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codingBat question: Returns true if for every '*' (star) in the string

i am working on this codingbat question:
Returns true if for every '*' (star) in the string, if there are chars both immediately before and after the star, they are the same.
Example:
sameStarChar("xy*yzz") → true
sameStarChar("xy*zzz") → false
sameStarChar("*xa*az") → true
My first attempt of code is :
public boolean sameStarChar(String str) {
//boolean flag = false;
for(int i =1;i< str.length()-1; i++){
if(str.charAt(i) == '*' && str.charAt(i-1) == str.charAt(i+1)){
return true;
}
}
return false;
}
and did not pass few case such: sameStarChar("12*2*3*") → false , sameStarChar("XYYYYYZ*") → false.
However, if changed the code to:
public boolean sameStarChar(String str) {
//boolean flag = false;
for(int i =1;i< str.length()-1; i++){
if(str.charAt(i) == '*' && str.charAt(i-1) != str.charAt(i+1)){
return false;
}
}
return true;
}
The result is corrected. I do not understand what is wrong with my first attempt.

In your second attempt you fixed the only problem you had with your code.
Your first attempt will be immediately terminate by returning true, if there is an * surrounded by matching chars. (Here you will miss out the upcoming *s with mismatching chars)
Your second case now will check whether your string contain an * which is surrounded by mismatching chars, and terminate it by returning false.
NOTE: This is an example where you align your checks to detect the negative scenario rather than traversing all and finding the positive ones.
In your case you can do two things (NOTE: we can safely remove the corner two, and you already done that),
Return true if : Every * in the string is surrounded by matching chars
OR
Return false if : There is at least one * which is surrounded by a mismatching chars.

Two issues with your first attempt:
It fails to consider the case of a string that doesn't contain any asterisks at all, for which the loop will run to completion and then false will incorrectly be returned.
It incorrectly returns true on the first pair of matched neighbours of an asterisk, regardless of any unmatched pairs that may follow.

In your first case, you are returning as soon as you find one valid match/occurrence which is wrong. Also, you are not checking for the cases where there are * at one or either ends.
In the second case, you are searching explicitly for mismatches. Also, the corner cases of * at ends naturally becomes a match/success.

finding the middle index of a substring when there are duplicates in the string

I was working on a Java coding problem and encountered the following issue.
Problem:
Given a string, does "xyz" appear in the middle of the string? To define middle, we'll say that the number of chars to the left and right of the "xyz" must differ by at most one
xyzMiddle("AAxyzBB") → true
xyzMiddle("AxyzBBB") → false
My Code:
public boolean xyzMiddle(String str) {
boolean result=false;
if(str.length()<3)result=false;
if(str.length()==3 && str.equals("xyz"))result=true;
for(int j=0;j<str.length()-3;j++){
if(str.substring(j,j+3).equals("xyz")){
String rightSide=str.substring(j+3,str.length());
int rightLength=rightSide.length();
String leftSide=str.substring(0,j);
int leftLength=leftSide.length();
int diff=Math.abs(rightLength-leftLength);
if(diff>=0 && diff<=1)result=true;
else result=false;
}
}
return result;
}
Output I am getting:
Running for most of the test cases but failing for certain edge cases involving more than once occurence of "xyz" in the string
Example:
xyzMiddle("xyzxyzAxyzBxyzxyz")
My present method is taking the "xyz" starting at the index 0. I understood the problem. I want a solution where the condition is using only string manipulation functions.
NOTE: I need to solve this using string manipulations like substrings. I am not considering using list, stringbuffer/builder etc. Would appreciate answers which can build up on my code.

There is no need to loop at all, because you only want to check if xyz is in the middle.
The string is of the form
prefix + "xyz" + suffix
The content of the prefix and suffix is irrelevant; the only thing that matters is they differ in length by at most 1.
Depending on the length of the string (and assuming it is at least 3):
Prefix and suffix must have the same length if the (string's length - the length of xyz) is even. In this case:
int prefixLen = (str.length()-3)/2;
result = str.substring(prefixLen, prefixLen+3).equals("xyz");
Otherwise, prefix and suffix differ in length by 1. In this case:
int minPrefixLen = (str.length()-3)/2;
int maxPrefixLen = minPrefixLen+1;
result = str.substring(minPrefixLen, minPrefixLen+3).equals("xyz") || str.substring(maxPrefixLen, maxPrefixLen+3).equals("xyz");
In fact, you don't even need the substring here. You can do it with str.regionMatches instead, and avoid creating the substrings, e.g. for the first case:
result = str.regionMatches(prefixLen, "xyz", 0, 3);

Super easy solution:
Use Apache StringUtils to split the string.
Specifically, splitByWholeSeparatorPreserveAllTokens.
Think about the problem.
Specifically, if the token is in the middle of the string then there must be an even number of tokens returned by the split call (see step 1 above).
Zero counts as an even number here.
If the number of tokens is even, add the lengths of the first group (first half of the tokens) and compare it to the lengths of the second group.
Pay attention to details,
an empty token indicates an occurrence of the token itself.
You can count this as zero length, count as the length of the token, or count it as literally any number as long as you always count it as the same number.
if (lengthFirstHalf == lengthSecondHalf) token is in middle.

Managing your code, I left unchanged the cases str.lengt<3 and str.lengt==3.
Taking inspiration from #Andy's answer, I considered the pattern
prefix+'xyz'+suffix
and, while looking for matches I controlled also if they respect the rule IsMiddle, as you defined it. If a match that respect the rule is found, the loop breaks and return a success, else the loop continue.
public boolean xyzMiddle(String str) {
boolean result=false;
if(str.length()<3)
result=false;
else if(str.length()==3 && str.equals("xyz"))
result=true;
else{
int preLen=-1;
int sufLen=-2;
int k=0;
while(k<str.lenght){
if(str.indexOf('xyz',k)!=-1){
count++;
k=str.indexOf('xyz',k);
//check if match is in the middle
preLen=str.substring(0,k).lenght;
sufLen=str.substring(k+3,str.lenght-1).lenght;
if(preLen==sufLen || preLen==sufLen-1 || preLen==sufLen+1){
result=true;
k=str.length; //breaks the while loop
}
else
result=false;
}
else
k++;
}
}
return result;
}

Prompting a String value and check if it matches with a specific pattern. How can I solve it by using for loop and substring method?

I am doing an assignment which asks me check the given phone number whether it matches the pattern(the pattern is ddd-ddd-dddd) and the numbers are within 0 to 9. I want to use for loop and substring method, but I have no idea how to use them. This code is what I've done so far.
public static boolean phoneNum(String s){
boolean a = false;
boolean b = false;
String phone = s.substring(1,4)+s.substring(5,8)+s.substring(9);
for(int i =0; i<phone.length(); i++){
y = Character.isDigit(s1.charAt(i));
}
if(s.charAt(4)=='-' && s.charAt(8)=='-' && b==true){
a = true;
return a;
}
return a;
}
If I can add for loop, where should it be?

You are asking how to improve the quality of this code, so let's give some ideas:
change its name to meet java conventions, like isValidPhoneNumber()
you don't need to "carry forward" the results of a previous check. Your method should simply return false immediately whenever it finds a condition to be violated. If all tests pass, you then simply return true in the end.
In other words: what makes your code hard to read and overly complicated are those single character named boolean variables. They mainly add confusion. And keep in mind that your code is ignoring that y variable for example. And even if y is just a typo that could still be simplified.
And to make that clear: avoid using names like a or b. These names mean nothing. They don't tell the reader anything about the purpose of that variable.

Need some help in writing a recursive implementation for my Palindrome app

First of all I am not asking for people to "do my homework" like I have seen others on here ask for. I have managed to code a working iterative version of a program that determines if a string is a palindrome or not. Spaces, punctuation and special characters are ignored while determining if the string is a palindrome. This version does work but when I try and apply recursive statements in the "isPalindrome()" method I get Stack Overflow errors. I know what these errors are, it's just that applying a recursive method in a program like this is quite hard for me to get my head around (I only got taught about them 2 weeks ago). Anyway here is the code I have managed to compile (and run) so far:
/** Palindrome.java: A sigle application class that determines if a word or a string
* is a palindrome or not. This application is designed to ignore spaces between
* chars, punctuation marks and special characters while determining if the word or
* string is a palindrome or not.
*
**/
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.regex.*;
public class Palindrome{
static String palindrome, str, str2, str3;
/** The main method of the Palindrome application. Takes input from the
* user, removes spaces from their input, turns their string input into
* lowercase and then all non letter characters are taken out of the user's
* input. Finally the recursive method determines if the string entered in
* by the user is a palindrome.
*
* #param args Takes in a string array of arguements
**/
public static void main(String[] args){
Scanner input = new Scanner(System.in);
while(input.hasNext()){
str = removeSpaces(input.nextLine());
str2 = str.toLowerCase();
str3 = normalise(str2);
}
System.out.println(isPalindrome(str3));
}
/** The default constructor
**/
public Palindrome(){
}
/** isPalindrome(): A boolean method that is passed through a String input
* and uses a for loop, two inner while loops and an if-else to determine
* whether the users input is a palindrome.
*
* #param s The string input to be tested
* #return true The users input is a palindrome
* #return false The users input isn't a palindrome
**/
public static boolean isPalindrome(String s){
int first, last;
for(first = 0, last = s.length()-1 ; first < last ; first++ , last-- ){
while( (int)s.charAt(first) < 'a' || (int)s.charAt(first) > 'z' ){
first++;
}
while( (int)s.charAt(last ) < 'a' || (int)s.charAt(last ) > 'z' ){
last--;
}
}
if( first > last || s.charAt(first) != s.charAt(last) ){
//return isPalindrome(s.substring(0, s.length()-1)) == false;
return false;
}
else{
//return isPalindrome(s.substring(0, s.length()-1)) == true;
return true;
}
}
/**
* This method takes out punctuation marks in the string parsed
* through, using Java's regular expressions (regex) and Java's
* inbuilt method replaceAll(). The regex expression is passed
* through the replaceAll() method to remove all non alpha-numeric
* characters from the string passed through the method's parameter.
*
* #param t The string that will have punctuation stripped from it.
*
* #return t The string has had all non alpha-numeric characters
* removed and the new string is then returned.
*/
public static String normalise(String t){
t = t.replaceAll("[^a-zA-Z0-9]", "");
return t;
}
/** removeSpaces(): A method that deletes spaces from the users input
* and then decrements the string length count so any indexes aren't missed
* when it is incremented.
*
* #param s The string which is going to have it's spaces removed.
* #return temp The new string is then returned after the spaces have been taken out.
**/
public static String removeSpaces(String s){
StringBuilder temp = new StringBuilder(s); //creates a new StringBuilder with the inputted String
for(int i = 0; i < temp.length(); i++){ //do this for the entire length of the StringBuilder
if(temp.charAt(i) == ' '){ //if the char at i is a space
temp.deleteCharAt(i); //remove the char
i--; //subtract 1 from the counter so we don't miss an index when we increment it
}
}
return temp.toString(); //return the new String
}
}
I have blanked out the recursive statements in the recursive method for now. If someone can tell me what exactly I have done wrong and also help me in implementing a solution that would be really good. I would rather stick with the iterative version because I understand the mechanics of it, but have been asked to do a recursive version (I have been Java coding since after my mid year break last year but am a relative novice at recursion) which is proving to be quite a challenge. If you alter the code and it ends up working with the recursive version please explain how, when, why etc with your alterations. Am not looking for someone to just do this for me, I'm wanting to learn and it seems that I have learned best by example (I did get a B pass last year by analysing examples and reading explanations of implementations). Many thanks :).
EDIT: I think I have got the recursion going ok now, just the logic is the thing confusing me at the moment. Here is the recoded version of the isPalindrome() method:
public static boolean isPalindrome(String s){
int first, last;
boolean isPalindr = true;
if (s.length() <= 1){
return true; // Base case
}
for(first = 0, last = s.length()-1 ; first < last ; first++ , last-- ){
// while( (int)s.charAt(first) < 'a' || (int)s.charAt(first) > 'z' ){
// first++;
// }
// while( (int)s.charAt(last ) < 'a' || (int)s.charAt(last ) > 'z' ){
// last--;
// }
// }
if( first == last || s.charAt(first) == s.charAt(last) ){
//return isPalindrome(s.substring(first, last));
return isPalindrome(s.substring(first, last)) == true;
//isPalindr = false;
}
else{
return isPalindrome(s.substring(first, last)) == false;
//isPalindr = true;
}
}
return isPalindr;
}
If someone can help me with the logic I think this will be fixed :).

Removing all of the code that has nothing to do with the problem leaves us with this:
public static boolean isPalindrome(String s){
for loop {
isPalindrome();
}
}
isPalindrome calls isPalindrome calls isPalindrome, etc... infinitum.
The difference between this and a proper recursive function is that a recursive function will have some sort of conditional statement, breaking the cycle of the function calling itself. The flow of execution will go like this:
isPalindrome(1) begins execution and calls isPalidrome(2)
isPalindrome(2) begins execution and calls isPalidrome(3)
isPalindrome(3) begins execution and calls isPalidrome(4)
isPalindrome(4) begins execution and calls isPalidrome(5)
isPalindrome(5) begins execution and returns to isPalindrome(4)
isPalindrome(4) resumes execution and returns to isPalindrome(3)
isPalindrome(3) resumes execution and returns to isPalindrome(2)
isPalindrome(2) resumes execution and returns to isPalindrome(1)
isPalindrome(1) resumes execution and returns.
If that explanation doesn't help, think of it like this. Suppose someone was handing you plates, one at a time, to see if you can hold 25 plates at a time. It would go something like this:
Plate 1 is given to you. Are there 25 plates? No. Add another plate.
Plate 2 is stacked on top of Plate 1. Are there 25 plates? No. Add another plate.
Plate 3 is stacked on top of Plate 2. Are there 25 plates? No. Add another plate.
...
Plate 24 is stacked on top of Plate 23. Are there 25 plates? No. Add another plate.
Plate 25 is stacked on top of Plate 24. Are there 25 plates? Yes. Mission Accomplished. Now, let's put the plates back.
Plate 25 is removed.
Plate 24 is removed.
...
Plate 3 is removed.
Plate 2 is removed.
Plate 1 is removed.
Here's how that might be coded:
bool stackPlates(int i){
plateStack.addPlate();
if (plateStack.wasDropped == true) { return false; } // Were the plates dropped? Return FALSE to indicate failure.
else if (i < 25) { return stackPlates(i+1); } // Are there 25 plates yet? If not, add another.
else { return true; } // There are 25 plates stacked. Return TRUE to indicate success.
plateStack.removePlate(i);
}
Here's stackPlates(int i) called from another function:
bool success = stackPlates(1);
if (success==TRUE) { cout << "CONGRATULATIONS! YOU STACKED 25 PLATES!"; }
else { cout << "YOU BROKE THE PLATES! BETTER LUCK NEXT TIME!"; }
What your function needs to do in order to work properly is do this:
bool isPalindrome(string s, int i) {
char first = s[i]; // REPLACE THIS WITH THE CODE TO SKIP SPACES & SPECIAL CHARACTERS
char last = s[(s.length -1) -i]; // REPLACE THIS WITH THE CODE TO SKIP SPACES & SPECIAL CHARACTERS
if ( first != last ) { return false; } // return false if mismatch letter
else if ( i >= (s.length/2) ) { return true; } // return true if string fully checked
else { return isPalindrome(s, i+1); } // string not fully checked; move to next letter
}

You're experiencing stack overflows because the else branch at the bottom of the function is executed when (first <= last && "characters are equals"), so you keep recurring on the case where your string is composed by one character.
By the way, I think your code is not using recursion cleanly: you should preprocess your string only one time before starting recurring on the string, and the code that performs the palindrome recursion should be far simpler.

For any given entry into isPalindrome, it's going to recursively call itself regardless because you have no condition on your else. So, if it meets the criteria "first > last || s.charAt(first) != s.charAt(last)", it's going to recursively call isPalindrome, then the next call is too, even if it hits the else.
I don't know what a Palindrome is or what the real solution to the problem is, but that's why you're getting the stack overflow error. I suspect you need to add another condition to your else such that it will stop recursively calling itself.

When writing a recursive function the best way to go about this is usually to decide on a base case (:like "" is a palindrome, though so is "a" ... ) and then devise a method to take any state and move it to the base case.
So in the case of the palindrome, it's the same basic idea as before, if the first character and the last character are the same you return true and check the rest of the string ( thus moving closer to the base case ) and if they are not then you return false.
Your stack overflow comes from calling isPalindrome in every case rather than when you need to continue solving the problem, don't forget that if two characters mean that something isn't a palindrome, the rest is rendered irrelevant ( and thus needn't be recursed on )

Your recoded version is a bit strange, because it's still using a loop when it doesn't need to. In particular, your code will never go beyond the first iteration in your loop, because in the embedded if-else statement, you're going to return a result no matter what, so your function will always exit during the first iteration (unless there are no iterations at all).
Recursion should be approached by
Identifying a base case, i.e. a simplest case that can be solved
Re-representing a larger problem as a partial solution followed by the same, but smaller problem.
The base case you've handled correctly; any String which is length 1 or less is automatically a Palindrome.
The next step is to consider a larger problem, perhaps some string abcwewe....ba. How can we break this down into a simpler problem? We know that we'd normally check whether something is a palindrome by checking the letters one by one in pairs, starting at the ends, but then we also realise that each time we check the letters, we just repeat the same problem again and solve it the same way.
In the string I gave above, we check and verify that the first letter a is the same as the last letter a, so that's kind of a partial solution. Now we we end up with is the smaller word bcwewe....b, and it's the same problem again: Is this new String a palindrome also?
Thus, all you have to do now is to invoke the recursive call, but this time with the substring beginning with the 2nd character to the 2nd to last character. You can code the answer in just two lines, as below:
public static boolean isPalindrome(String s) {
if (s.length() <= 1) return true; // base case
return s.charAt(0) == s.charAt(s.length()-1) && isPalin(s.substring(1,s.length()-1)); // recursive case
}
One point to note is that I'm using the short circuit &&, so if the first condition fails (checking first and last character), then Java will not invoke the recursion.

Determine if a string can be transformed to another string when only insert/remove/replace operations are allowed

I must write a function that takes two words (strings) as arguments, and determines if the first word can be transformed into the second word using only one first-order transformation.
First-order transformations alter only one letter in a word
The allowed transformations are: insert, remove and replace
insert = insert a letter at any position in the word
remove = delete a letter from any position in the word
replace = replace a letter with another one
Any suggestions? Any Java examples would be great!

Think: If you're only allowed a single transformation, then the difference in length between the "before" and "after" words should give you a very strong hint as to which of those three transformations has any chance of being successful. By the same token, you can tell at a glance which transformations will be simply impossible.
Once you've decided on which transformation, the rest of the problem becomes a job for Brute Force Man and his sidekick, Looping Lady.

This does look like homework so I'm not going to give you the answer, but any time you approach a problem like this the best thing to do is start sketching out some ideas. Break the problem down into smaller chunks, and then it becomes easier to solve.
For example, let's look at the insert operation. To insert an letter, what is that going to do to the length of the word in which we are inserting the letter? Increase it or decrease it? If we increase the length of the word, and the length of this new word is not equal to the length of the word we are trying to match, then what does that tell you? So one condition here is that if you are going to perform an insert operation on the first word to make it match the second word, then there is a known length that the first word must be.
You can apply similar ideas to the other 2 operations.
So once you establish these conditions, it becomes easier to develop an algorithm to solve the problem.
The important thing in any type of assignment like this is to think through it. Don't just ask somebody, "give me the code", you learn nothing like that. When you get stuck, it's ok to ask for help (but show us what you've done so far), but the purpose of homework is to learn.

If you need to check if there is one and exactly one edit from s1 to s2, then this is very easy to check with a simple linear scan.
If both have the same length, then there must be exactly one index where the two differ
They must agree up to a common longest prefix, then skipping exactly one character from both, they must then agree on a common suffix
If one is shorter than the other, then the difference in length must be exactly one
They must agree up to a common longest prefix, then skipping exactly one character from the longer one, they must then agree on a common suffix
If you also allow zero edit from s1 to s2, then simply check if they're equal.
Here's a Java implementation:
static int firstDifference(String s1, String s2, int L) {
for (int i = 0; i < L; i++) {
if (s1.charAt(i) != s2.charAt(i)) {
return i;
}
}
return L;
}
static boolean oneEdit(String s1, String s2) {
if (s1.length() > s2.length()) {
return oneEdit(s2, s1);
}
final int L = s1.length();
final int index = firstDifference(s1, s2, L);
if (s1.length() == s2.length() && index != L) {
return s1.substring(index+1).equals(s2.substring(index+1));
} else if (s2.length() == L + 1) {
return s1.substring(index).equals(s2.substring(index+1));
} else {
return false;
}
}
Then we can test it as follows:
String[][] tests = {
{ "1", "" },
{ "123", "" },
{ "this", "that" },
{ "tit", "tat" },
{ "word", "sword" },
{ "desert", "dessert" },
{ "lulz", "lul" },
{ "same", "same" },
};
for (String[] test : tests) {
System.out.printf("[%s|%s] = %s%n",
test[0], test[1], oneEdit(test[0], test[1])
);
}
This prints (as seen on ideone.com):
[1|] = true
[123|] = false
[this|that] = false
[tit|tat] = true
[word|sword] = true
[desert|dessert] = true
[lulz|lul] = true
[same|same] = false

You can use the Levenshtein distance and only allow distances of 1 (which means, one char must be altered). There are several implementations just google "Levenshtein java" or so.
The other "not so smart" but working thing would be the good old brute force. Just try out every situation with every char and you get what you want. :-)