Java > String-1 > withoutX (CodingBat Solution)

Problem:

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

withoutX("xHix") → "Hi"
withoutX("xHi") → "Hi"
withoutX("Hxix") → "Hxi"


Solution:

public String withoutX(String str) {  
  if (str.length() == 0)
    return str;
  if (str.length() == 1){
    if (str.charAt(0) == 'x')
      return "";
    else
      return str;
  }
  
  if (str.charAt(0) == 'x')
    str = str.substring(1,str.length());
    
  if (str.charAt(str.length()-1) == 'x')
    str = str.substring(0,str.length()-1);

  return str;
}

13 comments:

  1. AnonymousDecember 3, 2017 at 10:30 AM


    My fantasy world... f me.....


    public String withoutX(String str) {
    String sum="";

    if(!str.contains("x")){
    return str;
    }


    if(str.charAt(0)=='x'&&str.charAt(str.length()-1)!='x'){
    sum=sum+str.substring(1);
    }
    else if(str.charAt(0)!='x'&&str.charAt(str.length()-1)=='x'){

    sum=sum+str.substring(0,str.length()-1);
    }
    else if(str.charAt(0)=='x'&&str.charAt(str.length()-1)=='x'){
    for(int i=1;i<str.length()-1;i++){

    sum=sum+str.substring(i,i+1);
    }
    }
    else if(str.charAt(0)!='x'&&str.charAt(str.length()-1)!='x'){
    return str;

    }

    return sum;
    }


    ReplyDelete
  2. Navid AfzalMarch 8, 2018 at 11:10 PM

    I find the easiest-to-understand solutions minimize the number of return statements, this solution only uses one:

    public String withoutX(String str) {
    StringBuilder result = new StringBuilder(str);

    if (!str.isEmpty() && str.length() > 1 && str.charAt(str.length()-1) == 'x') {
    result = result.deleteCharAt(str.length()-1);
    }

    if (!str.isEmpty() && str.charAt(0) == 'x') {
    result = result.deleteCharAt(0);
    }

    return result.toString();

    }

    ReplyDelete
  3. UnknownApril 6, 2018 at 1:59 PM

    public String withoutX(String str) {

    if(str.length()==0 || str.length()==1)
    return "";
    else if(str.charAt(0) == 'x' && str.charAt(str.length()-1) == 'x')
    return str.substring(1,str.length()-1);
    else if(str.charAt(0) == 'x')
    return str.substring(1,str.length());
    else if(str.charAt(str.length()-1) == 'x')
    return str.substring(0,str.length()-1);

    else return str;
    }

    ReplyDelete
  4. UnknownOctober 30, 2018 at 3:05 PM

    public String withoutX(String str) {

    int len = str.length();
    String word = "";

    for (int i = 0; i < len; i++){
    if (i == 0 && str.charAt(0) != 'x') {
    word += str.charAt(0);
    } else if (i > 0 && i < len-1) {
    word += str.charAt(i);
    } else if (i == len-1 && str.charAt(len-1) != 'x') {
    word += str.charAt(len-1);
    }
    }
    return word;
    }

    ReplyDelete
  5. UnknownMarch 3, 2019 at 2:24 PM

    if (str.length() <= 1)
    return "";

    if (str.charAt(0) == 'x')
    str = str.substring(1, str.length());

    if (str.charAt(str.length()-1) == 'x')
    str = str.substring(0, str.length()-1);

    return str;

    ReplyDelete
  6. PiyushApril 27, 2019 at 4:52 PM

    public String withoutX(String str) {
    if(str.startsWith("x")) str = str.substring(1);
    if(str.endsWith("x")) str = str.substring(0,str.length()-1);
    return str;
    }

    ReplyDelete
  7. UnknownMay 30, 2020 at 4:47 PM

    if (str.length()>=1 && str.charAt(0) == 'x')
    str = str.substring(1,str.length());

    if (str.length()>=1 && str.charAt(str.length()-1) == 'x')
    str = str.substring(0,str.length()-1);

    return str;

    ReplyDelete
  8. AnonymousSeptember 13, 2020 at 3:27 PM

    public String withoutX(String str) {
    if(str.equals("x")){
    return "";
    }
    if(str.startsWith("x") && str.endsWith("x")){
    return str.substring(1, str.length() -1);
    }
    if(str.startsWith("x")){
    return str.substring(1);
    }
    if(str.endsWith("x")){
    return str.substring(0, str.length() - 1);
    }
    return str;
    }

    ReplyDelete
  9. AnonymousSeptember 13, 2020 at 4:49 PM

    public String repeatFront(String str, int n) {
    return (n == 0 || str.isEmpty())? "":
    str.substring(0, n) + repeatFront(str, n - 1);
    }

    ReplyDelete
  10. checknoDecember 19, 2020 at 2:13 PM

    public String withoutX(String str) {
    if(str.length()<2){
    return "";
    }
    else if(str.substring(0,1).equals("x")&&str.substring(str.length()-1,str.length()).equals("x")){
    return str.substring(1,str.length()-1);
    }
    else if(str.substring(0,1).equals("x")){
    return str.substring(1);
    }
    else if (str.substring(str.length()-1,str.length()).equals("x")){
    return str.substring(0,str.length()-1);
    }
    return str;
    }

    ReplyDelete
  11. AnonymousApril 25, 2021 at 3:36 AM

    public String withoutX(String str) {
    if(str.length()>1 && str.startsWith("x") && str.endsWith("x"))return str.substring(1,str.length()-1);
    if(str.length()>1 && str.startsWith("H") && str.endsWith("x"))return str.substring(0,str.length()-1);
    if(str.length()>1 && str.startsWith("x") && str.endsWith("i"))return str.substring(1,str.length());
    else if(str.length()<2)return "";
    return str;
    }

    ReplyDelete
  12. ewedaaNovember 13, 2021 at 2:03 PM

    public String withoutX(String str) {
    if (str.length()>=1 && str.charAt(0) == 'x')
    str = str.substring(1);

    if (str.length()>=1 && str.charAt(str.length()-1) == 'x')
    str = str.substring(0,str.length()-1);

    return str;

    }

    ReplyDelete
  13. Iqboljon AbdumalikovApril 12, 2022 at 8:46 PM

    In JavaScript:

    function withoutX(str){
    if(str.length == 1) {
    if(str[0] == 'x') {
    return '';
    } else {
    return str;
    }
    }
    if(str.startsWith('x') && str.endsWith('x')) {
    str = str.substring(1, str.length - 1)
    return str;
    } else if(str[str.length - 1] == 'x') {
    return str = str.substring(0, str.length - 1);
    } else if(str[0] == 'x') {
    return str = str.substring(1)
    } else {
    return str
    }
    }

    ReplyDelete