We Keep Coding

Recursive way to parse a string

I have a string "{x{y}{a{b{c}{d}}}}"
And want to print out recursively.
x
-y
-a
--b
---c
---d
This is what I have so far -
private static void printPathInChild2(String path) {
if (path.length() == 0) {
return;
}
if (path.charAt(0) == '{') {
for (int i = 0; i < path.length(); i++) {
if (path.charAt(i) == '{' && i != 0) {
String t1 = path.substring(0,i);
System.out.println(t1);
printPathInChild2(path.substring(i));
} else if (path.charAt(i) == '}') {
String t2 = path.substring(0, i+1);
System.out.println(t2);
printPathInChild2(path.substring(i+1));
}
}
}
}
Struggling with the termination logic

If you want to add '-' characters that depend on the depth of the nesting, you should pass a second argument to the recursive call, which keeps track of the prefix of '-' characters.
When you encounter a '{', you add a '-' to the prefix.
When you encounter a '}', you remove a '-' from the prefix.
When you encounter any other character, you print the prefix followed by that character.
private static void printPathInChild2(String path,String prefix) {
if (path.length() == 0) {
return;
}
if (path.charAt(0) == '{') {
printPathInChild2(path.substring(1),prefix + "-");
} else if (path.charAt(0) == '}') {
printPathInChild2(path.substring(1),prefix.substring(0,prefix.length()-1));
} else {
System.out.println (prefix.substring(1) + path.charAt(0));
printPathInChild2(path.substring(1),prefix);
}
}
When you call this method with:
printPathInChild2("{x{y}{a{b{c}{d}}}}","");
You get:
x
-y
-a
--b
---c
---d
(I see that in your expected output 'd' has 4 '-'s, but I think it's an error, since 'd' has the same nesting level as 'c', so it should have 3 '-'s).
The method can also be written as follows:
private static void printPathInChild2(String path,String prefix) {
if (path.length() == 0) {
return;
}
char c = path.charAt(0);
if (c == '{') {
prefix = prefix + '-';
} else if (c == '}') {
prefix = prefix.substring(0,prefix.length()-1);
} else {
System.out.println (prefix.substring(1) + c);
}
printPathInChild2(path.substring(1),prefix);
}

Verify if String matches real number

I'm trying to verify if a String s match/is a real number. For that I created this method:
public static boolean Real(String s, int i) {
boolean resp = false;
//
if ( i == s.length() ) {
resp = true;
} else if ( s.charAt(i) >= '0' && s.charAt(i) <= '9' ) {
resp = Real(s, i + 1);
} else {
resp = false;
}
return resp;
}
public static boolean isReal(String s) {
return Real(s, 0);
}
But obviously it works only for round numbers. Can anybody give me a tip on how to do this?
P.S: I can only use s.charAt(int) e length() Java functions.

You could try doing something like this. Added recursive solution as well.
public static void main(String[] args) {
System.out.println(isReal("123.12"));
}
public static boolean isReal(String string) {
boolean delimiterMatched = false;
char delimiter = '.';
for (int i = 0; i < string.length(); i++) {
char c = string.charAt(i);
if (!(c >= '0' && c <= '9' || c == delimiter)) {
// contains not number
return false;
}
if (c == delimiter) {
// delimiter matched twice
if (delimiterMatched) {
return false;
}
delimiterMatched = true;
}
}
// if matched delimiter once return true
return delimiterMatched;
}
Recursive solution
public static boolean isRealRecursive(String string) {
return isRealRecursive(string, 0, false);
}
private static boolean isRealRecursive(String string, int position, boolean delimiterMatched) {
char delimiter = '.';
if (position == string.length()) {
return delimiterMatched;
}
char c = string.charAt(position);
if (!(c >= '0' && c <= '9' || c == delimiter)) {
// contains not number
return false;
}
if (c == delimiter) {
// delimiter matched twice
if (delimiterMatched) {
return false;
}
delimiterMatched = true;
}
return isRealRecursive(string, position+1, delimiterMatched);
}

You need to use Regex. The regex to verify that whether a string holds a float number is:
^[-+]?[0-9]*\.?[0-9]+([eE][-+]?[0-9]+)?$

Can anybody give me a tip on how to do this?
Starting with your existing recursive matcher for whole numbers, modify it and use it in another method to match the whole numbers in:
["+"|"-"]<whole-number>["."[<whole-number>]]
Hint: you will most likely need to change the existing method to return the index of that last character matched rather than just true / false. Think of the best way to encode "no match" in an integer result.

public static boolean isReal(String str) {
boolean real = true;
boolean sawDot = false;
char c;
for(int i = str.length() - 1; 0 <= i && real; i --) {
c = str.charAt(i);
if('-' == c || '+' == c) {
if(0 != i) {
real = false;
}
} else if('.' == c) {
if(!sawDot)
sawDot = true;
else
real = false;
} else {
if('0' > c || '9' < c)
real = false;
}
}
return real;
}

Check if a string A can be created from string B chars (is abbcccd derived from abcd )?

I'm trying to write a recursive method that checks if all string A unique chars are a subset of string B.
Examples:
String A = "a"
isTrans(A,"a") == true
isTrans(A,"aa") == true
isTrans(A,"aaa") == true
isTrans(A,"aab") == false
String A = "acd"
isTrans(A,"addbc") == false
isTrans(A,"aacdd") == true
isTrans(A,"acccd") == true
isTrans(A,"aaaccdd") == true
That's my code:
public static boolean isTrans(String s,String t)
{
boolean isT=false;
if(t.length()>1)
{
if(s.substring(0,1).equals(t.substring(0,1))&&(t.substring(0,1)!=t.substring(1,2)))
{
return isTrans(s.substring(1,s.length()),t.substring(1,t.length()));
}
if((s.substring(0,1)).equals(t.substring(0,1))&&((t.substring(0,1)).equals(t.substring(1,2))))
{
return isTrans(s,t.substring(1,t.length()));
}
}
else
{
isT =(s.substring(0,1).equals(t.substring(0,1)))?true:false;
}
return isT;
}

You need to correct
t.substring(0,1)!=t.substring(1,2)
The == and != operators don't work on strings the same way they do on primitive types.
You could replace it with something like
!t.substring(0,1).equals(t.substring(1,2))
But a better approach would be to check characters directly.
Try the below snippet i modified to replace substring with charAt.
public static boolean isTrans(String s, String t) {
System.out.println(s + " - " + t);
boolean isT = false;
if (t.length() > 1) {
if (s.charAt(0) == t.charAt(0) && (t.charAt(0) != t.charAt(1))) {
return isTrans(s.substring(1, s.length()), t.substring(1, t.length()));
}
if (s.charAt(0) == t.charAt(0) && t.charAt(0) == t.charAt(1)) {
return isTrans(s, t.substring(1, t.length()));
}
} else {
isT = (s.charAt(0) == t.charAt(0)) ? true : false;
}
return isT;
}

You can achieve this with the help of regular expressions.
Try this:
String s = "a";
String t1 = "aa";
String t2 = "aab";
System.out.println("t1->" + t1.replaceAll("(.+?)\\1+", "$1").equals(s));
System.out.println("t2->" + t2.replaceAll("(.+?)\\1+", "$1").equals(s));
s = "acd";
t1="addc";
t2 = "aaaccddd";
System.out.println("t1->" + t1.replaceAll("(.+?)\\1+", "$1").equals(s));
System.out.println("t2->" + t2.replaceAll("(.+?)\\1+", "$1").equals(s));
Hope it helps.

String equation error checker not working

I have a method that checks to see if an equation written is correct.
This method check for:
Multiple Parentheses
Excess operators
Double Digits
q's
and any character in a string that is not and of these:
.
private static final String operators = "-+/*%_";
private static final String operands = "0123456789x";
It was working fine, but then I added in modular to the operators and now whenever my code reaches the part in the method that checks to the left and the right of an operand to see if it is neither the end of the string or the beginning I get an error saying
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 3
My method and all it's additional methods.
private static final String operators = "-+/*%_";
private static final String operands = "0123456789x";
public Boolean errorChecker(String infixExpr)
{
char[] chars = infixExpr.toCharArray();
StringBuilder out = new StringBuilder();
for (int i = 0; i<chars.length; i++)
{
System.out.print(infixExpr.charAt(i));
if (isOperator(infixExpr.charAt(i)))
{
if (i == 0 || i == infixExpr.length())
{
out.append(infixExpr.charAt(i));
}
else if (isOperator(infixExpr.charAt(i + 1)) && isOperator(infixExpr.charAt(i - 1)))
{
System.out.println("To many Operators.");
return false;
}
else if (isOperator(infixExpr.charAt(i + 1)))
{
if (infixExpr.charAt(i) != '-' || infixExpr.charAt(i + 1) != '-')
{
System.out.println("To many Operators.");
return false;
}
}
else if (isOperator(infixExpr.charAt(i - 1)))
{
if (infixExpr.charAt(i) != '-' || infixExpr.charAt(i - 1) != '-')
{
System.out.println("To many Operators.");
return false;
}
}
}
else if (isOperand(infixExpr.charAt(i)))
{
if (i == 0 || i == infixExpr.length())
{
out.append(infixExpr.charAt(i));
}//THE LINE RIGHT BELOW THIS COMMENT THROWS THE ERROR!!!!!
else if (isOperand(infixExpr.charAt(i + 1)) || isOperand(infixExpr.charAt(i - 1)))
{
System.out.println("Double digits and Postfix form are not accepted.");
return false;
}
}
else if (infixExpr.charAt(i) == 'q')
{
System.out.println("Your meow is now false. Good-bye.");
System.exit(1);
}
else if(infixExpr.charAt(i) == '(' || infixExpr.charAt(i) == ')')
{
int p1 = 0;
int p2 = 0;
for (int p = 0; p<chars.length; p++)
{
if(infixExpr.charAt(p) == '(')
{
p1++;
}
if(infixExpr.charAt(p) == ')')
{
p2++;
}
}
if(p1 != p2)
{
System.out.println("To many parentheses.");
return false;
}
}
else
{
System.out.println("You have entered an invalid character.");
return false;
}
out.append(infixExpr.charAt(i));
}
return true;
}
private boolean isOperator(char val)
{
return operators.indexOf(val) >= 0;
}
private boolean isOperand(char val)
{
return operands.indexOf(val) >= 0;
}
My main portion that runs the method:
Boolean meow = true;
while(meow)
{
System.out.print("Enter infix expression: ");
infixExpr = scan.next();//THE LINE RIGHT BELOW THIS COMMENT THROWS THE ERROR!!!!!
if(makePostfix.errorChecker(infixExpr) == true)
{
System.out.println("Converted expressions: "
+ makePostfix.convert2Postfix(infixExpr));
meow = false;
}
}
It was working fine before, but now it won't even pass 1+2 which was previously working and I changed NONE of that you see. What's wrong!?!?

What looks like what's happening is that you check for the character at index (i + 1) several times in your code. Lets say you input a string with a length of five characters. The program goes through and reaches the line:
else if (isOperator(infixExpr.charAt(i + 1)) && isOperator(infixExpr.charAt(i - 1)))
If i == 4, this will cause the code:
infixExpr.charAt(i + 1)
to throw an index error.
In essance, you're checking for a character at index five (the sixth character) in a string with a maximum index index of four which is five characters in length. Also, your checking for
if(i==0 || i == infixExpr.length)
won't work as is. Maybe check for (i==infixExpr.length-1).

Return true if string cointains "xyz" not preceeded by a period?

I'm trying to solve this CodingBat problem:
Return true if the given string contains an appearance of "xyz" where the xyz is not directly preceeded by a period (.). So "xxyz" counts but "x.xyz" does not.
xyzThere("abcxyz") → true
xyzThere("abc.xyz") → false
xyzThere("xyz.abc") → true
My attempt:
public boolean xyzThere(String str) {
boolean res = false;
if(str.contains(".xyz") == false && str.contains("xyz")){
res = true;
}
return res;
}
The problem is that is passes all the tests except the one below because it contains two instances of xyz:
xyzThere("abc.xyzxyz")
How can I make it pass all tests?

public static boolean xyzThere(String str) {
int i = -1;
while ((i = str.indexOf("xyz", i + 1 )) != -1) {
if (i == 0 || (str.charAt(i-1) != '.')) {
return true;
}
}
return false;
}

Alternatively, you could replace all occurrences of ".xyz" in the string with "", then use the .contains method to verify that the modified string still contains "xyz". Like so:
return str.replace(".xyz", "").contains("xyz");

public boolean xyzThere(String str) {
return(!str.contains(".xyz") && str.contains("xyz"));
}
Edit: Given that ".xyzxyz" should return true, the solution should be:
public boolean xyzThere(String str) {
int index = str.indexOf(".xyz");
if(index >= 0) {
return xyzThere(str.substring(0, index)) || xyzThere(str.substring(index + 4));
} else return (str.contains("xyz"));
}

The below code worked fine for me:
if '.xyz' in str:
return xyz_there(str.replace('.xyz',''))
elif 'xyz' in str:
return True
return False

Ok, I know everyone is eager to share their expertise but straight giving the kid the answer does little good.
#EnTHuSiAsTx94
I was able to pass all of the tests with three statements. Here is a hint: Try using the string replace method. Here is the method signature:
String replace(CharSequence target, CharSequence replacement)
On a minor note, the first condition in your if statement can be simplified from:
str.contains(".xyz") == false
to:
!str.contains(".xyz")
The contains method already returns true or false, so there is no need for the explicit equals comparison.

public boolean xyzThere(String str) {
return str.startsWith("xyz") || str.matches(".*[^.]xyz.*");
}

You can use the equivalent java code for the following solution:
def xyz_there(str):
pres = str.count('xyz')
abs = str.count('.xyz')
if pres>abs:
return True
else:
return False

Ok, let's translate your question into a regexp:
^ From the start of the string
(|.*[^\.]) followed by either nothing or any amount of any chars and and any char except .
xyz and then xyz
Java code:
public static boolean xyzThere(String str) {
return str.matches("^(|.*[^\\.])xyz");
}

boolean flag = false;
if(str.length()<=3){
flag = str.contains("xyz");
}
for (int i = 0; i < str.length()-3; i++) {
if (!str.substring(i, i+3).equals("xyz") &&
str.substring(i, i+4).equals(".xyz")) {
flag=false;
}else{
if(str.contains("xyz")) flag=true;
}
}
return flag;

public boolean xyzThere(String str) {
boolean res=false;
if(str.length()<3)res=false;
if(str.length()==3){
if(str.equals("xyz"))res=true;
else res=false;
}
if(str.length()>3){
for(int i=0;i<str.length()-2;i++){
if(str.charAt(i)=='x' && str.charAt(i+1)=='y' && str.charAt(i+2)=='z'){
if(i==0 || str.charAt(i-1)!='.')res=true;
}
}
}
return res;
}

public class XyzThereDemo {
public static void main(String[] args) {
System.out.println(xyzThere("abcxyz"));
System.out.println(xyzThere("abc.xyz"));
System.out.println(xyzThere("xyz.abc"));
}
public static boolean xyzThere(String str) {
int xyz = 0;
for (int i = 0; i < str.length() - 2; i++) {
if (str.charAt(i) == '.') {
i++;
continue;
}
String sub = str.substring(i, i + 3);
if (sub.equals("xyz")) {
xyz++;
}
}
return xyz != 0;
}
}

Another method
public boolean xyzThere(String str) {
if(str.startsWith("xyz")) return true;
for(int i=0;i<str.length()-3;i++) {
if(str.substring(i+1,i+4).equals("xyz") && str.charAt(i)!='.') return true;
}
return false;
}

public boolean xyzThere(String str) {
if (str.startsWith("xyz")){
return true;
}
for (int i = 0; i < str.length()-2; i++) {
if (str.subSequence(i, i + 3).equals("xyz") && !(str.charAt(i-1) == '.')) {
return true;
}
}
return false;
}

This is the best possible and easiest way to solve this question with very simple logic:
def xyz_there(str):
for i in range(len(str)):
if str[i-1]!= '.' and str[i:i+3]=='xyz' :
return True
return False

public boolean xyzThere(String str) {
boolean flag = false;
if (str.startsWith("xyz"))
{
return true;
}
for (int i = 0; i < str.length() - 3; i++)
{
if (str.charAt(i) != '.' && str.charAt(i + 1) == 'x'
&& str.charAt(i + 2) == 'y' && str.charAt(i + 3) == 'z')
{
flag = true;
break;
}
}
return flag;
}

def xyz_there(str1):
for i in range(len(str1)):
if str1[i-1] != '.' and str1[i:i+3] == 'xyz':
return True
else:
return False

def xyz_there(str):
if '.xxyz' in str:
return'.xxyz' in str
if '.' in str:
a=str.replace(".xyz","")
return 'xyz' in a
if '.' not in str:
return 'xyz' in str

'''python
def xyz_there(str):
dot=str.find('.') # if period is present in str if not dot==-1
if dot==-1: # if yes dot will show position of period
return 'xyz' in str
elif dot!=-1: #if period is present at position dot
if 'xyz' in str[:dot]:
return True
while str[dot+1:].find('.')!=-1: #while another period is present
if '.xyz' in str[dot+1:]==False: # .xyz will not be counted
return True
else:
dot=dot+str[dot+1:].find('.')+2 #now dot=previous dot+new dot+2
else:
return 'xyz' in str[dot+2:]
'''

def xyz_there(str):
list = [i for i in range(len(str)) if str.startswith('xyz', i)]
if list == []:
return False
else:
found = 0
for l in list:
if str[l-1:l+3] != ".xyz":
found += 1
if found >=1:
return True
else:
return False

simple solution just by replace and check the "xyz " in a thats it
def xyz_there(str):
a=str.replace('.xyz','')
return 'xyz' in a