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Attributed Value of Ternary Operator Resets After Each Loop

I would like to be able to go through an inputted string and count the amount of times "good" is written and compare it to how many times "bad" is written. If the good and the bad match, then goodVbad==0 and it returns true. Otherwise it returns false.
The code worked fine when I was using if statements inside the for-loop, but when using the ternary operator it doesn't. While debugging, I realized that each time the for-loop moves onto the next element 'goodVbad' becomes zero again. Kind of stumped, would love some advice. Thanks!
public static boolean goodbadClean(String word) {
String [] wordS;
int goodVbad=0;
String good="good";
String bad="bad";
word=word.toLowerCase();
word=word.replaceAll(good, " good ");
word=word.replaceAll(bad, " bad ");
wordS=word.split(" ");
for(String i:wordS) {
goodVbad=i.equals(good)?goodVbad++
:i.equals(bad) ?goodVbad--
:goodVbad;
}
if(goodVbad==0) {
return true;
}
return false;
}

The problem is the postfix ++ operator returns the old value, which you are assigning the variable, then increments. ie
goodVBad = goodVBad++; // returns the old value, so does nothing
so you should use the prefixed ++ operator:
goodVBad = ++goodVBad; // increments first, returning the new value
But both of these are hard to read and brittle.
If you must use ternaries, change your code to:
goodVbad += i.equals(good) ? 1 : (i.equals(bad) ? -1 : 0);
However, nested ternaries are generally a style smell. I recommend instead:
if (i.equals(good)) {
goodVBad++;
} else if (i.equals(bad)) {
goodVBad--;
}

Assuming that the OP needs to count the frequency of a string pattern in a given string, then you could do something like this with Java 8 or older:
public class CountMatches {
public static void main(String[] args) {
String phrase1 = "goodbadbadgoodgoodbad"; // equal amount of good vs bad.
String phrase2 = "goodbadbadgoodgoodbadbad"; // more bad than good.
String phrase3 = "goodbadbadgoodgoodbadgood"; // more good than bad.
// create capturing groups for "good" and "bad"
String GOOD_REGEX = "(good)";
String BAD_REGEX = "(bad)";
Pattern gPattern = Pattern.compile(GOOD_REGEX);
Pattern bPattern = Pattern.compile(BAD_REGEX);
Matcher countGood = gPattern.matcher(phrase1);
Matcher countBad = bPattern.matcher(phrase1);
int count = 0;
while (countBad.find()) {
count--;
}
while (countGood.find()) {
count++;
}
System.out.println(count == 0);
}
}
With Java 9 or later:
public class CountMatches {
public static void main(String[] args) {
String phrase1 = "goodbadbadgoodgoodbad"; // equal amount of good vs bad.
String phrase2 = "goodbadbadgoodgoodbadbad"; // more bad than good.
String phrase3 = "goodbadbadgoodgoodbadgood"; // more good than bad.
String GOOD_REGEX = "(good)";
String BAD_REGEX = "(bad)";
Pattern gPattern = Pattern.compile(GOOD_REGEX);
Pattern bPattern = Pattern.compile(BAD_REGEX);
Matcher countGood = gPattern.matcher(phrase1);
Matcher countBad = bPattern.matcher(phrase1);
long cCount = countGood.results().count();
long bCount = countBad.results().count();
System.out.println(cCount - bCount == 0);
}
}
My assumption is based on this line of code word=word.replaceAll(good, " good ");. This tells me that the expected input is something similar to the phrase variables I used for my testing.
By the way, this solution should work even if the words "good" and/or "bad" are preceded or followed by spaces.
UPDATE: Integrated looping to evaluate all expressions against all phrases.
public static void main(String[] args) {
List<String> expressions = List.of("(good)", "(bad)");
List<String> phrases = List.of("goodbadbadgoodgoodbad", "goodbadbadgoodgoodbadbad", "goodbadbadgoodgoodbadgood", " good bad bad good good bad ");
for (String phrase : phrases) {
List<Long> itemCount = new ArrayList<>();
for (String regex : expressions) {
Pattern gPattern = Pattern.compile(regex);
Matcher matcher = gPattern.matcher(phrase);
long count = matcher.results().count();
System.out.println("Pattern \"" + regex + "\" appears " + count + (count == 1 ? " time" : " times"));
itemCount.add(count);
}
long count = itemCount.stream().reduce((value1, value2) -> value1 - value2).get();
System.out.println(count == 0);
}
}
This outputs:
Pattern "(good)" appears 3 times
Pattern "(bad)" appears 3 times
true
Pattern "(good)" appears 3 times
Pattern "(bad)" appears 4 times
false
Pattern "(good)" appears 4 times
Pattern "(bad)" appears 3 times
false
Pattern "(good)" appears 3 times
Pattern "(bad)" appears 3 times
true

Email format type JAVA

I have a task to write a return type method which takes String as an argument and returns boolean. If the String matches with the requirements it returns true else false.
So it should not have any space, and more then 1 '#'and format must be 2>chars#2>chars.2>chars so it should be xyz#xyz.com is true but if any part of it less then 3 it should return to false. I checked so many forums but all i can find regex and we didn't learn anything about it. I could able to do this much but i just couldn't figure out how can i specifically set length of each part and set it as true or false. I'm missing so many and this is all i able to complete;
public static boolean emailAddress(String str) {
if (str.contains(" "))
return false;
boolean flag = true;
if(str3.length()> 11)
for (int i = 0; i < str3.length(); i++) {
for (int j = i + 1; j < str3.length(); j++) {
if (str3.charAt(i) == str3.charAt(j)) {
flag = false;
break;
}
}
if(flag)
return true;
}
return false;
}

This is how I would approach it:
public static boolean isEmailAddress(String str)
{
if(str.trim().equals("")) // Checking if it is an empty string
{
return false;
}
boolean flag = true;
int atIndex = str.indexOf("#"); // Get the index of #
if (atIndex == -1)
{
return false;
}
// Slice your email str and get only the characters before the #
// Check whether the length and the characters are as you want them.
// Do the same thing about the string between # and .
// and then do the same thing about the sliced string after .
}
Check substring() and indexOf().
For anyone who needs it.
This should be one of the Regex to match your requirement:
[a-zA-Z_0-9]{2,}#[a-zA-Z_0-9]{2,}\.[a-zA-Z_0-9]{2,}
[a-zA-Z_0-9] it requires one of the characters, from a-z,A-Z or 0-9
{2,} it requires two or more from one of the characters inside.

A few hints instead of complete solution:
Check if the input string...
has # and . (there could be two or three period's as some domain uses .co.in)
does not have any white spaces (' ')
does not start with special characters ('#', '^')
If possible, do not check for length of each part, as sometimes .in also valid domain.

how to compare two strings to find common substring

i get termination due to timeout error when i compile. Please help me
Given two strings, determine if they share a common substring. A substring may be as small as one character.
For example, the words "a", "and", "art" share the common substring "a" . The words "be" and "cat" do not share a substring.
Input Format
The first line contains a single integer , the number of test cases.
The following pairs of lines are as follows:
The first line contains string s1 .
The second line contains string s2 .
Output Format
For each pair of strings, return YES or NO.
my code in java
public static void main(String args[])
{
String s1,s2;
int n;
Scanner s= new Scanner(System.in);
n=s.nextInt();
while(n>0)
{
int flag = 0;
s1=s.next();
s2=s.next();
for(int i=0;i<s1.length();i++)
{
for(int j=i;j<s2.length();j++)
{
if(s1.charAt(i)==s2.charAt(j))
{
flag=1;
}
}
}
if(flag==1)
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
n--;
}
}
}
any tips?

Below is my approach to get through the same HackerRank challenge described above
static String twoStrings(String s1, String s2) {
String result="NO";
Set<Character> set1 = new HashSet<Character>();
for (char s : s1.toCharArray()){
set1.add(s);
}
for(int i=0;i<s2.length();i++){
if(set1.contains(s2.charAt(i))){
result = "YES";
break;
}
}
return result;
}
It passed all the Test cases without a time out issue.

The reason for the timeout is probably: to compare two strings that each are 1.000.000 characters long, your code needs 1.000.000 * 1.000.000 comparisons, always.
There is a faster algorithm that only needs 2 * 1.000.000 comparisons. You should use the faster algorithm instead. Its basic idea is:
for each character in s1: add the character to a set (this is the first million)
for each character in s2: test whether the set from step 1 contains the character, and if so, return "yes" immediately (this is the second million)
Java already provides a BitSet data type that does all you need. It is used like this:
BitSet seenInS1 = new BitSet();
seenInS1.set('x');
seenInS1.get('x');

Since you're worried about execution time, if they give you an expected range of characters (for example 'a' to 'z'), you can solve it very efficiently like this:
import java.util.Arrays;
import java.util.Scanner;
public class Whatever {
final static char HIGHEST_CHAR = 'z'; // Use Character.MAX_VALUE if unsure.
public static void main(final String[] args) {
final Scanner scanner = new Scanner(System.in);
final boolean[] characterSeen = new boolean[HIGHEST_CHAR + 1];
mainloop:
for (int word = Integer.parseInt(scanner.nextLine()); word > 0; word--) {
Arrays.fill(characterSeen, false);
final String word1 = scanner.nextLine();
for (int i = 0; i < word1.length(); i++) {
characterSeen[word1.charAt(i)] = true;
}
final String word2 = scanner.nextLine();
for (int i = 0; i < word2.length(); i++) {
if (characterSeen[word2.charAt(i)]) {
System.out.println("YES");
continue mainloop;
}
}
System.out.println("NO");
}
}
}
The code was tested to work with a few inputs.
This uses a fast array rather than slower sets, and it only creates one non-String object (other than the Scanner) for the entire run of the program. It also runs in O(n) time rather than O(n²) time.
The only thing faster than an array might be the BitSet Roland Illig mentioned.
If you wanted to go completely overboard, you could also potentially speed it up by:
skipping the creation of a Scanner and all those String objects by using System.in.read(buffer) directly with a reusable byte[] buffer
skipping the standard process of having to spend time checking for and properly handling negative numbers and invalid inputs on the first line by making your own very fast int parser that just assumes it's getting the digits of a valid nonnegative int followed by a newline

There are different approaches to solve this problem but solving this problem in linear time is a bit tricky.
Still, this problem can be solved in linear time. Just apply KMP algorithm in a trickier way.
Let's say you have 2 strings. Find the length of both strings first. Say length of string 1 is bigger than string 2. Make string 1 as your text and string 2 as your pattern. If the length of the string is n and length of the pattern is m then time complexity of the above problem would be O(m+n) which is way faster than O(n^2).
In this problem, you need to modify the KMP algorithm to get the desired result.
Just need to modify the KMP
public static void KMPsearch(char[] text,char[] pattern)
{
int[] cache = buildPrefix(pattern);
int i=0,j=0;
while(i<text.length && j<pattern.length)
{
if(text[i]==pattern[j])
{System.out.println("Yes");
return;}
else{
if(j>0)
j = cache[j-1];
else
i++;
}
}
System.out.println("No");
return;
}
Understanding Knuth-Morris-Pratt Algorithm

There are two concepts involved in solving this question.
-Understanding that a single character is a valid substring.
-Deducing that we only need to know that the two strings have a common substring — we don’t need to know what that substring is.
Thus, the key to solving this question is determining whether or not the two strings share a common character.
To do this, we create two sets, a and b, where each set contains the unique characters that appear in the string it’s named after.
Because sets 26 don’t store duplicate values, we know that the size of our sets will never exceed the letters of the English alphabet.
In addition, the small size of these sets makes finding the intersection very quick.
If the intersection of the two sets is empty, we print NO on a new line; if the intersection of the two sets is not empty, then we know that strings and share one or more common characters and we print YES on a new line.
In code, it may look something like this
import java.util.*;
public class Solution {
static Set<Character> a;
static Set<Character> b;
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
for(int i = 0; i < n; i++) {
a = new HashSet<Character>();
b = new HashSet<Character>();
for(char c : scan.next().toCharArray()) {
a.add(c);
}
for(char c : scan.next().toCharArray()) {
b.add(c);
}
// store the set intersection in set 'a'
a.retainAll(b);
System.out.println( (a.isEmpty()) ? "NO" : "YES" );
}
scan.close();
}
}

public String twoStrings(String sOne, String sTwo) {
if (sOne.equals(sTwo)) {
return "YES";
}
Set<Character> charSetOne = new HashSet<Character>();
for (Character c : sOne.toCharArray())
charSetOne.add(c);
Set<Character> charSetTwo = new HashSet<Character>();
for (Character c : sTwo.toCharArray())
charSetTwo.add(c);
charSetOne.retainAll(charSetTwo);
if (charSetOne.size() > 0) {
return "YES";
}
return "NO";
}
This must work. Tested with some large inputs.

Python3
def twoStrings(s1, s2):
flag = False
for x in s1:
if x in s2:
flag = True
if flag == True:
return "YES"
else:
return "NO"
if __name__ == '__main__':
q = 2
text = [("hello","world"), ("hi","world")]
for q_itr in range(q):
s1 = text[q_itr][0]
s2 = text[q_itr][1]
result = twoStrings(s1, s2)
print(result)

static String twoStrings(String s1, String s2) {
for (Character ch : s1.toCharArray()) {
if (s2.indexOf(ch) > -1)
return "YES";
}
return "NO";
}

Get all combinations of Strings

I have requirement that need to get all combinations of given Strings.
For example
I have a String of digits and some special charcters
String chars="0123456789##$%&";
String guessedPw="authentic";
So I want to get combinations like this
$authentic%4
authentic##
5&authentic
authentic8
How should I improve my method to get all combinations?
This is the code that I have written.
but it doesn't give me all combinations.
private static String combination(String prefix, String s, String pw, String md5Pw) {
String pwFound = "";
if (s.length() > 0) {
// System.out.println(prefix + s.charAt(0) + pw);
String tempPw1 = prefix + s.charAt(0) + pw;
System.out.println("pw1 : " + tempPw1);
if (md5(tempPw1).equals(md5Pw)) {
// String tempPw1;
pwFound = tempPw1;
return pwFound;
}
String tempPw2 = pw + prefix + s.charAt(0);
if (md5(tempPw2).equals(md5Pw)) {
// String tempPw1;
pwFound = tempPw2;
return pwFound;
}
pwFound = combination(prefix + s.charAt(0), s.substring(1), pw, md5Pw);
pwFound = combination(prefix, s.substring(1), pw, md5Pw);
}
return pwFound;
}

Just if you don't want write own algo, you can use google.
Try to search for: "Generate Permutations".
For example, on this link: Generate all combinations from multiple lists there is a algo which you can use it (but with list).
void GeneratePermutations(List<List<Character>> Lists, List<String> result, int depth, String current)
{
if(depth == Lists.size())
{
result.add(current);
return;
}
for(int i = 0; i < Lists.get(depth).size(); ++i)
{
GeneratePermutations(Lists, result, depth + 1, current + Lists.get(depth).get(i));
}
}
But sure, there are a lot of other ways.

If you want to code everything by your own then this is how you should approach your problem =>
Have a special character say ~ to denote guessedPw, and create another string say str = chars + "~".
Now first you need to look for all possible combinations of str, finding which will take exponential time , and then for each found combination, you should generate all it's permutation , which is again factorial time complex. And then in the final string , you should replace all the occurrence of '~' with guessedPw, and get the answer string.
Here you go for the link for generating permutations and combinations :
Link : http://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/
To find out combinations , you can use bit-masking if you have at max. 64 characters in the string , combinations from which you want to extract.

To make new combination, you can add one char as prefix or add one char as suffix. To avoid repeats, stop adding prefix when suffix exists
Pseudocode:
GenPass(CurrentString, boolAllowPrefix)
check CurrentString, return it if good result
check length limit, exit if too long
for c in Chars
if (boolAllowPrefix)
GenPass(Chars[i] + CurrentString, True)
GenPass(s + Chars[i], False)
Delphi implementation to check
procedure GenPass(s: string; bPrefix: Boolean);
var
i: integer;
begin
List.Add(s);
if Length(s) = MaxLen then
Exit;
if bPrefix then
for i := 1 to Length(Chars) do
GenPass(Chars[i] + s, True);
for i := 1 to Length(Chars) do
GenPass(s + Chars[i], False);
end;

Java codingbat help - withoutString

I'm using codingbat.com to get some java practice in. One of the String problems, 'withoutString' is as follows:
Given two strings, base and remove, return a version of the base string where all instances of the remove string have been removed (not case sensitive).
You may assume that the remove string is length 1 or more. Remove only non-overlapping instances, so with "xxx" removing "xx" leaves "x".
This problem can be found at: http://codingbat.com/prob/p192570
As you can see from the the dropbox-linked screenshot below, all of the runs pass except for three and a final one called "other tests." The thing is, even though they are marked as incorrect, my output matches exactly the expected output for the correct answer.
Here's a screenshot of my output:
And here's the code I'm using:
public String withoutString(String base, String remove) {
String result = "";
int i = 0;
for(; i < base.length()-remove.length();){
if(!(base.substring(i,i+remove.length()).equalsIgnoreCase(remove))){
result = result + base.substring(i,i+1);
i++;
}
else{
i = i + remove.length();
}
if(result.startsWith(" ")) result = result.substring(1);
if(result.endsWith(" ") && base.substring(i,i+1).equals(" ")) result = result.substring(0,result.length()-1);
}
if(base.length()-i <= remove.length() && !(base.substring(i).equalsIgnoreCase(remove))){
result = result + base.substring(i);
}
return result;
}

Your solution IS failing AND there is a display bug in coding bat.
The correct output should be:
withoutString("This is a FISH", "IS") -> "Th a FH"
Yours is:
withoutString("This is a FISH", "IS") -> "Th a FH"
Yours fails because it is removing spaces, but also, coding bat does not display the correct expected and run output string due to HTML removing extra spaces.
This recursive solution passes all tests:
public String withoutString(String base, String remove) {
int remIdx = base.toLowerCase().indexOf(remove.toLowerCase());
if (remIdx == -1)
return base;
return base.substring(0, remIdx ) +
withoutString(base.substring(remIdx + remove.length()) , remove);
}
Here is an example of an optimal iterative solution. It has more code than the recursive solution but is faster since far fewer function calls are made.
public String withoutString(String base, String remove) {
int remIdx = 0;
int remLen = remove.length();
remove = remove.toLowerCase();
while (true) {
remIdx = base.toLowerCase().indexOf(remove);
if (remIdx == -1)
break;
base = base.substring(0, remIdx) + base.substring(remIdx + remLen);
}
return base;
}

I just ran your code in an IDE. It compiles correctly and matches all tests shown on codingbat. There must be some bug with codingbat's test cases.
If you are curious, this problem can be solved with a single line of code:
public String withoutString(String base, String remove) {
return base.replaceAll("(?i)" + remove, ""); //String#replaceAll(String, String) with case insensitive regex.
}
Regex explaination:
The first argument taken by String#replaceAll(String, String) is what is known as a Regular Expression or "regex" for short.
Regex is a powerful tool to perform pattern matching within Strings. In this case, the regular expression being used is (assuming that remove is equal to IS):
(?i)IS
This particular expression has two parts: (?i) and IS.
IS matches the string "IS" exactly, nothing more, nothing less.
(?i) is simply a flag to tell the regex engine to ignore case.
With (?i)IS, all of: IS, Is, iS and is will be matched.
As an addition, this is (almost) equivalent to the regular expressions: (IS|Is|iS|is), (I|i)(S|s) and [Ii][Ss].
EDIT
Turns out that your output is not correct and is failing as expected. See: dansalmo's answer.

public String withoutString(String base, String remove) {
String temp = base.replaceAll(remove, "");
String temp2 = temp.replaceAll(remove.toLowerCase(), "");
return temp2.replaceAll(remove.toUpperCase(), "");
}

Please find below my solution
public String withoutString(String base, String remove) {
final int rLen=remove.length();
final int bLen=base.length();
String op="";
for(int i = 0; i < bLen;)
{
if(!(i + rLen > bLen) && base.substring(i, i + rLen).equalsIgnoreCase(remove))
{
i +=rLen;
continue;
}
op += base.substring(i, i + 1);
i++;
}
return op;
}
Something things go really weird on codingBat this is just one of them.

I am adding to a previous solution, but using a StringBuilder for better practice. Most credit goes to Anirudh.
public String withoutString(String base, String remove) {
//create a constant integer the size of remove.length();
final int rLen=remove.length();
//create a constant integer the size of base.length();
final int bLen=base.length();
//Create an empty string;
StringBuilder op = new StringBuilder();
//Create the for loop.
for(int i = 0; i < bLen;)
{
//if the remove string lenght we are looking for is not less than the base length
// and the base substring equals the remove string.
if(!(i + rLen > bLen) && base.substring(i, i + rLen).equalsIgnoreCase(remove))
{
//Increment by the remove length, and skip adding it to the string.
i +=rLen;
continue;
}
//else, we add the character at i to the string builder.
op.append(base.charAt(i));
//and increment by one.
i++;
}
//We return the string.
return op.toString();
}

Taylor's solution is the most efficient one, however I have another solution that is a naive one and it works.
public String withoutString(String base, String remove) {
String returnString = base;
while(returnString.toLowerCase().indexOf(remove.toLowerCase())!=-1){
int start = returnString.toLowerCase().indexOf(remove.toLowerCase());
int end = remove.length();
returnString = returnString.substring(0, start) + returnString.substring(start+end);
}
return returnString;
}

#Daemon
your code works. Thanks for the regex explanation. Though dansalmo pointed out that codingbat is displaying the intended output incorrectly, I through in some extra lines to your code to unnecessarily account for the double spaces with the following:
public String withoutString(String base, String remove){
String result = base.replaceAll("(?i)" + remove, "");
for(int i = 0; i < result.length()-1;){
if(result.substring(i,i+2).equals(" ")){
result = result.replace(result.substring(i,i+2), " ");
}
else i++;
}
if(result.startsWith(" ")) result = result.substring(1);
return result;
}

public String withoutString(String base, String remove){
return base.replace(remove,"");
}