String Manipulation
Modules - Data Structures
Teacher’s Notes:
========================================================================
String Manipulation is an important skill that is used throughout programming. This assignment challenges students to solve several typical problems that require the manipulation of strings. The solutions to these will depend on the language used since different programming languages offer different types of string representation and support.
This was written as a project for repl.it and worked well for students there.
The language for this assignment is in JAVA so some adjustments to the base code might be needed.
Many of the problems have multiple solutions.
Assignment:
========================================================================
Look at the following code and fill in the given method stubs. There will be a description of what each method should do above each stub. The methods do not need to be completed in the order given.
import java.util.Scanner;
class Main {
//This method returns the length of the String, st
public static int getLength(String st) {
return -1;
}
//This method returns the number of times a char, ch, appears in the String, st
public static int letterCount(String st, char ch){
return -1;
}
//This method returns the reverse of the String, st
public static String reverse(String st) {
return "Not Done Yet.";
}
//This method returns true if st1 and st2 are anagrams of each other.
//If they are not anagrams, it returns false
//The method is case sensitive
public static boolean isAnagram(String st1, String st2){
return false;
}
/*This method will return the pig latin version of a word. The pig latin
version of a word is found by moving the first letter of the word to the
end of the word and then adding "ay" to it.
Example: Robin = obinRay;
Example: Raven = avenRay
Example: Cyborg = yborgCay;
*/
public static String pigLatin(String st) {
return "Not Done Yet.";
}
/*This method will remove all duplicate characters from a String so that it
is only made of unique characters. You can assume all characters in the String
will have ASCII values between 32 and 126 (inclusive)
Sample Input:
banana
Sample Output:
ban
*/
public static String removeDuplicates(String st) {
return "Not Done Yet.";
}
/*The following method takes a string of 1's and 0's and compresses it by
making a new string in the following way:
1. The new string will start with the first character (either 1 or 0) in the
string
2. The next character will be a number showing how many of that digit appeared
in a row before the opposite digit appeared.
3. The next character will be a number showing how many of the opposite digit
appeared in a row before the first digit appeared again.
4. Go to step 2 and keep repeating until the whole string is compressed.
Sample input: 111111111111000000001111111110000110001111000
Sample output: 1 12 8 9 4 2 3 4 3
*/
public static String compression(String st) {
return "Not Done Yet.";
}
//!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
/////////////////////////////////////////
//DO NOT EDIT ANY CODE BELOW THIS LINE!//
/////////////////////////////////////////
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int testNumber = input.nextInt();
if(testNumber == 1){
String st = input.next();
System.out.println( getLength(st) );
}
else if(testNumber == 2){
String st = input.next();
System.out.println( letterCount(st, 'p'));
System.out.println( letterCount(st, 'b'));
System.out.println( letterCount(st, 'a'));
}
else if(testNumber == 3){
String st = input.next();
System.out.println( reverse(st) );
}
else if(testNumber == 4){
String s1 = input.next();
String s2 = input.next();
System.out.println( isAnagram(s1, s2) );
}
else if(testNumber == 5){
String st = input.next();
System.out.println( pigLatin(st) );
}
else if (testNumber == 6){
String st = input.next();
System.out.println( removeDuplicates(st) );
}
else if(testNumber == 7){
String st = input.next();
System.out.println( compression(st) );
}
}
}
Solution (JAVA):
========================================================================
import java.util.Scanner;
class Main {
//This method returns the length of the String, st
public static int getLength(String st) {
return st.length();
}
//This method returns the number of times a char, ch, appears in the String, st
public static int letterCount(String st, char ch){
int count = 0;
for(int x = 0; x < st. length(); x++)
if(st.charAt(x) == ch)
count++;
return count;
}
//This method returns the reverse of the String, st
public static String reverse(String st) {
String word = "";
for(int x = st. length()-1; x >= 0 ; x--)
word += st.charAt(x);
return word;
}
//This method returns true if st1 and st2 are anagrams of each other.
//If they are not anagrams, it returns false
//The method is case sensitive
public static boolean isAnagram(String st1, String st2){
char[] s1 = st1.toCharArray();
char[] s2 = st2.toCharArray();
boolean letterFound = false;
for(int x = 0; x < s1.length; x++)
{
letterFound = false;
for(int y = 0; y < s2.length; y++){
if(s1[x] == s2[y]){
s2[y] = '~';
letterFound = true;
break;
}
}
if(!letterFound)
return false;
}
return true;
}
/*This method will return the pig latin version of a word. The pig latin
version of a word is found by moving the first letter of the word to the
end of the word and then adding "ay" to it.
Example: Robin = obinRay;
Example: Raven = avenRay
Example: Cyborg = yborgCay;
*/
public static String pigLatin(String st) {
return st.substring(1) + st.charAt(0) + "ay";
}
/*This method will remove all duplicate characters from a String so that it
is only made of unique characters. You can assume all characters in the String
will have ASCII values between 32 and 126 (inclusive)
Sample Input:
banana
Sample Output:
ban
*/
public static String removeDuplicates(String st) {
String word = "";
int[] letterCount = new int[127];
for(int x = 0; x < st.length(); x++)
if(letterCount[st.charAt(x)]++ == 0)
word+=st.charAt(x);
return word;
}
/*The following method takes a string of 1's and 0's and compresses it by
making a new string in the following way:
1. The new string will start with the first character (either 1 or 0) in the
string
2. The next character will be a number showing how many of that digit appeared
in a row before the opposite digit appeared.
3. The next character will be a number showing how many of the opposite digit
appeared in a row before the first digit appeared again.
4. Go to step 2 and keep repeating until the whole string is compressed.
Sample input: 111111111111000000001111111110000110001111000
Sample output: 1 12 8 9 4 2 3 4 3
*/
public static String compression(String st) {
String word = "" + st.charAt(0);
char currentChar = st.charAt(0);
int count = 1;
for(int x = 1; x < st.length(); x++){
if(st.charAt(x) == currentChar)
count++;
else{
word += " " + count;
count = 1;
currentChar = st.charAt(x);
}
}
word += " " + count;
return word;
}
/////////////////////////////////////////
//DO NOT EDIT ANY CODE BELOW THIS LINE!//
/////////////////////////////////////////
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int testNumber = input.nextInt();
if(testNumber == 1){
String st = input.next();
System.out.println( getLength(st) );
}
else if(testNumber == 2){
String st = input.next();
System.out.println( letterCount(st, 'p'));
System.out.println( letterCount(st, 'b'));
System.out.println( letterCount(st, 'a'));
}
else if(testNumber == 3){
String st = input.next();
System.out.println( reverse(st) );
}
else if(testNumber == 4){
String s1 = input.next();
String s2 = input.next();
System.out.println( isAnagram(s1, s2) );
}
else if(testNumber == 5){
String st = input.next();
System.out.println( pigLatin(st) );
}
else if (testNumber == 6){
String st = input.next();
System.out.println( removeDuplicates(st) );
}
else if(testNumber == 7){
String st = input.next();
System.out.println( compression(st) );
}
}
}
