24 thoughts on “CodingBat: Java. String-2, Part II”

1. MaximNovember 13, 2014 at 12:32 am

   I don’t think you’re supposed to use more than 1 loop in the String-2 problem section. This is my solution for plusOut:

   public String plusOut(String str, String word) {
   String res = “”;
   int found = str.indexOf(word);

   for (int i = 0; i < str.length(); i++) {
   if (i == found) {
   res += word;
   found = str.indexOf(word, found + word.length());
   i += word.length()-1;
   } else {
   res += "+";
   }
   }

   return res;
   }

   Reply ↓
   1. Simon LahnMarch 16, 2017 at 8:38 pm

      Here’s my solution using the equals method. It requires one less variable.

      public String plusOut(String str, String word) {
      String result = “”;

      for (int i = 0; i < str.length(); i++) {
      //ensure there is no out of bounds and compare substring with word
      if(i + word.length() <= str.length() && str.substring(i, i + word.length()).equals(word)){
      result += word;

      //increment i by word length minus one since the for loop already increments i by 1
      i += word.length() – 1;
      }
      //Otherwise add a "+" and continue to increment i
      else
      result += "+";
      }

      return result;
      }

      Reply ↓
2. StephanOctober 26, 2015 at 1:28 am

   I don’t see an answer for the program oneTwo.
   Here is what i got:
   public static String oneTwo(String str) {
   String result = “”;

   for (int i = 0; i < str.length() – 2; i++) {
   if (i == 0 || i % 3 == 0) {
   result += str.substring(i + 1, i + 3);
   result += str.charAt(i);
   }
   }
   return result;

   Reply ↓
   1. Chau NguyenAugust 17, 2016 at 3:50 pm

      You can do this instead!

      public String oneTwo(String str) {
      if (str.length() < 3) return ""; // this part doesn't matter too much

      String result = "";

      for (int i = 0; i < str.length() – 2; i += 3)
      result = result + str.substring(i + 1, i + 3) + str.substring(i, i + 1);

      return result;
      }

      Reply ↓
      1. Chau NguyenAugust 17, 2016 at 3:51 pm

         woops, instead of “str.substring(i, i + 1)” just use “str.charAt(i)”

         Reply ↓
   2. FulopApril 3, 2018 at 10:01 am

      Amazing.
      i really like that one.
      thank you for sharing.

      Reply ↓
3. DaveDecember 4, 2015 at 5:34 am

   for prefix again i just used one return function:

   return(str.lastIndexOf(str.substring(0,n)) != str.indexOf(str.substring(0,n)));

   Reply ↓
   1. First LastMay 25, 2016 at 5:48 pm

      This can be shortened to:

      return (str.indexOf(str.substring(0,n), 1) != -1);

      Reply ↓
      1. Gregor Ulm Post authorMay 26, 2016 at 9:49 am

         This is about as unreadable than the version of Dave. When you start out with programming, you may want to avoid picking up bad habits, such as questionable conciseness.

         Reply ↓
4. nick woodwardJune 5, 2016 at 9:27 pm

   Hi,

   Would you mind explaining your solution to starOut a little please? I’m struggling to follow it.

   Here’s my version:

   public String starOut(String str) {
   String temp = “”;

   for(int x = 0; x < str.length(); x++){
   //check previous
   if(x != 0 && str.charAt(x-1) == '*') continue;
   //check next
   if(x+1 < str.length() && str.charAt(x+1)=='*') continue;
   //check current
   temp += str.charAt(x)=='*'? "":str.charAt(x);
   }
   return temp;
   }

   Regards,

   Reply ↓
5. nick woodwardJune 6, 2016 at 5:15 pm

   just working through your ‘wordEnds’ solution too. nice.

   any thoughts on this way? (other than the tertiary operator being a little less legible)

   public String wordEnds(String str, String word) {
   String temp = “”;

   for(int x = 0; x < str.length(); x++){
   //check previous
   if(x != 0 && x + word.length() <= str.length())
   temp += str.substring(x,x+word.length()).equals(word)? str.substring(x-1,x):"";

   if(x + word.length()+1 <= str.length()){
   //check next
   temp += str.substring(x,x+word.length()).equals(word)? str.substring(x+ word.length(),x+word.length()+1):"";
   }
   } return temp;
   }

   regards,

   nick

   Reply ↓
   1. GeloJune 17, 2016 at 3:40 pm

      we have almost same idea. here’s my solution:
      public String wordEnds(String str, String word) {

      String snip = “”;
      for(int i=0; i<str.length()-word.length()+1; i++){
      String s = str.substring(i, i+word.length());

      if(i != 0 && s.equals(word)){
      String s2 = str.substring(i-1, i);
      snip += s2;
      }

      if(i != str.length()-word.length() && s.equals(word)){
      String s3 = str.substring(i+word.length(), i+word.length()+1);
      snip += s3;
      }
      }
      return snip;
      }

      Reply ↓
6. SurajAugust 14, 2016 at 7:55 am

   I have another solution to plusOut
   public String plusOut(String str, String word) {

   char[] arr = str.toCharArray();
   Arrays.fill(arr,’+’);

   for(int i = 0 ; i < str.length() – word.length() +1 ; i++){
   if(str.substring(i , i + word.length()).equals(word)){
   int p = i;
   for(int k = 0 ; k < word.length() ; k++)
   arr[p++] = word.charAt(k);
   }
   }
   return new String(arr);
   }

   Reply ↓
   1. Gregor Ulm Post authorAugust 14, 2016 at 10:34 am

      At this point in Coding Bat, arrays have not yet been introduced.

      Reply ↓
7. SurajAugust 14, 2016 at 8:21 am

   starOut : Another solution

   public String starOut(String str) {
   return str.replaceAll(“.?\\*+.?”,””);
   }

   Note : .?\\*+.?
   .? => zero or one character (for boundary cases)
   \\*+ => one or more star

   Reply ↓
   1. Gregor Ulm Post authorAugust 14, 2016 at 10:30 am

      If I recall correctly, replaceAll() is not part of the methods Nick Parlante mentions on the Coding Bat website. Thus, using it defeats the purpose of the exercise.

      Reply ↓
8. KyranSeptember 3, 2016 at 6:34 pm

   For prefixAgain, couldn’t you just avoid looping altogether with this? Bit of a newb, but I figure it’d be quicker to run (marginally)

   public boolean prefixAgain(String str, int n) {
   boolean res = false;
   String s = str.substring(0,n);

   if (str.lastIndexOf(s)>=n)
   {
   res = true;
   }
   return res;
   }

   Reply ↓
   1. Gregor Ulm Post authorSeptember 4, 2016 at 9:24 am

      How do you think lastIndexOf() might be implemented?

      Reply ↓
9. SimonMarch 6, 2017 at 9:37 pm

   I look at these solutions only after I’ve figured it out on my own only to realize I tend to over complicate the problem. What takes me 10+ lines, you manage to do in four lines. Thanks for the solutions.

   Reply ↓
10. WojtekApril 6, 2017 at 7:52 pm

    public String repeatSeparator(String word, String sep, int count) {
    String result = “”;
    for(int i=0; i<count; i++)
    result = String.join(sep,word,result);
    if(result.endsWith(sep)) return result.substring(0,result.length()-sep.length());
    return result;
    }

    Reply ↓
11. WojtekApril 6, 2017 at 8:22 pm

    My solution

    public boolean prefixAgain(String str, int n) {
    String end = str.substring(n);
    return end.contains(str.substring(0,n));
    }

    Reply ↓
12. WojtekApril 6, 2017 at 9:44 pm

    My solution

    public String getSandwich(String str) {
    if(str.length()<=10) return "";
    return str.substring(str.indexOf("bread")+5,str.lastIndexOf("bread"));
    }

    Reply ↓
13. kevinNovember 13, 2017 at 9:48 pm

    Can anybody explain to me what is happening in the code for repeatfront. It would be very helpful.

    Reply ↓
14. LeeNovember 27, 2017 at 12:05 am

    public Boolean prefixAgain(String str, int n) {
    return str.substring(n).contains(str.substring(0,n));
    }

    Reply ↓

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