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- DescriptionA researcher is analyzing DNA. A DNA can be represented as a string composed of the characters A, G, C, or T.One day, researchers found a strange DNA, which is Smooth Repeated DNA. The DNA is represented by a string that has infinite length. The string has a repeating pattern, i.e. the DNA string 0 is repeated an infinite number of times. For example, if0 = "????", then = "???????????? . . . ".According to researchers, a DNA is said to be special if it contains substrings . Determine whetheris a substring of . Squad FormatA line containing the two strings 0 and . Output FormatA line that determines whether it is a substring of . Issue “YES” ifis a substring of . Output “NO” otherwise. Example Input and Output Input Example Example Output AGCT GC YES AGCT TA YES AGCT GT No AGCT TAGCTAGCT YES AGGACCTA CTAA YES Explanation ExampleIn the first to fourth test case examples, is worth "???????????? . . . ". The part in bold is one of the…public static String[] getTokens(String raw, String delimiter){ // Line #1 goes here... int numTokens = st.countTokens(); // Line #3 goes here... for(int i = 0; i < numTokens; i++){ // Line #5 goes here } // Line #7 goes here } Using the code above, which one of these should be the code for "Line #7": ret[i] = st.nextToken(); return ret; String[] ret = new String[numTokens]; StringTokenizer st = new StringTokenizer(raw, delimiter);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",…
- 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(); A bit convoluted, but does what is needed without fuss or muss. Maybe one day the Java strings will come with the reverse method built in, just like the string data types of all sensible…int main(){ string str; int count = 0; // user input cout << "Enter string: "; cin >> str; int n = str.length(); // iterating over the string for (int i = 0; i < n - 1; i++) { // calling the fucntion to find the palidrome strings from the user input if (isPalindrome(0, i, str) && isPalindrome(i + 1, n - 1, str)) { // if 2 substring are palindrome then print them and increment count for (int x = 0; x < i + 1; x++) cout << str[x]; cout << " "; for (int x = i + 1; x < n; x++) cout << str[x]; count++; // break the loop after printing the palindrom string break; } } // if count is 0 then no palindrome subtring pair found if (count == 0) cout << "NO";} change this code to stdio.h string.h! ... #include using namespace std; int main() { int low=0, high=10, i; bool NN = true; while (low high) { NN = true; if (low == 0 || low = 1) { NN = false; } else { for (i = 2; i <= low / 2; ++i) { if (low % i == 0) { NN = false; break; } } if (NN) cout << low << "". docs.google.com. ++low; } return 0; 2468 1357 2579 2357 using namespace std; int div 2(int a){ 14 + ← ۹:۱۳ G P ↑
- Consider the following code segment: struct Dog { string name; int age; };…Dog x = { "Snoopy", 5 };Dog y = { "Bolt", 3 }; Which of the following is an invalid expression? Group of answer choices if (x == y) y = x; x.name = y.name; if (x.age < y.age)This is the issue and must be written in java:Write an application that accepts three specific User strings and shows either You have not entered the words alphabetically or You have entered the strings alphabetically depending on whether the strings were in alphabetic order.#include <stdio.h>int main(){int d;int s[20],i, r, p, lg=0,m;char c;printf("Enter number of salesman(max 20): ");scanf("%d", &d);for(i=0; i<d; i++){ printf("\n salesman %d sales: ");scanf("%d", &s[i]);}for(i=0; i<d; i++){for(r=i+1; j<d; r++){if(s[i] > s[r]){p= s[i];s[i] = s[r];s[j] = p;}}}printf("\nsalesman lowest to highest: ");for(i=0; i<d; i++){printf("%d\t", s[i]);}for(i=0;i<m;i++){printf("\n highest sales: %d ",lg);if(lg<=s[i])lg=s[i];break;}getch();} >>> the upper part output should be like this enter number of salesman (max 20): 5 salesman 1 500 salesman 2 300 salesman 3 1000 salesman 4 200 salesman 5 1000 in the lower part the lowest to highest the output should become like this salesman 4 200 salesman 2 300 salesman 1 500 salesman 3 1000 salesman 5 1000 highest total sales : 2000
- Х3: inlTol0 Write a function in Java that implements the following logic: Given a number n, return true if n is in the range 1..10, inclusive. Unless "outsideMode" is true, in which case return true if the number is less or equal to 1, or greater or equal to 10. Your Answer: 1 public boolean in1To10(int n, boolean outsideMode) 2 { 3 } H N Musing Java. Part B – String Classifier (30%) This program will classify a string as one of the following: a palindrome, a repeated word, or “nothing special”. A word is considered repeated if it repeats only once in a string (e.g., “pizzapizza”). Some palindrome examples are: “racecar”, “aa”, and “aaa”. To determine if it is a palindrome, recommend you make use of the below method. Please make sure this program works for all cases and conditions. /** * This method will return a reversed version of the input string */ static String reverseString(String inStr) { String outStr = ""; if (inStr != null) { int startIndex = inStr.length()-1; for (int i=startIndex; i >= 0; i--) { outStr = outStr + inStr.charAt(i); } } return outStr; }Code that need debugging fix with no errors // Gets a String from user // Converts the String to lowercase, and // displays the String's length // as well as a count of letters import java.util.*; public class DebugSeven4 { public static void main(String[] args) { Scanner kb = new Scanner(System.in); String aString ; int numLetters = 0; int stringLength; System.out.println("Enter a String. Include"); System.out.println("some uppercase letters, lowercase"); System.out.print("letters, and numbers >> "); aString = kb.nextLine(); stringLength = aString.length(); System.out.print("In all lowercase, the String is: "); for(int i = 0; i <= stringLength; i++) { char ch = Character.toLowerCase(aStringcharAt(i)); System.out.print(ch); if(!Character.isLetter(ch)) numLetters++; } System.out.println(); System.out.println ("The number of CHARACTERS…