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

Question

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?

Why are you using a `for` loop and completely ignoring `i`? You're just going the same thing over and over, for every character in the string. And what _should_ `xyzThere("abc.xyzxyz")` return? — Matt Ball, Apr 08 '13 at 19:53
what should be the result for the test case which you mentioned? Does it pass or not? — nommyravian, Apr 08 '13 at 19:55
Depending on your philosophical bent, answer to `xyzThere("abc.xyzxyz")` may be `true` or `false` unless the question states if only the first occurrence of `xyz`/'.xyz' to check. Also, instead of `str.contains(".xyz") == false`, use `!str.contains(".xyz")`. And, drop the `for` loop. — Karan Goel, Apr 08 '13 at 19:55
My bad, was trying to do it some other way, forgot to delete that line. — Manas Bajaj, Apr 08 '13 at 19:56
@nommyravian, it should return 'true', but returns false in my case. — Manas Bajaj, Apr 08 '13 at 19:57

Xavier Delamotte · Answer 1 · 2013-04-08T20:13:07.230

4

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;
}

edited Apr 08 '13 at 20:13

answered Apr 08 '13 at 19:55

Xavier Delamotte

3,519
19
30

R. Pehrson · Answer 2 · 2016-07-29T23:34:52.330

3

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");

edited Jul 29 '16 at 23:34

answered Jul 28 '16 at 21:10

R. Pehrson

31
3

Zim-Zam O'Pootertoot · Accepted Answer · 2013-04-08T20:08:00.050

2

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"));
}

edited Apr 08 '13 at 20:08

answered Apr 08 '13 at 19:56

Zim-Zam O'Pootertoot

17,888
4
41
69

No, the thing is that your code returns the same output as mine, i.e. returning false for xyzThere("abc.xyzxyz"), however, it should return true. – Manas Bajaj Apr 08 '13 at 20:04
Ah, gotcha. I'll edit my answer accordingly. The new answer will strip off all instances of ".xyz" and will then return true of the string still contains an "xyz" – Zim-Zam O'Pootertoot Apr 08 '13 at 20:05
Try again, I've changed "str.substring(index)" to "str.substring(index + 4)" which should eliminate the stack overflows – Zim-Zam O'Pootertoot Apr 08 '13 at 20:12

score 2 · Answer 4 · answered Aug 29 '21 at 15:44

2

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

answered Aug 29 '21 at 15:44

Neel

43
5

1

Welcome to Stack Overflow. Code dumps without any explanation are rarely helpful. Stack Overflow is about learning, not providing snippets to blindly copy and paste. Please [edit] your question and explain how it answers the specific question being asked. See [answer]. This is particularly important when answering old questions (this one is over eight years old) with existing answers. – ChrisGPT was on strike Aug 29 '21 at 23:21

score 1 · Answer 5 · answered Apr 08 '13 at 20:24

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.

score 1 · Answer 6 · answered Dec 20 '13 at 03:48

1

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

answered Dec 20 '13 at 03:48

user3119954

26
1

score 1 · Answer 7 · edited Oct 29 '19 at 11:28

1

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

edited Oct 29 '19 at 11:28

Chris32

4,716
2
18
30

answered Oct 29 '19 at 10:50

designDemon

21
2

score 0 · Answer 8 · answered Apr 08 '13 at 20:20

0

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");
}

answered Apr 08 '13 at 20:20

Johannes Kuhn

14,778
4
49
73

2

If you have the regexp hammer, every problem looks like a nail. Except [`(HT|X)ML`](http://stackoverflow.com/a/1732454/845414) – Johannes Kuhn Apr 08 '13 at 20:22
Why wouldn't you just use a negative lookbehind? `"(?<!\\.)xyz"` – Matt Ball Apr 08 '13 at 20:28
Also possible, but not necessary. But probably better (to explain). – Johannes Kuhn Apr 08 '13 at 20:34

score 0 · Answer 9 · edited Jul 28 '16 at 23:00

0

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;

edited Jul 28 '16 at 23:00

Emil Sierżęga

1,785
2
31
38

answered Sep 11 '15 at 06:37

Pradeep

1
1
4

coder kemp · Answer 10 · 2018-01-01T01:10:39.557

0

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;
}

edited Jan 01 '18 at 01:10

answered Dec 29 '17 at 01:01

coder kemp

361
2
13

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - [From Review](/review/low-quality-posts/18376696) – Mamun Dec 29 '17 at 02:08
Well this does run all the test cases. Do test it before commenting and ignorantly down-voting. – coder kemp Dec 29 '17 at 18:49

score 0 · Answer 11 · answered Feb 06 '19 at 12:17

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;
    }
}

score 0 · Answer 12 · answered Aug 17 '19 at 12:02

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;
}

answered Aug 17 '19 at 12:02

Rares

1
2

score 0 · Answer 13 · edited May 17 '20 at 17:37

0

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;
}

edited May 17 '20 at 17:37

Jeremy Caney

7,102
69
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answered May 17 '20 at 15:14

Rahul Gheewala

11
2

score 0 · Answer 14 · answered Aug 02 '20 at 13:02

0

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

answered Aug 02 '20 at 13:02

YashBora

1

score 0 · Answer 15 · answered Sep 28 '20 at 09:59

 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;
}

score 0 · Answer 16 · answered May 01 '21 at 04:02

0

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

answered May 01 '21 at 04:02

Farha Rubena

1
1

the `[i:i+3]` will be out of range towards the end of the loop – Christian Garbin May 02 '21 at 18:35

score 0 · Answer 17 · answered Jul 08 '21 at 13:44

0

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

answered Jul 08 '21 at 13:44

Jayant Verma

1

SANROOKIE · Answer 18 · 2020-01-09T09:56:32.573

'''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:]

'''

Could you provide any clarification or explanation of your code? — Gen Wan, Jan 08 '20 at 17:58

score -1 · Answer 19 · edited Apr 13 '20 at 15:24

-1

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

edited Apr 13 '20 at 15:24

Zoe

27,060
21
118
148

answered Apr 13 '20 at 14:55

Mungunbayar

1
2

score -2 · Answer 20 · answered Jun 15 '19 at 17:13

-2

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

answered Jun 15 '19 at 17:13

Thaneesh reddy

1
1

1

This solution does not pass the user's first 3 tests. – MForMarlon Jun 15 '19 at 17:44

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

20 Answers20