Solving the error: return true if the string "cat" and "dog" appear the same number of times in the given string.

Question

Here is the program I run:

public boolean catDog(String str) 
 { 
  int catAnswer = 0;
  int dogAnswer = 0;
  int cat_Count = 0;
  int dog_Count = 0;
 
  for (int i=0; i< str.length()-1; i++) 
  {
  String sub = str.substring(i, i+2);
 
 
  if ((sub.equals("cat"))) cat_Count++;
  if ((sub.equals("dog"))) dog_Count++;
  catAnswer = cat_Count; 
  dogAnswer = dog_Count;
 
  } //end for
 
  if(dogAnswer == catAnswer ) {return true;}
  // else
  return (dogAnswer != catAnswer);
 }

After I run, it returns an error:

Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: 7 (line number:10) - hence i use i + 2 (no errors are reported with it

Does anyone have any suggestions for the problem below: return true if the string "cat" and "dog" appear the same number of times in the given string in the java - How to correct it?

dttutoria · Accepted Answer

The cause: Run time exception is to blame for the error. StringIndexOutOfBoundsException is issued because you are accessing the element at a position that is bigger than its length. Solution: Using split is "all correct" answer for the error: return true if the string “cat” and “dog” appear the same number of times in the given string. To retain trailing empty strings, use the -1.

int count(String needle, String haystack) {
return haystack.split(needle, -1).length - 1;
}

public boolean catDog(String str) {
return count("dog", str) == count("cat", str);
}

Enoch Evans · Answer

Although you're close to finding a solution, there are some critical issues in your code.

A string of 2 is returned by substring() when you call substring(). This string cannot ever be cat or dog. To get 3 characters, change the second parameter of the method call i + 3.
Calling substring() will result in IndexOutOfRangeException as i nears the end of the input string. You should not ask for substrings that exceed the length of your input string. This bug can be fixed by changing the for conditional expression to i < str.length() - 2.
Your method's return value will always be true. If dogAnswer != catAnswer is used, you will return the exact same expression. This will resolve to true. One solution is to merge the return statements into return dogAnswer == catAnswer.

There are also a few things that you can do to make your code easier:

It is not necessary to convert cat_Count into catAnswer or dog_Count into dogAnswer. You can discard two variables and only use one pair.
~~Your loop can only be optimized to consider the third position of the input string if it is restricted to cat or dog. Change i++ into i += 3.~~ (Update: After reviewing the test data at CodingBat (I can assure you that this isn't the case.

After I had implemented fix #1, #2, and #3, as well as suggestion 1, I ran a test using the test bench. The result was quite satisfying.

All right

Hunter Morton · Answer

Although you're close to finding a solution, there are some critical issues in your code.

A string of 2 is returned by substring() when you call substring(). This string cannot ever be cat or dog. To get 3 characters, change the second parameter of the method call i + 3.
Calling substring() will result in IndexOutOfRangeException as i nears the end of the input string. You should not ask for substrings that exceed the length of your input string. This bug can be fixed by changing the for conditional expression to i < str.length() - 2.
Your method's return value will always be true. If dogAnswer != catAnswer is used, you will return the exact same expression. This will resolve to true. One solution is to merge the return statements into return dogAnswer == catAnswer.

There are also a few things that you can do to make your code easier:

It is not necessary to convert cat_Count into catAnswer or dog_Count into dogAnswer. You can discard two variables and only use one pair.
~~Your loop can only be optimized to consider the third position of the input string if it is restricted to cat or dog. Change i++ into i += 3.~~ (Update: After reviewing the test data at CodingBat (I can assure you that this isn't the case.

After I had implemented fix #1, #2, and #3, as well as suggestion 1, I ran a test using the test bench. The result was quite satisfying.

All right

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