Java > String-1 > lastChars (CodingBat Solution)

Problem:

Given 2 strings, a and b, return a new string made of the first char of a and the last char of b, so "yo" and "java" yields "ya". If either string is length 0, use '@' for its missing char.

lastChars("last", "chars") → "ls"
lastChars("yo", "java") → "ya"
lastChars("hi", "") → "h@"


Solution:

public String lastChars(String a, String b) {
  if(a.length() == 0)
    a = "@";
  if(b.length() == 0)
    b = "@";
  
  return a.substring(0,1) + b.substring(b.length()-1, b.length());
}


15 comments :

  1. you can just say "b.substring(b.length()-1)" on line 7 as one parameter in the substring method means to start at that index and go to the end of the string.

    ReplyDelete
    Replies
    1. You're exactly right. Who wrote these solutions?

      Delete
  2. haha I am such a N00B!
    public String lastChars(String a, String b) {
    String result = "";
    if(a.length() >= 2 && b.length() >= 2){
    result = a.substring(0,1) + b.substring(b.length()-1);
    }if(a.length() < 2 && b.length() >= 2){
    result = "@" + b.substring(b.length()-1);
    }if(a.length() >= 2 && b.length() <= 1){
    result = a.substring(0,1) + "@";
    }if(a.length() < 1 && b.length() < 1){
    result = "@@";
    }
    return result;
    }

    ReplyDelete
  3. public String lastChars(String a, String b) {
    String output = "";
    if (a.length() > 0) {
    output += a.substring(0,1);
    } else {
    output += "@";
    }

    if (b.length() > 0) {
    output += b.substring(b.length()-1);
    } else {
    output += "@";
    }
    return output;
    }

    ReplyDelete
  4. Wow... I did a much more complicated version
    if(a.length() > 0 && b.length() > 0)
    return "" + a.charAt(0) + b.charAt(b.length() - 1);
    else if(a.length() > 0 && b.length() == 0)
    return "" + a.charAt(0) + new String(new char[1 - Math.min(b.length(), 1)]).replace("\0", "@");
    else if(a.length() == 0 && b.length() > 0)
    return "" + new String(new char[1 - Math.min(a.length(), 1)]).replace("\0", "@") + b.charAt(b.length() - 1);
    return new String(new char[2]).replace("\0", "@");

    ReplyDelete
  5. I alrdy forgot, what i did right there, but yeah, it works at least...lol


    public String lastChars(String a, String b) {


    String z= "yo";
    String u= "java";
    String x= "ya";
    String doe="@";


    if(a.substring(0).equals(z)&&b.substring(0).equals(u))
    return x;

    if(a.length()==0&&b.length()!=0)
    return doe + b.substring (b.length()-1);

    if(a.length()!=0&&b.length()==0)
    return a.charAt(0) + doe;

    if(a.length()==0&&b.length()==0)
    return doe+doe;

    String s = a.charAt(0) + b.substring (b.length()-1);

    return s;

    }

    ReplyDelete
  6. public String lastChars(String a, String b) {
    String sum="";
    if(a.length()==0&&b.length()==0){
    return "@"+"@";
    }


    if(a.length()==0){
    sum=sum+ "@"+b.charAt(b.length()-1);
    }
    else if(b.length()==0){
    sum=sum+ a.charAt(0)+"@";
    }
    else if (a.length()==0&&b.length()==0){
    sum=sum+"@"+"@";
    }
    else if(a.length()!=0&&b.length()!=0){
    sum=sum+a.charAt(0)+b.charAt(b.length()-1);
    }
    return sum;
    }

    ReplyDelete
  7. if (a.length() == 0)
    return "@" + a.substring(0, 1);
    if (b.length() == 0)
    return a.substring(0, 1) + "@";
    if (b.length() == 0 && a.length() == 0)
    return "@@";
    else
    return a.substring(0, 1) + b.substring(b.length() - 1, b.length());

    ReplyDelete
  8. public String lastChars(String a, String b) {
    return a.length()>0&&b.length()>0?a.substring(0,1)+b.substring(b.length()-1):
    a.length()==0&&b.length()!=0?"@"+b.substring(b.length()-1):
    b.length()==0&&a.length()!=0?a.substring(0,1)+"@":"@@";
    }

    ReplyDelete
  9. public String lastChars(String a, String b) {
    if(a.length()<1 && b.length()<1){
    return "@"+"@";
    }if(a.length()>1 && b.length()>1){
    return a.substring(0,1)+b.substring(b.length()-1,b.length());
    }
    if(a.length()<1 && b.length()>1){
    return "@"+b.substring(b.length()-1,b.length());
    }
    if(a.length()>1 && b.length()<1){
    return a.substring(0,1)+"@";
    }if(a.length()==1 && b.length()>1){
    return a.substring(0)+b.substring(b.length()-1);
    }
    return a+b;
    }

    ReplyDelete
  10. public String lastChars(String a, String b) {
    if(a.length()==0 && b.length()==0)
    return "@"+""+"@";
    else if(a.length()==0)
    return "@"+""+b.charAt(b.length()-1);
    else if(b.length()==0)
    return a.charAt(0)+""+"@";
    else
    return a.charAt(0)+""+b.charAt(b.length()-1);
    }

    ReplyDelete
  11. public String lastChars(String a, String b) {
    if (a.length()>0 && b.length()>0 ){
    return a.substring(0,1)+ b.substring(b.length()-1);
    }
    if (a.length()>0 && b.length()==0 ){
    return a.substring(0,1)+ "@";
    }
    if (a.length()==0 && b.length()>0 ){
    return "@"+ b.substring(b.length()-1);
    }
    else
    return "@"+"@";
    }

    ReplyDelete
  12. String first = "@";
    String second = "@";

    if(a.length() > 0) first = a.substring(0,1);
    if(b.length() > 0) second = b.substring(b.length()-1);


    return first+second;

    ReplyDelete
  13. In JavaScript, we can solve it, like this:


    ```function lastChars(a, b){
    if(a == '' && b == '') {
    return '@' + '@'
    } else if(b == 0) {
    return a[0] + '@'
    } else if(a == 0) {
    return '@' + b[b.length - 1]
    } else {
    return a[0] + b[b.length - 1]
    }
    }```

    ReplyDelete
  14. In JavaScript, we can solve it, like this:


    function lastChars(a, b) {
    if(a == '' && b == '') {
    return '@' + '@'
    } else if(b == 0) {
    return a[0] + '@'
    } else if(a == 0) {
    return '@' + b[b.length - 1]
    } else {
    return a[0] + b[b.length - 1]
    }
    }

    ReplyDelete

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