Develop Reference

How to optimize the longest subsequence generation code in java?

import java.util.Scanner;
public class POD9 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String tito_has = sc.nextLine();
String tito_wants = sc.nextLine();
String remaining = "";
int strips = 0;
while(true){
String longest_common = getLongestCommonSubstring(tito_has,tito_wants);
strips++;
tito_has = tito_has.replaceAll(longest_common,"*");
remaining = remaining + longest_common;
if(remaining.length() == tito_wants.length()){
break;
}
}
System.out.println(strips-1);
}
public static String getLongestCommonSubstring(String str1, String str2){
int m = str1.length();
int n = str2.length();
int max = 0;
int[][] dp = new int[m][n];
int endIndex=-1;
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
if(str1.charAt(i) == str2.charAt(j) && str1.charAt(i) != Character.valueOf('*'))
{
if(i==0 || j==0){
dp[i][j]=1;
}else{
dp[i][j] = dp[i-1][j-1]+1;
}
if(max < dp[i][j])
{
max = dp[i][j];
endIndex=i;
}
}
}
}
return str1.substring(endIndex-max+1,endIndex+1);
}
}
This is my code used to retrieve the minimum number of paper cuts that are needed to form string 2 from string 1. This is the solution to the problem: Paper Cut Problem
I approached the problem by finding the longest substring between what Tito wants and what he has.
Thereafter, I replace the found sequences with '' as the input will only have letters. Also, in the cases, say, it has: "abbap" and tito wants:"bbaap", then this gives 3 cuts a|bb|a|p and not
2 (found bb->remove it (cut 1)->got (aap) -> found aa(cut2)-> remove it-> found b). So using "" helps in not making wrong substrings match.
I am pretty sure that my approach will solve most of the test cases. But I am getting the TIMELIMIT error here.
Please help me find an optimum way to achieve this?

Java Poem Palindrome Checker: Iterate through array matching elements in order

Currently trying to program a poem Palindrome checker. This is not for palindromes specifically, but that the array has words in the same order both ways. For example the following is a poem palindrome
Life-
imitates nature,
always moving, traveling continuously.
Continuously traveling, moving always,
nature imitates
life
My issue is iterating through the array to match the first and last elements, as currently it compares things in the wrong order.
My code is as follows:
import java.util.Scanner;
import java.io.*;
public class WordPalindromeTest {
public static void main(String[] args) {
System.out.println("This program determines if an entered sentence/word poem is a palindrome.");
Scanner input = new Scanner(System.in);
System.out.println("Please enter a string to determine if it is a palindrome: ");
while(input.hasNextLine()) {
String palin = input.nextLine();
if(palin.equals("quit")) {
break;
}
else {
boolean isPalin = isWordPalindrome(palin);
if(isPalin == true) {
System.out.println(palin + " is a palindrome!");
}
else
System.out.println(palin + " is NOT a palindrome!");
}
}
System.out.println("Goodbye!");
input.close();
}
public static boolean isWordPalindrome(String s) {
boolean isWordPal = false;
String lowerCase = s.toLowerCase();
String replaced = lowerCase.replaceAll("[^a-zA-Z0-9\\s]", "");
String words[] = replaced.split(" ");
for(int i = 0; i < words.length; i++) {
for(int j = 0; j < words.length; j++) {
if (words[i].equals(words[j]) && i != j) {
isWordPal = true;
}
else
isWordPal = false;
}
}
return isWordPal;
}
}
With the specific point in question being
public static boolean isWordPalindrome(String s) {
boolean isWordPal = false;
String lowerCase = s.toLowerCase();
String replaced = lowerCase.replaceAll("[^a-zA-Z0-9\\s]", "");
String words[] = replaced.split(" ");
for(int i = 0; i < words.length; i++) {
for(int j = 0; j < words.length; j++) {
if (words[i].equals(words[j]) && i != j) {
isWordPal = true;
}
else
isWordPal = false;
}
}
return isWordPal;
}
I am confused on how to properly set up the loop to compare the right elements. It should compare the first element to the last, the second to the second to last, etc. until the loop is finished. I realize I have it compare the first to the entire array before moving on.

This seems like a homework assignment so I won't give you a working solution. But this of it like this:
-You don't need two loops. You only need to compare the first to the last, the second to the second to last, etc. (Hint: if you subtract i-1 from the length of the Array you'll get the corresponding element to i that you need to compare to). Also you only need to iterate over half of the length of the Array
-If ever isWordPal becomes false, you need to return false. Otherwise it might get overwritten and at the end it will return true.

Return word specified by the integer

I know I'm missing some things and that's what I really need help with. The code doesn't work in all cases and am looking for help improving/fixing it.
Assignment:
The code I have so far:
public String word(int num, String words)
{
int l = words.indexOf(" ");
int r = words.indexOf(" ", l+1);
for(int i = 3; i <= num; i++){
l = r;
r = words.indexOf(" ", l+1);
//if(i != num)
// l = r;
}
String theword = words.substring(l,r);
return theword;
}
}

As this is clearly homework, I will give you text only.
Your approach may work eventually, but it is laborious and overly complicated, so it's hard to debug and hard to get right.
make use of String's API by using the split() method
after splitting the sentence into an array of word Strings, return the element at num less one (array are indexed starting at zero
check the length of the array first, in case there are less words than num, and take whatever action you think is appropriate in that case
For part 2, a solution in a simple form may be:
create a new blank string for the result
iterate over the characters of the given string adding the character to the front of the result string
make use of String's toUpperCase() method

Since this is homework and you have showed some effort. This is how you can do part 1 of your question. This code is pretty evident.
1) I am returning null if number is greater than the number of words in string as we dont want user to enter 5 when there are only 2 words in a string
2) Splitting the string by space and basically returning the array with the number mentioned by user
There are more conditions which you must figure out such as telling the user to enter a number of the string length since it would not give him any result and taking input from Scanner instead of directy adding input in method.
public static String word(int num, String words)
{
String wordsArr[] = words.split(" ");
if(num <= 0 || num > wordsArr.length) return null;
return (wordsArr[num-1]);
}
the second part of your question must be attempted by you.

Well... not often you see people coming here with homework AND showing effort at the same time so bravo :).
This is example of how you can split the string and return the [x] element from that string
public class SO {
public static void main(String[] args) throws Exception {
int number = 3;
String word = "Hello this is sample code";
SO words = new SO();
words.returnWord(number, word);
}
private void returnWord(int number, String word) throws Exception {
String[] words = word.split("\\s+");
int numberOfWords = words.length;
if(numberOfWords >= number) {
System.out.println(words[number-1]);
} else {
throw new Exception("Not enought words!!!");
}
}
}
Yes it is a working example but do not just copy and paste that for your homework - as simple question from teacher - What is this doing, or how this works and your out :)! So understand the code, and try to modify it in a way that you are familiar what is doing what. Also its worth getting some Java book - and i recommend Head first Java by O'Really <- v.good beginner book!
if you have any questions please do ask!. Note that this answer is not 100% with what the textbook is asking for, so you can modify this code accordingly.
As of part 2. Well what Bohemian said will also do, but there is a lot quicker solution to this.
Look at StringBuilder(); there is a method on it that will be of your interest.
To convert String so all letter are upper case you can use .toUpperCase() method on this reversed string :)

You can try:
public class trial {
public static void main(String[] args)
{
System.out.println(specificword(0, "yours faithfully kyobe"));
System.out.println(reverseString("derrick"));}
public static String specificword(int number, String word){
//split by space
String [] parts = word.split("\\ ");
if(number <= parts.length){
return parts[number];
}
else{
return "null String";
}
}
public static String reverseString(String n){
String c ="";
for(int i = n.length()-1; i>=0; i--){
char m = n.charAt(i);
c = c + m;
}
String m = c.toUpperCase();
return m;
}
}

For the first problem, I'll give you two approaches (1. is recommended):
Use the String.split method to split the words up into an array of words, where each element is a word. Instead of one string containing all of the words, such as "hello my name is Michael", it will create an array of the words, like so [hello, my, name, is, Michael] and that way you can use the array to access the words. Very easy:
public static String word(int num, String words)
{
// split words string into array by the spaces
String[] wordArray = words.split(" "); // or = words.split("\\s+");
// if the number is within the range
if (num > 0 && num <= wordArray.length) {
return wordArray[num - 1]; // return the word from the word array
} else { // the number is not within the range of words
return null;
}
}
Only use this if you cannot use arrays! Loop through the word until you have found enough spaces to match the word you want to find:
public static String word(int num, String words)
{
for (int i = 0; i < words.length(); i++) { // every character in words
if (words.substring(i, i+1).equals(" ")) { // if word is a space
num = num - 1; // you've found the next word, so subtract 1 (number of words left is remaining)
}
if (num == 1) { // found all words
// return this word
int lastIndex = i+1;
while (lastIndex < words.length()) { // until end of words string
if (words.substring(lastIndex, lastIndex+1).equals(" ")) {
break;
}
lastIndex = lastIndex + 1; // not a space so keep moving along the word
}
/*
// or you could use this to find the last index:
int lastIndex = words.indexOf(" ", i + 1); // next space after i+1
if (lastIndex == -1) { // couldn't find another space
lastIndex = words.length(); // so just make it the last letter in words
}*/
if (words.substring(i, i+1).equals(" ")) { // not the first word
return words.substring(i+1, lastIndex);
} else {
return words.substring(i, lastIndex);
}
}
}
return null; // didn't find word
}
As for the second problem, just iterate backwards through the string and add each letter to a new string. You add each letter from the original string to a new string, but just back to front. And you can use String.toUpperCase() to convert the string to upper case. Something like this:
public static String reverse(String str) {
String reversedString = ""; // this will be the reversed string
// for every character started at the END of the string
for (int i = str.length() - 1; i > -1; i--) {
// add it to the reverse string
reversedString += str.substring(i, i+1);
}
return reversedString.toUpperCase(); // return it in upper case
}

write a code to find the number of words in a string using methods

I've been looking and I can't find anywhere how to write a word count using 3 methods. Here is what the code looks like so far. I'm lost on how to use the methods. I can do this without using different methods and just using one. Please help!!!
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
System.out.print("Enter a string: ");
String s = in.nextLine();
if (s.length() > 0)
{
getInputString(s);
}
else
{
System.out.println("ERROR - string must not be empty.");
System.out.print("Enter a string: ");
s = in.nextLine();
}
// Fill in the body with your code
}
// Given a Scanner, prompt the user for a String. If the user enters an empty
// String, report an error message and ask for a non-empty String. Return the
// String to the calling program.
private static String getInputString(String s) {
int count = getWordCount();
while (int i = 0; i < s.length(); i++)
{
if (s.charAt(i) == " ")
{
count ++;
}
}
getWordCount(count);
// Fill in the body
// NOTE: Do not declare a Scanner in the body of this method.
}
// Given a String return the number of words in the String. A word is a sequence of
// characters with no spaces. Write this method so that the function call:
// int count = getWordCount("The quick brown fox jumped");
// results in count having a value of 5. You will call this method from the main method.
// For this assignment you may assume that
// words will be separated by exactly one space.
private static int getWordCount(String input) {
// Fill in the body
}
}
EDIT:
I have changed the code to
private static String getInputString(String s) {
String words = getWordCount(s);
return words.length();
}
private static int getWordCount(String s) {
return s.split(" ");
}
But I can't get the string convert to integer.

You have read the name of the method, and look at the comments to decide what should be implemented inside the method, and the values it should return.
The getInputString method signature should be:
private static String getInputString(Scanner s) {
String inputString = "";
// read the input string from system in
// ....
return inputString;
}
The getWordCount method signature should be:
private static int getWordCount(String input) {
int wordCount = 0;
// count the number of words in the input String
// ...
return wordCount;
}
The main method should look something like this:
public static void main(String[] args) {
// instantiate the Scanner variable
// call the getInputString method to ... you guessed it ... get the input string
// call the getWordCount method to get the word count
// Display the word count
}

count=1 //last word must be counted
for(int i=0;i<s.length();i++)
{
ch=s.charAt(i);
if(ch==' ')
{
count++;
}
}

Use trim() and split() on 1-n whitespace chars:
private static int getWordCount(String s) {
return s.trim().split("\\s+").length;
}
The call to trim() is necessary, otherwise you'll get one extra "word" if there is leading spaces in the string.
The parameter "\\s+" is necessary to count multiple spaces as a single word separator. \s is the regex for "whitespace". + is regex for "1 or more".

What you need to do is, count the number of spaces in the string. That is the number of words in the string.
You will see your count will be off by 1, but after some pondering and bug hunting you will figure out why.
Happy learning!

You can do this by :
private static int getWordCount(String input) {
return input.split("\\s+").length;
}

Use String.split() method like :
String[] words = s.split("\\s+");
int wordCount = words.length;

I'm not sure what trouble you're having with methods but I dont think you need more than one, try this: it uses split to split up the words in a string, and you can chose the delimeters
String sentence = "This is a sentence.";
String[] words = sentence.split(" ");
for (String word : words) {
System.out.println(word);
}
then you can do:
numberOfWords = words.length();
if you want to use 3 methods, you can call a method from your main() method that does this for you, for example:
public String getInputString() {
Scanner in = new Scanner (System.in);
System.out.print("Enter a string: ");
String s = in.nextLine();
if (s.length() > 0) {
return s;
} else {
System.out.println("ERROR - string must not be empty.");
System.out.print("Enter a string: ");
return getInputString();
}
}
public int wordCount(String s) {
words = splitString(s)
return words.length();
}
public String[] splitString(String s) {
return s.split(" ");
}

Based on your code i think this is what you're trying to do:
private static int getWordCount(String input) {
int count = 0;
for (int i = 0; i < input.length(); i++) {
if (input.charAt(i) == ' ') {
count++;
}
}
return count;
}
Here's what I've done:
I've moved the code you were 'playing' with into the right method (getWordCount).
Corrected the loop you were trying to use (I think you have for and while loops confused)
Fixed your check for the space character (' ' not " ")
There is a bug in this code which you'll need to work out how to fix:
getWordCount("How are you"); will return 2 when it should be 3
getWordCount(""); will return 0
getWordCount("Hello"); will return 0 when it should be 1
Good luck!

Better use simple function of spilt() with arguments as space
int n= str.split(" ").length;

public static int Repeat_Words(String arg1,String arg2)
{
//It find number of words can be formed from a given string
if(arg1.length() < 1 || arg2.length() < 1)
return 0;
int no_words = 99999;
char[] str1 = arg1.toCharArray();
char[] str2 = arg2.toCharArray();
for(int x = 0; x < str1.length; x++)
{
int temp = 0;
for(int y = 0; y < str2.length; y++)
{
if(str1[x] == str2[y])
temp++;
}
if(temp == 0)
return 0;
if(no_words > temp)
no_words = temp;
temp = 0;
}
return no_words;
}

java Run-length encoding

I have no idea how to start my assignment.
We got to make a Run-length encoding program,
for example, the users enters this string:
aaaaPPPrrrrr
is replaced with
4a3P5r
Can someone help me get started with it?

Hopefully this will get you started on your assignment:
The fundamental idea behind run-length encoding is that consecutively occurring tokens like aaaa can be replaced by a shorter form 4a (meaning "the following four characters are an 'a'"). This type of encoding was used in the early days of computer graphics to save space when storing an image. Back then, video cards supported a small number of colors and images commonly had the same color all in a row for significant portions of the image)
You can read up on it in detail on Wikipedia
http://en.wikipedia.org/wiki/Run-length_encoding
In order to run-length encode a string, you can loop through the characters in the input string. Have a counter that counts how many times you have seen the same character in a row. When you then see a different character, output the value of the counter and then the character you have been counting. If the value of the counter is 1 (meaning you only saw one of those characters in a row) skip outputting the counter.

public String runLengthEncoding(String text) {
String encodedString = "";
for (int i = 0, count = 1; i < text.length(); i++) {
if (i + 1 < text.length() && text.charAt(i) == text.charAt(i + 1))
count++;
else {
encodedString = encodedString.concat(Integer.toString(count))
.concat(Character.toString(text.charAt(i)));
count = 1;
}
}
return encodedString;
}
Try this one out.

This can easily and simply be done using a StringBuilder and a few helper variables to keep track of how many of each letter you've seen. Then just build as you go.
For example:
static String encode(String s) {
StringBuilder sb = new StringBuilder();
char[] word = s.toCharArray();
char current = word[0]; // We initialize to compare vs. first letter
// our helper variables
int index = 0; // tracks how far along we are
int count = 0; // how many of the same letter we've seen
for (char c : word) {
if (c == current) {
count++;
index++;
if (index == word.length)
sb.append(current + Integer.toString(count));
}
else {
sb.append(current + Integer.toString(count));
count = 1;
current = c;
index++;
}
}
return sb.toString();
}
Since this is clearly a homework assignment, I challenge you to learn the approach and not just simply use the answer as the solution to your homework. StringBuilders are very useful for building things as you go, thus keeping your runtime O(n) in many cases. Here using a couple of helper variables to track where we are in the iteration "index" and another to keep count of how many of a particular letter we've seen "count", we keep all necessary info for building our encoded string as we go.

Try this out:
private static String encode(String sampleInput) {
String encodedString = null;
//get the input to a character array.
// String sampleInput = "aabbcccd";
char[] charArr = sampleInput.toCharArray();
char prev=(char)0;
int counter =1;
//compare each element with its next element and
//if same increment the counter
StringBuilder sb = new StringBuilder();
for (int i = 0; i < charArr.length; i++) {
if(i+1 < charArr.length && charArr[i] == charArr[i+1]){
counter ++;
}else {
//System.out.print(counter + Character.toString(charArr[i]));
sb.append(counter + Character.toString(charArr[i]));
counter = 1;
}
}
return sb.toString();
}

Here is my solution in java
public String encodingString(String s){
StringBuilder encodedString = new StringBuilder();
List<Character> listOfChars = new ArrayList<Character>();
Set<String> removeRepeated = new HashSet<String>();
//Adding characters of string to list
for(int i=0;i<s.length();i++){
listOfChars.add(s.charAt(i));
}
//Getting the occurance of each character and adding it to set to avoid repeated strings
for(char j:listOfChars){
String temp = Integer.toString(Collections.frequency(listOfChars,j))+Character.toString(j);
removeRepeated.add(temp);
}
//Constructing the encodingString.
for(String k:removeRepeated){
encodedString.append(k);
}
return encodedString.toString();
}

import java.util.Scanner;
/**
* #author jyotiv
*
*/
public class RunLengthEncoding {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println("Enter line to encode:");
Scanner s=new Scanner(System.in);
String input=s.nextLine();
int len = input.length();
int i = 0;
int noOfOccurencesForEachChar = 0;
char storeChar = input.charAt(0);
String outputString = "";
for(;i<len;i++)
{
if(i+1<len)
{
if(input.charAt(i) == input.charAt(i+1))
{
noOfOccurencesForEachChar++;
}
else
{
outputString = outputString +
Integer.toHexString(noOfOccurencesForEachChar+1) + storeChar;
noOfOccurencesForEachChar = 0;
storeChar = input.charAt(i+1);
}
}
else
{
outputString = outputString +
Integer.toHexString(noOfOccurencesForEachChar+1) + storeChar;
}
}
System.out.println("Encoded line is: " + outputString);
}
}
I have tried this one. It will work for sure.