This is my first post so go easy on me. This IS a homework question, but I have spent about 7 hours working through various means to complete this goal and have had no success. I am building various methods for an assignment, and I need to figure out how to split a String into several int variables.
Ex: given the String "100 200 300" I need to change it to three int of 100, 200, 300. I have to use indexOf(), and cannot use split() or arrays.
String scores="100 200 300";
int n=scores.indexOf(" ");
String sub=scores.substring(0,n);
Integer.parseInt(sub);
This lets me get the first string "100" and parse it. However, I do not know how to continue the code so it will get the next ones. For my method, I will need the new int variables for later arguments.
EDIT: I think I need to use a for loop: something like:
for(int i=0; i<=scores.length; i++)
{//I do not know what to put here}
Joe, indexOf() is overloaded, check out this version:
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf(int,%20int)
You need two things:
a loop;
being able to run indexOf() from where it left off (hint: read the Javadoc).
public static void main(String[] args) {
String scores = "100 200 300";
ArrayList<Integer> numbers = new ArrayList<Integer>();
int n = 0;
while (n != -1) {
String sub = "";
n = scores.indexOf(" ");
if (n != -1) {
sub = scores.substring(0, n);
scores = scores.substring((n + 1));
} else {
sub = scores;
}
numbers.add(Integer.parseInt(sub));
}
for (int i : numbers) {
System.out.println("" + i);
}
}
Try something like this to loop through and add numbers to arraylist. The arraylist numbers will contain all your numbers.
try this:
String scores="100 200 300";
int offset = 0;
int space;
int score;
scores = scores.trim(); //clean the string
do
{
space= scores.indexOf(" ", offset);
if(space > -1)
{
score = Integer.parseInt(scores.substring(offset , space));
}
else
{
score = Integer.parseInt(scores.substring(offset));
}
System.out.println(score);
offset = space + 1;
}while(space > -1);
Your 'n' variable is the important part. You get your first String by slicing from 0 to 'n', so your next string starts not at 0, but at
n + " ".size()
Ok, so here is what I have come up with:
Since I needed to compare the newly parsed ints with a different variable, as well as ensure that the amount of ints was equal to a different variable, I created this while loop:
public boolean isValid()
{
int index=0;
int initialindex=0;
int ntotal=0;
int ncount=0;
boolean flag=false;
while (index!=-1)
{
index=scores.indexOf(" ");
String temp=scores.substring(initialindex,index);
int num=Integer.parseInt(temp);
ntotal+=num;
ncount++;
initialindex=index;
}
if (ntotal==total && ncount==count)
{
flag=true;
}
return flag;
}
Related
Suppose we have a string of binary values in which some portions may correspond to specific letters, for example:
A = 0
B = 00
C = 001
D = 010
E = 0010
F = 0100
G = 0110
H = 0001
For example, if we assume the string "00100", we can have 5 different possibilities:
ADA
AF
CAA
CB
EA
I have to extract the exact number of combinations using Dynamic programming.
But I have difficulty in the formulation of subproblems and in the composition of the corresponding vector of solutions.
I appreciate any indications of the correct algorithm formulation.
class countString {
static int count(String a, String b, int m, int n) {
if ((m == 0 && n == 0) || n == 0)
return 1;
if (m == 0)
return 0;
if (a.charAt(m - 1) == b.charAt(n - 1))
return count(a, b, m - 1, n - 1) +
count(a, b, m - 1, n);
else
return count(a, b, m - 1, n);
}
public static void main(String[] args) {
Locale.setDefault(Locale.US);
ArrayList<String> substrings = new ArrayList<>();
substrings.add("0");
substrings.add("00");
substrings.add("001");
substrings.add("010");
substrings.add("0010");
substrings.add("0100");
substrings.add("0110");
substrings.add("0001");
if (args.length != 1) {
System.err.println("ERROR - execute with: java countString -filename- ");
System.exit(1);
}
try {
Scanner scan = new Scanner(new File(args[0])); // not important
String S = "00100";
int count = 0;
for(int i=0; i<substrings.size(); i++){
count = count + count(S,substrings.get(i),S.length(),substrings.get(i).length());
}
System.out.println(count);
} catch (FileNotFoundException e) {
System.out.println("File not found " + e);
}
}
}
In essence, Dynamic Programming is an enhanced brute-force approach.
Like in the case of brute-force, we need to generate all possible results. But contrary to a plain brute-force the problem should be divided into smaller subproblems, and previously computed result of each subproblem should be stored and reused.
Since you are using recursion you need to apply so-called Memoization technic in order to store and reuse the intermediate results. In this case, HashMap would be a perfect mean of storing results.
But before applying the memoization in order to understand it better, it makes sense to start with a clean and simple recursive solution that works correctly, and only then enhance it with DP.
Plain Recursion
Every recursive implementation should contain two parts:
Base case - that represents a simple edge-case (or a set of edge-cases) for which the outcome is known in advance. For this problem, there are two edge-cases: the length of the given string is 0 and result would be 1 (an empty binary string "" results into an empty string of letters ""), another case is when it's impossible to decode a given binary string and result will be 0 (in the solution below it resolves naturally when the recursive case is being executed).
Recursive case - a part of a solution where recursive calls a made and when the main logic resides. In the recursive case, we need to find each binary "binary letter" at the beginning of the string and then call the method recursively by passing the substring (without the "letter"). Results of these recursive calls need to be accumulated in the total count that will returned from the method.
In order to implement this logic we need only two arguments: the binary string to analyze and a list of binary letters:
public static int count(String str, List<String> letters) {
if (str.isEmpty()) { // base case - a combination was found
return 1;
}
// recursive case
int count = 0;
for (String letter: letters) {
if (str.startsWith(letter)) {
count += count(str.substring(letter.length()), letters);
}
}
return count;
}
This concise solution is already capable of producing the correct result. Now, let's turn this brute-force version into a DP-based solution, by applying the memoization.
Dynamic Programming
As I've told earlier, a HashMap will be a perfect mean to store the intermediate results because allows to associate a count (number of combinations) with a particular string and then retrieve this number almost instantly (in O(1) time).
That how it might look like:
public static int count(String str, List<String> letters, Map<String, Integer> vocab) {
if (str.isEmpty()) { // base case - a combination was found
return 1;
}
if (vocab.containsKey(str)) { // result was already computed and present in the map
return vocab.get(str);
}
int count = 0;
for (String letter: letters) {
if (str.startsWith(letter)) {
count += count(str.substring(letter.length()), letters, vocab);
}
}
vocab.put(str, count); // storing the total `count` into the map
return count;
}
main()
public static void main(String[] args) {
List<String> letters = List.of("0", "00", "001", "010", "0010", "0100", "0110", "0001"); // binary letters
System.out.println(count("00100", letters, new HashMap<>())); // DP
System.out.println(count("00100", letters)); // brute-force recursion
}
Output:
5 // DP
5 // plain recursion
A link to Online Demo
Hope this helps.
Idea is to create every possible string with these values and check whether input starts with the value or not. If not then switch to another index.
If you have test cases ready with you you can verify more.
I have tested only with 2-3 values.
public int getCombo(String[] array, int startingIndex, String val, String input) {
int count = 0;
for (int i = startingIndex; i < array.length; i++) {
String matchValue = val + array[i];
if (matchValue.length() <= input.length()) {
// if value matches then count + 1
if (matchValue.equals(input)) {
count++;
System.out.println("match Found---->" + count); //ommit this sysout , its only for testing.
return count;
} else if (input.startsWith(matchValue)) { // checking whether the input is starting with the new value
// search further combos
count += getCombo(array, 0, matchValue, input);
}
}
}
return count;
}
In main Method
String[] arr = substrings.toArray(new String[0]);
int count = 0;
for (int i = 0; i < arr.length; i++) {
System.out.println("index----?> " + i);
//adding this condition for single inputs i.e "0","010";
if(arr[i].equals(input))
count++;
else
count = count + getCombo(arr, 0, arr[i], input);
}
System.out.println("Final count : " + count);
My test results :
input : 00100
Final count 5
input : 000
Final count 3
I am trying to take the input and if there is an # symbol in the input then it finds the maximum of the integers before and after the # symbol. The maximum part I have no problem with but I do not know how to access and find the values before and after the # symbol.
import java.util.Scanner;
public class Max_Min {
public static void main(String[] args) {
//gets keyboard
Scanner keyboard = new Scanner(System.in);
//puts input into string
String inputString = keyboard.nextLine();
//splits string between characters
String[] splitInput = inputString.split("");
for (String s : splitInput) {
if(s.equals("#")){
//computes the maximum of the two integers before and after the #
}
}
//close keyboard
keyboard.close();
I did do a search to find something simliar (and im sure there is something) but could not find anything. If someone could help that would be great!
Try with this:
for (int i = 0; i < splitInput.length; i++){
if (splitInput[i].equals("#") && i != 0 && i != splitInput.length -1){
int max = Math.max(Integer.parseInt(splitInput[i - 1]), Integer.parseInt(splitInput[i + 1]));
}
//...
}
You could try:
String[] splitInput = inputString.split("#");
which would split your string at the #s.
Then you can do a iteration over your splitInput array and do a .length on each index.
You have written the simple for loop, with which you can only access the string, but not its index in the array. If you had the index, you could write:
int possibleMax = Integer.parseInt(splitInput[i - 1]) + Integer.parseInt(splitInput[i + 1]);
To get the index, there are two ways:
for (int i = 0; i < splitInput.length; i++) {
String s = splitInput[i];
...
}
Or:
int i = 0;
for (String s : splitInput) {
…
i++;
}
I don't like either version because both are more complicated than absolutely necessary, in terms of written code. If you would use Kotlin instead of Java, it would be:
splitInput.forEachIndexed { i, s ->
…
}
In Java this could be written:
forEachIndexed(
splitInput,
(i, s) -> …
);
The problem in Java is that the code inside the … cannot update the variables of the enclosing method. I'm not sure whether this will ever change. It would be possible but needs a lot of work by the language committee.
A simple way to do this would be
String input = "12#23";
String [] arr = input.split("#");
if (arr.length == 2) {
System.out.println("Max is "+Math.max(Integer.valueOf(arr[0]),Integer.valueOf(arr[1])));
}
Good morning all,
Today is my first time trying to make a recursion method. Just when I thought it would work I got the error; missing return statement. This confuses me because it literally has a return value ( the total string from the Stringbuilder ) after n < 1
This is what I got:
import java.util.Scanner;
public class P5_4MethodRepeatString {
public void main(String[] args){
Scanner sc = new Scanner(System.in);
System.out.println("Enter a string followed by the times you want
it repeated:");
String input = sc.nextLine();
int count = sc.nextInt();
String total = repeat(input, count);
System.out.println(total);
}
public static String repeat(String str, int n) {
StringBuilder full = new StringBuilder();
if(n < 1) {return full.toString();}
repeat(str, n - 1);
for(int i = 0; i < n; i++){
full.append(str);
}
}
}
if(n < 1) {return full.toString();} but if n >= 1 you don't return anything. It should be return repeat(str, n - 1); but then you need to move the for... part
The immediate cause of your problem is that not all conditional flows in your repeat() method have a return value. But I am not sure if even making that change would result in working code. Consider the following refactor:
public void repeat(StringBuilder result, String str, int n) {
if (n == 0) return;
result.append(str);
repeat(result, str, n - 1);
return;
}
Here we are passing in a StringBuilder object which we want to contain our repeated string results. We also pass the string to be repeated, and the number of remaining turns. The base case occurs where there are no more turns, in which case we simply return from the recursive method. Otherwise, we add one copy of the string and make a recursive call.
Your original recursive attempt had the following for loop:
for (int i=0; i < n; i++) {
full.append(str);
}
This does not look recursive, and looping to repeatedly add the string defeats the point of doing recursion (though I would probably use a loop in real life if I had to do this).
My Question is :
First I have String variable of 1000 character, and I have another set of String variable of 1000 character
1] In First Set of Variable contains "1110000XXXXX0001111...." like this and so on, till 1000
2] In The second Set of Variable contains "1110000101010001111..." like this and so on till 1000
3] I need to get the position of X in first Variable and replace the Value of similar position from the second variable
For ex : 1st Variable of data "000XXX000X0"
2nd Variable of data "00011000010"
The X should be replaced by the values which is in the position in 2nd set of data.
NOTE : TO BE DONE WITHOUT LOOP
because if we put loop its runs 1000 times in a loop and 'X' may be anywhere in 1000 characters in the String
For ex: 1 Record 1000 Times
if 100K Records means 1000*100K (PERFORMANCE FAILS)
So need solution for it.
Kindly Help me out with this.
My Code is :
String sInputStr="0X11XXXXX000000000000000000000000000000000000000000000000X000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011";
String sDbStr="0111111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011";
int iLength=sInputStr.length();
for(int i=0;i<iLength;i++){
if(sInputStr.charAt(i)=='X'){
}else{
if(i>sDbStr.length()){
break;
}else{
sChar[i] = sInputStr.charAt(i);
}
}
}//End of For
sVal=String.valueOf(sChar);
System.out.println("sVal == " +sVal);
Help Me friends
All you need is something like this
class FirstApp {
public static void main(String[] args) {
String sDbStr="0111111110000001234000000000000011";
StringBuilder sNewStr= new StringBuilder("011111111000000XXXX00000000000001112");
String findStr = "X";
int lastIndex = 0;
System.out.println("Starting");
long startTime = System.currentTimeMillis();
String result = replaceValues("X", sDbStr, sNewStr);
long endTime = System.currentTimeMillis();
System.out.println("Result");
System.out.println(result);
System.out.println(String.valueOf(endTime-startTime));
}
public static String replaceValues(String toReplace, String fromStr, StringBuilder toStr) {
int lastIndex = toStr.indexOf(toReplace);
if(lastIndex != -1){
toStr.replace(lastIndex,lastIndex+1,Character.toString(fromStr.charAt(lastIndex)));
System.out.println(toStr);
return replaceValues(toReplace, fromStr, toStr);
} else {
return toStr.toString();
}
}
}
sample result:
Starting
0111111110000001XXX00000000000001112
01111111100000012XX00000000000001112
011111111000000123X00000000000001112
011111111000000123400000000000001112
Result
011111111000000123400000000000001112
UPDATE Updated solution to ensure less execution time using stringBuilder and recursion
If X point to one value, like as mentioned X replace by 1. Go for string.replaceAll function.
ie.
String oriString="000XXX000X0";
String replaceOne=oriString.replace('X','1');
System.out.println(replaceOne);
If I understood the problem correctly then you want to replace the values of X in your first array with the values from seconds array at the same positions. For Example, array1: 000XXX000 & array2: 100101001. Then array1 should finally be 000101000.
Here is a simple code snippet to achieve this:
char[] arr1 = sInputStr.toCharArray();
char[] arr2 = sDbStr.toCharArray();
for(int i = 0; i < arr1.size(); i++)
if(arr1[i] == 'X')
arr1[i] == arr2[i];
The idea is to search for the index of the occurrence of the character say 'X' and copy all the characters from second string into the first as long as we find 'X'. Repeat the process till the last occurrence of 'X'.
public class Test
{
public static void main(String args[])
{
String s1 = "0X11XXXXX000000000000000000000000000000000000000000000000X000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011";
String s2= "0111111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011";
char a[] = s1.toCharArray();
int i = s1.indexOf('X', 0);
while(i!=-1)
{
while(a[i] == 'X'){
a[i] = s2.charAt(i);
i++;
}
i = s1.indexOf('X',i+1);
}
s1 = new String(a);
System.out.println("result: "+s1);
}
}
I have to create a program that takes a user input (a number) and then the program should have that number and apply a search to the array and output the corresponding title by matching the index and the number the user inputted. However during run time, nothing happens. I have set breakers in my code and noticed a problem with the for loop (search algorithm). Please help me and let me know what is wrong is my search algorithm. What I am trying to do is use the number of that the user inputs to match a index and then output the book title that is stored in the index.
private void btnFindActionPerformed(java.awt.event.ActionEvent evt) {
// TODO add your handling code here:
// declares an array
String[] listOfBooks = new String [101];
// assigns index in array to book title
listOfBooks[1] = "The Adventures of Tom Sawyer";
listOfBooks[2] = "Huckleberry Finn";
listOfBooks[4] = "The Sword in the Stone";
listOfBooks[6] = "Stuart Little";
listOfBooks[10] = "Treasure Island";
listOfBooks[12] = "Test";
listOfBooks[14] = "Alice's Adventures in Wonderland";
listOfBooks[20] = "Twenty Thousand Leagues Under the Sea";
listOfBooks[24] = "Peter Pan";
listOfBooks[26] = "Charlotte's Web";
listOfBooks[31] = "A Little Princess";
listOfBooks[32] = "Little Women";
listOfBooks[33] = "Black Beauty";
listOfBooks[35] = "The Merry Adventures of Robin Hood";
listOfBooks[40] = "Robinson Crusoe";
listOfBooks[46] = "Anne of Green Gables";
listOfBooks[50] = "Little House in the Big Woods";
listOfBooks[52] = "Swiss Family Robinson";
listOfBooks[54] = "The Lion, the Witch and the Wardrobe";
listOfBooks[54] = "Heidi";
listOfBooks[66] = "A Winkle in Time";
listOfBooks[100] = "Mary Poppins";
// gets user input
String numberInput = txtNumberInput.getText();
int number = Integer.parseInt(numberInput);
// Linear search to match index number and user input number
for(int i = 0; i < listOfBooks.length - 1; i++) {
if (listOfBooks.get(i) == number) {
txtLinearOutput.setText(listOfBooks[i]);
break;
}
}
*There is a problem with the listOfBooks.get in the if statement. Also I need to apply a binary search that would search the same array just using the binary method. Need help to apply this type of binary search.
How could I make a statement that checks if the int number is equal to an index?
Note that the following code is just an example of what I have to apply. Variables are all for example purposes:
public static Boolean binarySearch(String [ ] A, int left, int right, String V){
int middle;
if (left > right) {
return false;
}
middle = (left + right)/2;
int compare = V.compareTo(A[middle]);
if (compare == 0) {
return true;
}
if (compare < 0) {
return binarySearch(A, left, middle-1, V);
} else {
return binarySearch(A, middle + 1, right, V);
}
}
you can avoid for loop and check condition by just giving number like this: txtLinearOutput.setText(listOfBooks[number-1]);
remove your code
// Linear search to match index number and user input number
for(int i = 0; i < listOfBooks.length - 1; i++) {
if (listOfBooks.get(i) == number) {
txtLinearOutput.setText(listOfBooks[i]);
break;
}
with
try{
int number = Integer.parseInt(numberInput);
if(number>0 && number<101){
txtLinearOutput.setText(listOfBooks[number-1]);
}else{
// out of range
}
}catch(Exception e){
// handle exception here
}
You are comparing if (listOfBooks.get(i) == number) it is wrong, you should compare: if (i == number), becouse you need compare element position.
This isn't a binary search answer. Just an implementation of HashMap. Have a look at it.
HashMap<String, Integer> books = new HashMap();
books.put("abc", 1);
books.put("xyz", 2);
books.put("pqr", 3);
books.put("lmo", 4);
System.out.println(books.getValue("abc");
Using the inbuilt BinarySearch.
String []arr = new String[15];
arr[0] = "abc";
arr[5] = "prq";
arr[7] = "lmo";
arr[10] = "xyz";
System.out.println(Arrays.binarySearch(arr, "lmo"));
How to compare Strings using binary search.
String[] array = new String[4];
array[0] = "abc";
array[1] = "lmo";
array[2] = "pqr";
array[3] = "xyz";
int first, last, middle;
first = 0;
last = array.length - 1;
middle = (first + last) / 2;
String key = "abc";
while (first <= last) {
if (compare(array[middle], key))
first = middle + 1;
else if (array[middle].equals(key)) {
System.out.println(key + " found at location " + (middle) + ".");
break;
} else {
last = middle - 1;
}
middle = (first + last) / 2;
}
if (first > last)
System.out.println(key + " is not found.\n");
}
private static boolean compare(String string, String key) {
// TODO Auto-generated method stub
for (int i = 0; i < Math.min(string.length(), key.length()); ++i)
if (string.charAt(i) < key.charAt(i))
return true;
return false;
}
Your linear search code looks something like this
try{
txtLinearOutput.setText(listOfBooks[yourNumber]);
}
catch(IndexOutOfBoundsException ie){
// prompt that number is not an index
}
catch(Exception e){
// if any other exception is caught
}
What you are doing here:
if (listOfBooks.get(i) == number) {
is that you are matching the content of the array with the input number, which is irrelevant.
You can directly use the input number to fetch the value stored at the index.
For example:
txtLinearOutput.setText(listOfBooks[number-1]);
Additionally, int number = Integer.parseInt(numberInput); should be placed within try-catch block to validate input number parsing. And you can check if the input number is within the range of the array to avoid exceptions like:
try{
int number = Integer.parseInt(numberInput);
// Linear search to match index number and user input number
if (number > 0 && number <=100) {
txtLinearOutput.setText(listOfBooks[number-1]);
} else {
// Display error message
}
} catch(Exception e) {
// Handle exception and display error message
}
And for using binary search, the string array need to be sorted. You can use Arrays.sort() method for sorting it.
And regarding using binary search, you can use Java Arrays Binary Search method