Java > String-1 > deFront (CodingBat Solution)

Problem:

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'. The string may be any length. Harder than it looks.

deFront("Hello") → "llo"
deFront("java") → "va"
deFront("away") → "aay"


Solution:

public String deFront(String str) {    
  int len = str.length();
  String temp = "";
  
  for (int i = 0; i < len; i++) {
    if (i == 0 && str.charAt(i) == 'a')
      temp += 'a';
    else if (i == 1 && str.charAt(i) == 'b')
      temp += 'b';
    else if (i > 1)
      temp += str.charAt(i);
  }
    return temp;
}


Labels: , ,

32 comments :

  1. UnknownOctober 2, 2015 at 8:36 AM

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

    }

    ReplyDelete
    Replies
    1. AnonymousNovember 4, 2015 at 3:46 AM

      good solution. but still lacks some mechanism to prevent "IndexOutOfBoundsException" when the string has 1 or less length.
      in the exercice it says "The string may be any length.", but actually they only provide strings with 2 or more characters long in the test battery. so, your solution works for this, and does not involves any loop.

      Delete
    2. satyabratDecember 27, 2019 at 9:09 AM

      public String deFront(String str) {

      if(str.substring(0,1).equals("a")&&str.substring(1,2).equals("b"))
      {
      return str;
      }
      else if(str.substring(0,1).equals("a"))
      {
      return str.substring(0,1)+str.substring(2);
      }
      else if(str.substring(1,2).equals("b"))
      {
      return str.substring(1);
      }
      else if(str.length()>=2)
      {
      return str.substring(2);
      }
      else
      return str;
      }

      Delete
    3. satyabratDecember 27, 2019 at 9:09 AM

      public String deFront(String str) {

      if(str.substring(0,1).equals("a")&&str.substring(1,2).equals("b"))
      {
      return str;
      }
      else if(str.substring(0,1).equals("a"))
      {
      return str.substring(0,1)+str.substring(2);
      }
      else if(str.substring(1,2).equals("b"))
      {
      return str.substring(1);
      }
      else if(str.length()>=2)
      {
      return str.substring(2);
      }
      else
      return str;
      }

      Delete
  2. AnonymousOctober 4, 2015 at 9:51 AM

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

    }

    Can anyone explain to me why this doesn't return an out of bounds error for a string that's two characters long?

    ReplyDelete
    Replies
    1. AnonymousNovember 4, 2015 at 3:42 AM

      because if the string has 2 characters, the substring will return a empty string since the begin index (inclusive) is the same as the end index (exclusve).
      this is from the String API: "IndexOutOfBoundsException - if the beginIndex is negative, or endIndex is larger than the length of this String object, or beginIndex is larger than endIndex."

      Delete
  3. AnonymousNovember 4, 2015 at 3:37 AM

    hi. the article's proposed solution involves loop, which is not the context of the exercise. so, the comment solution is the better one.

    ReplyDelete
  4. AnonymousMay 8, 2017 at 6:57 AM

    public String deFront(String str) {

    if (str.length()==1&&str.equals("a")){return "a";}

    else if (!(str.substring(0,1).equals("a")||str.substring(1,2).equals("b"))){return str.substring(2);}
    else if (str.substring(0,1).equals("a")&&str.substring(1,2).equals("b")){return "ab"+str.substring(2);}
    else if (str.substring(0,1).equals("a")){return "a"+str.substring(2);}
    else if (str.substring(1,2).equals("b")){return "b"+str.substring(2);}
    else return "";
    }

    ReplyDelete
  5. Bryan MDAugust 13, 2017 at 5:32 PM

    public String deFront(String str)
    {
    //this 2 String vars are meant to "boolean" the return
    //so below the if's statements will assign the "true/false"
    //I covered some other extra conditions

    String a = "";
    String b = "";

    if (str.length() <= 0) //if 0, nothing to check
    return "";
    if (str.substring(0,1).equals("a")) //check if (0,1) = "a"
    a = "a";
    if (str.length() == 1) //avoid checking 1,2 outofbounds
    return a;
    if (str.substring(1,2).equals("b")) // check if(1,2) = "b"
    b = "b";
    if (str.length() == 2) //if 2, no need to substring
    return a + b;

    return a + b + str.substring (2, str.length()); //default
    }

    ReplyDelete
  6. UnknownAugust 2, 2018 at 7:07 PM

    public String deFront(String str) {

    String a=str.substring(0,1);
    String b=str.substring(1,2);

    if(str.startsWith("a") && !b.equals("b")) return "a" +str.substring (2);
    else if(!str.startsWith("a") && b.equals("b")) return str.substring(1);
    else if( !str.startsWith("a") && !b.equals("b")) return str.substring(2);
    return str;
    }

    ReplyDelete
  7. UnknownAugust 2, 2018 at 7:07 PM

    public String deFront(String str) {

    String a=str.substring(0,1);
    String b=str.substring(1,2);

    if(str.startsWith("a") && !b.equals("b")) return "a" +str.substring (2);
    else if(!str.startsWith("a") && b.equals("b")) return str.substring(1);
    else if( !str.startsWith("a") && !b.equals("b")) return str.substring(2);
    return str;
    }

    ReplyDelete
  8. RED CARPET Fashion GroupFebruary 22, 2019 at 12:57 AM

    public String deFront(String str) {

    if((str.charAt(0)=='a')&&(str.charAt(1)=='b'))
    return str;
    if(str.charAt(0)=='a')
    return 'a'+str.substring(2);
    if(str.charAt(1)=='b')
    return 'b'+str.substring(2);
    return str.substring(2);
    }

    ReplyDelete
  9. UnknownMarch 3, 2019 at 10:25 AM

    if ((str.substring(0, 1).equals("a")) &&
    (str.substring(1, 2).equals("b")))
    return str.substring(0, str.length());

    if (str.substring(0, 1).equals("a"))
    return "a" + str.substring(2, str.length());

    if (str.substring(1, 2).equals("b"))
    return "b" + str.substring(2, str.length());

    else
    return str.substring (2, str.length());

    ReplyDelete
  10. AnonymousMay 4, 2019 at 4:44 AM

    public String deFront(String str) {
    if(str.startsWith("ab")){
    return str;
    }
    if(str.charAt(0) == 'a'){
    return String.valueOf(str.charAt(0)).concat(str.substring(2,str.length()));
    }
    if(str.charAt(1) == 'b'){
    return String.valueOf(str.charAt(1)).concat(str.substring(2,str.length()));
    }
    return str.substring(2,str.length());
    }

    ReplyDelete
  11. KnownJune 13, 2019 at 4:03 AM

    public String deFront(String str) {
    if(str.length() >= 2) {
    boolean a = str.charAt(0) == 'a';
    boolean b = str.charAt(1) == 'b';

    if(a && b)
    return str;
    if(a && !b)
    return "a" + str.substring(2);
    if(!a && b)
    return "b" + str.substring(2);
    if(!a && !b)
    return str.substring(2);

    }
    else {
    if(str.length() == 1) {
    if(str.charAt(0) == 'a') return "a";
    if(str.charAt(0) == 'b') return "b";
    }
    }
    return str;
    }

    ReplyDelete
  12. UnknownMay 1, 2020 at 5:25 PM

    boolean aAtFirst = str.charAt(0) == 'a';
    boolean bAtSecond = str.charAt(1) == 'b';

    if (!aAtFirst && !bAtSecond)
    return str.substring(2);
    else if (!bAtSecond)
    return str.substring(0,1).concat(str.substring(2));
    else if (!aAtFirst)
    return str.substring(1);
    else
    return str;

    ReplyDelete
  13. DogoJuly 7, 2020 at 1:42 PM

    public String deFront(String str) {
    boolean existA = false;
    boolean existB = false;
    String a = "a";
    String b = "b";
    if (str.length() == 0) {
    return str;
    }

    if (str.charAt(0) == 'a') {
    existA=true;
    }
    if (str.charAt(1) == 'b') {
    existB=true;
    }

    String str2 = str.substring(2,str.length());

    if (existA && existB) {
    return a+b+str2;
    }
    if (existA && !existB) {
    return a+str2;
    }
    if (!existA && existB) {
    return b+str2;
    }
    if(!existA && !existB) {
    return str2;
    }
    return str;
    }

    ReplyDelete
  14. AnarAghaAugust 13, 2020 at 10:39 PM

    public String deFront(String str) {

    if(str.charAt(0)=='a' && str.charAt(1)!='b'){
    return str.substring(0,1) + str.substring(2);
    }
    if(str.charAt(0)!='a' && str.charAt(1)=='b'){
    return str.substring(1);
    }
    if(str.startsWith("ab")){
    return str;
    }
    return str.substring(2);

    }

    ReplyDelete
  15. UNKNOWNAugust 25, 2020 at 8:27 PM

    public String deFront(String str) {
    if (str.substring(0,2).equals("ab")) {
    return str;
    }
    String result = str.substring(2);
    if (str.charAt(0) == 'a') {
    return "a" + result;
    }
    if (str.charAt(1) == 'b') {
    return "b" + result;
    }
    return result;
    }

    ReplyDelete
  16. UnknownNovember 5, 2020 at 10:45 PM

    public String deFront(String str) {
    if(str.startsWith("a") && str.startsWith("ab"))
    return str;
    else if (str.startsWith("a"))
    return str.substring(0,1)+ str.substring(2);
    else if(str.substring(1,2).equals("b"))
    return str.substring(1,2)+ str.substring(2);

    return str.substring(2);
    }

    ReplyDelete
  17. UnknownDecember 11, 2020 at 4:53 AM

    good question spent more than two hours to solve it

    ReplyDelete
  18. Anmol ParikhMarch 11, 2021 at 11:42 PM

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

    ReplyDelete
  19. UnknownJune 10, 2021 at 6:03 AM

    public String deFront(String str) {
    if(str.substring(0,1).equals("a")&&str.substring(1,2).equals("b"))
    return str;
    else if(str.startsWith("a"))
    return "a"+str.substring(2);
    else if (str.substring(1,2).equals("b"))
    return str.substring(1);
    else
    return str.substring(2);
    }

    ReplyDelete
  20. GabeJuly 2, 2021 at 3:23 PM

    String result = "";
    int len = str.length();
    if(len == 0) return result;

    if(len >= 1 && str.charAt(0) == 'a')
    result += str.charAt(0);
    if(len >= 2 && str.charAt(1) == 'b')
    result += str.charAt(1);

    result += str.substring(2);
    return result;

    ReplyDelete
  21. UnknownOctober 27, 2021 at 4:38 AM

    public String deFront(String str) {
    if(str.charAt(0)=='a' && str.charAt(1)=='b')
    return str;
    if(str.charAt(0)=='a')
    return str.charAt(0)+str.substring(2,str.length());
    if(str.charAt(1)=='b')
    return str.charAt(1)+str.substring(2,str.length());
    else return str.substring(2,str.length());
    }

    ReplyDelete
  22. UnknownOctober 27, 2021 at 4:39 AM

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

    ReplyDelete
  23. ewedaaNovember 13, 2021 at 1:54 PM

    public String deFront(String str){
    String result = str.substring(2);
    if (str.charAt(1)=='b') result = "b" + result;
    if (str.charAt(0)=='a') result = "a" + result;
    return result;}

    ReplyDelete
  24. AvengerDecember 9, 2021 at 9:39 AM

    public String deFront(String str)
    {
    String data = "";

    if(str.charAt(0) == 'a')
    data += 'a';

    if(str.charAt(1) == 'b')
    data += 'b';


    data += str.substring(2);
    return data;
    }

    ReplyDelete
  25. Cema2009June 18, 2022 at 4:45 PM

    Another way

    public String deFront(String str) {
    if (str.startsWith("ab")) return str;
    if (str.startsWith("a")) return "a" + str.substring(2);
    if (str.substring(1, 2).equals("b")) return "b" + str.substring(2);
    return str.substring(2);
    }

    ReplyDelete
  26. JoxonNovember 20, 2022 at 8:18 AM


    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'. The string may be any length. Harder than it looks.


    deFront("Hello") → "llo"
    deFront("java") → "va"
    deFront("away") → "aay"

    ReplyDelete
  27. JoxonNovember 20, 2022 at 8:19 AM

    public String deFront(String str) {
    if (str.substring(0,2).equals("ab"))
    return "ab" + str.substring(2);

    if (str.substring(1,2).equals("b"))
    return "b" + str.substring(2);

    if (str.substring(0,1).equals("a"))
    return "a" + str.substring(2);
    return str.substring(2);

    }

    ReplyDelete
  28. javaKingDecember 25, 2023 at 10:56 AM

    public String deFront(String str) {
    if (str.startsWith("ab")){
    return str;
    }
    if(str.charAt(0) == 'a'){
    return "a"+str.substring(2);
    } else if (str.charAt(1) == 'b'){
    return "b"+str.substring(2);
    }

    return str.substring(2);
    }

    ReplyDelete

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