In Java please:

public class DNASequencing extends Objects{

/*Use a for loop to visit each element in the array input parameter. Remember that array indices start at 0, as do the indices of each character in a String. Prints out the contents of an array, one item per line. Parameters: strings - The array of String objects to be printed.*/ public static void printArray(String[] strings){ }

/*After completing and testing printArray, work on the findLongest or the findFrequency methods; they are similar. Within both methods, use a loop to visit each element in the array. In the findLongest method, you must keep track of which String object in the array contains the most characters, whereas in the findFrequency method, you must keep track of how many times a specific String object is found in the array. Make sure you finish and test both of these before moving to the next step. Finds the first occurrence of the longest item in an array of String objects. The length of a String object is determined by its number of characters Parameters: strings - The array of String objects to search. Returns: The longest String object. If there is more than one, then the first occurrence is returned. The first occurrence has the lowest index position in the array.*/ public static String findLongest(String[] strings){ }

/*Determines the number of times a particular item occurs within an array of String objects. Parameters: target - A string that is equivalent to the string(s) we are looking for. strings - The array of String objects to search. Returns: The number of times an equivalent string to the target occurs within the array.*/ public static int findFrequency(String target, String[] strings){ }

/*This method examines the contents of the sequence array, looking for a specific sequence and its mutations. By the time you have done all the other methods, this should be reasonably simple. You may want to create a helper method, but it is not required Determines the number of times a particular DNA sequence occurs within an array of DNA sequences. A typical DNA sequence contains a sequence of two to four of the following characters: {A,C,G,T}. A mutated sequence contains adjacent repeated characters , so TCA, TTCCCCA, and TCCAA are all equivalent, TTCCCCA and TCCAA being mutations of TCA. Parameters: target - The exact sequence we are looking for, in its non-mutated form. The string is a valid DNA sequence as defined in the CSC110 Assignment 5 instructions. seqArray - An array of DNA sequences, as described in the CSC110 Assignment 5 instructions. Returns: The number of times an equivalent DNA sequence to the target DNA sequence occurs within the array.*/ public static int findFreqWithMutations(String target, String[] seqArray){ }

/*Implement this method by building on the technique for findFrequency and adding calls to hasMutation. Counts the number of mutated DNA sequences that occur within an array. Note that DNA sequences contain two to four of the following characters: {A,C,G,T} in any order. A mutated sequence contains adjacent repeated characters , so ACCG, TTCCA, and AACC are all mutated sequences. Parameters: seqArray - An array of DNA sequences, as described in the CSC110 Assignment 5 instructions. Returns: The number of mutated DNA sequences found within the array.*/ public static int countTotalMutations(String[] seqArray){ }

/*This method is probably the easier one to implement, so you can start with that. For those of you who like an extra challenge, consider writing a more general method that deals with any string and returns true if there are consecutive characters that are identical. Experienced programmers often look for ways to make code that is re-usable, easy to maintain, and modular. For example, suppose that you wish to create a more general method (for possible future use) that deals with any string. And suppose it is called hasCharacterRun. After fully testing it, it is stored in your personal library (or toolkit) of code. Now, the hasMutation method is very simple and requires no testing Determines if a given DNA sequence has a mutation. A mutated sequence contains adjacent repeated characters, so ACCG, TTCCA, and AACC are all mutated sequences. Parameters: sequence - A DNA sequence that may or may not have consecutive repeated bases. Returns: True if the sequence has a consecutive repeated base.*/ public static boolean hasMutation(String sequence){ } //An internal tester to check the methods. public static void main(String[] args){ }

}