Coding Bat Java String 1 solutions 26 to 33

This post contains the solutions to Java String-1 section of codingbat.com for problems 26 to 33. If you are new here, you can check my previous post learn-coding-computer-programming-for-beginners. Solutions to problems 1 to 18 are available in the link java-string1-part-1, java-string1-part2.Solutions to previous sections are also available check the links to java-warmup-1 , java-warmup-2.

26.       Problem Name : frontAgain 

Solution:

Hint: Given a string, return true if the first 2 chars in the string also appear at the end of the string, such as with "edited". 

public boolean frontAgain(String str)        {                 if(str.length()>2)                 {                        if(str.substring(0,2).equals(str.substring(str.length()-2,str.length())))                              return true;                        else                              return false;                 }                 else if(str.length()==2)                        return true;                 else                        return false;        }

27.       Problem Name : minCat 

Solution:

Hint: Given two strings, append them together (known as "concatenation") and return the result

Constraint: However, if the strings are different lengths, omit chars from the longer string so it is the same length as the shorter string. So "Hello" and "Hi" yield "loHi". The strings may be any length. 

public String minCat(String a, String b)        {               if(a.length()>b.length())                      return(a.substring(a.length()-b.length(),a.length())+b);               else                      return(a+b.substring(b.length()-a.length(),b.length()));        }

28.       Problem Name : extraFront 

Solution:

Hint: Given a string, return a new string made of 3 copies of the first 2 chars of the original string.

Constraint: The string may be any length. If there are fewer than 2 chars, use whatever is there. 

public String extraFront(String str)        {                 String result="";int i=0;                 if(str.length()>2)                 {                        for(i=0;i<3;i++)                              result+=str.substring(0,2);                 }                 else                 {                        for(i=0;i<3;i++)                              result+=str.substring(0,str.length());                 }                 return result;        }

29.       Problem Name : without2 

Solution:

Hint: Given a string, if a length 2 substring appears at both its beginning and end, return a string without the substring at the beginning, so "HelloHe" yields "lloHe".

Constraint: The substring may overlap with itself, so "Hi" yields "". Otherwise, return the original string unchanged. 

public String without2(String str)        {               int len=str.length();                 if(str.length()>2)                 {                       if(str.substring(0,2).equals(str.substring(len-2,len)))                             return(str.substring(2,len));                       else                             return str;                 }                 else if(str.length()==1)                        return str;                 else                        return "";        }

30.       Problem Name : deFront 

Solution:

Hint: Given a string, return a version without the first 2 chars. except keep the first char if it is 'a' and keep the second char if it is 'b'.

Constraint: The string may be any length.

public String deFront(String str)        {                    int len=str.length();                 int flag=0;                 String result="";                                 if(str.charAt(1)=='b')                 {                        result=(str.charAt(1)+str.substring(2,len));                        flag=1;                 }                                 if(str.charAt(0)=='a')                 {                   if(flag==1)                   {                      result=(str.charAt(0)+result);                   }                   else                      {                        result=str.charAt(0)+str.substring(2,len);                      }                   flag=1;                 }                                 if(flag!=1)                                 {                 result= str.substring(2,len);                 }                 return result;        }

31.       Problem Name : startWord 

Solution:

Hint: Given a string and a second "word" string, we'll say that the word matches the string if it appears at the front of the string, except its first char does not need to match exactly.

Constraint: On a match, return the front of the string, or otherwise return the empty string. So, so with the string "hippo" the word "hi" returns "hi" and "xip" returns "hip". The word will be at least length 1. 

public String startWord(String str, String word)        {               int len=word.length();               if(str.length()==0)                      return new String("");               if(len==1 ){                      return str.substring(0, 1);               }               String str1="";               if(str.length()>=word.length())               {                str1=str.substring(1,len);               }               else               {                str1="";               }               word=word.substring(1);               if(word.equals(str1))                      return str.substring(0, len);               else                      return new String("");        }

32.       Problem Name : withoutX 

Solution:

Hint: Given a string, if the first or last chars are 'x', return the string without those 'x' chars, and otherwise return the string unchanged. 

public String withoutX(String str)        {                 int len=str.length();                 String result="";                 if(len>2)                 {                        if(str.charAt(0)=='x' && str.charAt(len-1)=='x')                              result=str.substring(1,len-1);                        else if(str.charAt(0)=='x' || str.charAt(len-1)=='x')                        {                              if(str.charAt(0)=='x')                                     result=str.substring(1,len);                              if(str.charAt(len-1)=='x')                                     result=str.substring(0,len-1);                        }                        else                              result=str;                                 }                 else                 {                 if(!str.contains("x"))                        result=str;                                 }                 return result;        }

33.       Problem Name : withoutX2 

Solution:

Hint: Given a string, if one or both of the first 2 chars is 'x', return the string without those 'x' chars, and otherwise return the string unchanged.

public String withoutX2(String str)        {                 int len=str.length();                 String result="";                 if(str.length()>1)                 {                        if(str.charAt(0)=='x' && str.charAt(1)=='x')                        {                              str=str.substring(1,1)+str.substring(2,str.length());                               result=str;                        }                        else if(str.charAt(0)=='x')                        {                              str=str.substring(1,2)+str.substring(2,str.length());                               result=str;                        }                        else if(str.charAt(1)=='x')                        {                              str=str.substring(0,1)+str.substring(2,str.length());                               result=str;                        }                        else                        {                              result=str;                        }                 }                 else                 {                        result="";                 }                   return result;               }

Please check my posts for solutions to other codingbat.com sections.