respect to that variable evaluation. If the values are closer than a specified tolerance tolerance, you consider ? and ? equal (for that variable valuation). Otherwise, you can stop and return that ? and ? are different. If you can repeat the process num_sample times, and ? and ? are considered equal every time, then you declare them equal. WRITE CODE IN PYTHONImplement a function called tautology? that takes as input a fully parenthesized formula and returns true if it is a tautology and false otherwise. As was the case in the previous lab, the formula will contain at least one set of parentheses for each operator, but may contain more. The best strategy for doing is to use an insight called Quine's method. It is based on the observation that if a formula, such as (p or (not p)), is a tautology, the result of substituting p with true is a tautology and the result of substituting p with false is also a tautology. This implies a computational strategy: to evaluate whether a formula F is a tautology1. Collect all of the propositional variables.2. Substitute every occurrence of the first variable with #t and every occurrence ofthe same variable with #f3. Make two recursive calls to Tautology? “And” the results.4. When all possible substitutions have been made, i.e., the formula contains onlytruth values, evaluate the formula and return the…Implement a function called tautology? that takes as input a fully parenthesized formula and returns true if it is a tautology and false otherwise. As was the case in the previous lab, the formula will contain at least one set of parentheses for each operator, but may contain more. The best strategy for doing is to use an insight called Quine's method. It is based on the observation that if a formula, such as (p or (not p)), is a tautology, the result of substituting p with true is a tautology and the result of substituting p with false is also a tautology. This implies a computational strategy: to evaluate whether a formula F is a tautology 1. Collect all of the propositional variables. 2. Substitute every occurrence of the first variable with #t and every occurrence of the same variable with #f 3. Make two recursive calls to Tautology? “And” the results. 4. When all possible substitutions have been made, i.e., the formula contains only truth values, evaluate the formula and return…
- solve in C please. Implement the following two functions that get a string, and compute an array of non-emptytokens of the string containing only lower-case letters. For example:● For a string "abc EFaG hi", the list of tokens with only lower-case letters is ["abc", "hi"].● For a string "ab 12 ef hi ", the list of such tokens is ["ab","ef","hi"].● For a string "abc 12EFG hi ", the list of such tokens is ["abc","hi"].● For a string " abc ", the list of such tokens is ["abc"].● For a string "+*abc!! B" the list of such tokens is empty.That is, we break the string using the spaces as delimiters (ascii value 32), and look only at thetokens with lower-case letters only .1. The function count_tokens gets a string str, and returns the number ofsuch tokens.int count_tokens(const char* str);For example● count_tokens("abc EFaG hi") needs to return 2.● count_tokens("ab 12 ef hi") needs to return 3.● count_tokens("ab12ef+") needs to return 0.2. The function get_tokens gets a string str, and…Please implement in Java implement a keyed bag in which the items to be stored are strings (perhaps people’s names) and the keys are numbers (perhaps Social Security or other identification numbers). So, the in- sertion method has this specification: public void insert(String entry, int key);// Precondition: size( ) < CAPACITY, and the // bag does not yet contain any item// with the given key.// Postcondition: A new copy of entry has// been added to the bag, with the given key. When the programmer wants to remove or retrieve an item from a keyed bag, the key of the item must be specified rather than the item itself. The keyed bag should also have a boolean method that can be used to determine whether the bag has an item with a specified key. In a keyed bag, the pro- grammer using the class specifies a particular key when an item is inserted. Here’s an implementation idea: A keyed bag can have two private arrays, one that holds the string data and one that holds the corresponding…def reverse_sentence(s: str) -> str:"""Given a sentence <s>, we define a word within <s> to be a continuoussequence of characters in <s> that starts with a capital letter andends before the next capital letter in the string or at the end ofthe string, whichever comes first. A word can include a mixture ofpunctuation and spaces.This means that in the string 'ATest string!', there are in fact only twowords: 'A' and 'Test string!'. Again, keep in mind that words start with acapital letter and continue until the next capital letter or the end of thestring, which is why we consider 'Test string!' as one word.This function will reverse each word found in the string, and return a newstring with the reversed words, as illustrated in the doctest below.>>> reverse_sentence('ATest string!')'A!gnirts tseT' """ do it on python, and do not use any list, import or split, or append or join.
- def reverse_sentence(s: str) -> str:"""Given a sentence <s>, we define a word within <s> to be a continuoussequence of characters in <s> that starts with a capital letter andends before the next capital letter in the string or at the end ofthe string, whichever comes first. A word can include a mixture ofpunctuation and spaces.This means that in the string 'ATest string!', there are in fact only twowords: 'A' and 'Test string!'. Again, keep in mind that words start with acapital letter and continue until the next capital letter or the end of thestring, which is why we consider 'Test string!' as one word.This function will reverse each word found in the string, and return a newstring with the reversed words, as illustrated in the doctest below.>>> reverse_sentence('watchNow')'watchwoN' >>> reverse_sentence('hot')'hot' """ do it on python, and do not use any list, import or split, or append or join.def reverse_sentence(s: str) -> str:"""Given a sentence <s>, we define a word within <s> to be a continuoussequence of characters in <s> that starts with a capital letter andends before the next capital letter in the string or at the end ofthe string, whichever comes first. A word can include a mixture ofpunctuation and spaces.This means that in the string 'ATest string!', there are in fact only twowords: 'A' and 'Test string!'. Again, keep in mind that words start with acapital letter and continue until the next capital letter or the end of thestring, which is why we consider 'Test string!' as one word.This function will reverse each word found in the string, and return a newstring with the reversed words, as illustrated in the doctest below.>>> reverse_sentence('ATest string!')'A!gnirts tseT'""" passdef reverse_sentence(s: str) -> str:"""Given a sentence <s>, we define a word within <s> to be a continuoussequence of characters in <s> that starts with a capital letter andends before the next capital letter in the string or at the end ofthe string, whichever comes first. A word can include a mixture ofpunctuation and spaces.This means that in the string 'ATest string!', there are in fact only twowords: 'A' and 'Test string!'. Again, keep in mind that words start with acapital letter and continue until the next capital letter or the end of thestring, which is why we consider 'Test string!' as one word.This function will reverse each word found in the string, and return a newstring with the reversed words, as illustrated in the doctest below.>>> reverse_sentence('ATest string!')'A!gnirts tseT' """
- Complete the method below that prints all elements that are in both Set<String> s and Set<String> t. Each matching element should be followed by a space. What can I change to make it work? import java.util.Set;import java.util.Iterator; public class Sets{ public static void printMatches(Set<String> s, Set<String> t) { System.out.print("{ "); while(in.hasNext()){ String next = in.next(); { if((s.contains(next))&& (t.contains(next))){ System.out.print(next+" "); } } System.out.println("}"); }} import java.util.Set;import java.util.TreeSet;import java.util.Arrays; public class SetsTester{ public static void main(String[] args) { Set<String> set1 = new TreeSet<>(Arrays.asList("Bob", "Bill", "Sally")); Set<String> set2 = new TreeSet<>(Arrays.asList("Sam", "Bob", "Frank")); Sets.printMatches(set1, set2);…The map file holds a bunch of rectangles. Some of the rectangles are solid, and you can think of them like crates in a warehouse. They can be enclosed by a non-solid rectangle, and those rectangles are just there to make theintersection testing and such go faster. If a line doesn't cross a non-solid rectangle, then it cannot hit any rectangle inside it. Nice! The format is:Solid(1)/Nonsolid(0) x_min y_min x_max y_max (list of children iff not solid) The list of children are line numbers (the lines start at 0) of solid rectangles inside the non-solid rectangle. Solid rectangles do not have children. All non-solid rectangles must have children. Here's a sample file:0 0 0 100 100 1 21 1.1 1.0 1.2 1.31 10 10 11 110 50 60 70 80 40 55 65 65 75 51 56 66 64 74 The first line (line 0) is a non-solid rectangle with two children (the next two rows, lines 1 and 2). The first rectangle ranges from (0,0) to (100,100). The children of it are both small solid rectangles, the first is only .1 x .3,…Improve the given code below, make sure that the code will require input from the user. Problem: Write a JavaScript function that would accept a string and return the reverse of that string. const reverse = (str) => {const arr = []; //stores character in rev orderconst len = str.length; //finds the length of the stringlet i;for(i = len; i >= 0; i--)arr.push(str[i]); //adds the chars of str to arrreturn arr.join('');} let stringTest, reverseString; //for dummy stringstringTest = "john gabriel";reverseString = reverse(stringTest); console.log("Num3");console.log(`Reverse of the string "${stringTest}" is: "${reverseString}"`);console.log("\n");