Java > String-2 > prefixAgain (CodingBat Solution)

Problem:

Given a string, consider the prefix string made of the first N chars of the string. Does that prefix string appear somewhere else in the string? Assume that the string is not empty and that N is in the range 1..str.length().

prefixAgain("abXYabc", 1) → true
prefixAgain("abXYabc", 2) → true
prefixAgain("abXYabc", 3) → false


Solution:

public boolean prefixAgain(String str, int n) {
  int len = str.length();
  String prefix = str.substring(0,n);
  
  for (int i = n; i < len; i++) {
    if(n+i <= len) { 
      if (prefix.equals(str.substring(i,n+i)))
        return true;
    }
  }
  return false;
}

20 comments:

  1. JakeDecember 30, 2014 at 1:58 AM

    Smaller:

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

    ReplyDelete
  2. UnknownJanuary 5, 2015 at 3:26 PM

    or:

    public boolean prefixAgain(String str, int n) {
    return (str.substring(n).indexOf(str.substring(0,n)) != -1);
    }

    ReplyDelete
  3. AlvaroJanuary 8, 2016 at 1:24 AM

    public boolean prefixAgain(String str, int n) {
    String substr = str.substring(n);
    String pattern = str.substring(0, n);

    return substr.contains(pattern);
    }

    ReplyDelete
  4. UnknownApril 15, 2017 at 12:34 PM

    public boolean prefixAgain(String str, int n) {
    return str.matches(str.substring(0, n) + "(.*)" + str.substring(0, n) + "(.*)");
    }

    ReplyDelete
  5. AnonymousApril 22, 2017 at 10:07 AM

    public boolean prefixAgain(String str, int n) {
    return str.indexOf(str.substring(0, n), n) > -1;
    }

    ReplyDelete
  6. AnonymousJune 8, 2017 at 5:53 AM

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

    ReplyDelete
  7. AnonymousAugust 29, 2017 at 8:10 AM

    public boolean prefixAgain(String str, int n) {

    for(int i=0; i<str.length()-n; i++)
    {
    if(str.substring(0,n).equals(str.substring(i+1,i+n+1)))
    return true;
    }

    return false;
    }

    ReplyDelete
  8. AelionFebruary 16, 2018 at 2:34 PM

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

    ReplyDelete
  9. Kamoliddin UsmonovMay 11, 2019 at 12:27 AM

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

    ReplyDelete
  10. AbudéSeptember 2, 2019 at 12:29 AM

    public boolean prefixAgain(String str, int n) {
    int sep = 0;
    for (int i = n ; i <= str.length() - n ; i++){
    if(str.substring(0,n).equals(str.substring(i,i+n))){
    sep = str.substring(n,i+n).length();
    return str.substring(0,n).equals(str.substring(sep, sep + n));
    }
    }
    return false;

    }

    ReplyDelete
  11. manda007December 11, 2019 at 1:48 AM

    public boolean prefixAgain(String str, int n) {
    int len = str.length();
    String prefix = str.substring(0,n);

    if(str.substring(n,len).contains(prefix))
    return true;

    return false;
    }

    ReplyDelete
  12. UnknownApril 12, 2020 at 3:18 AM

    public boolean prefixAgain(String str, int n) {
    String number = str.substring(0,n);
    String newStr = str.substring(n);

    if(newStr.contains(number)){
    return true;
    } else {
    return false;
    }
    }

    ReplyDelete
  13. jJuly 15, 2020 at 2:49 AM

    public boolean prefixAgain(String str, int n) {
    String checkString = str.substring(0, n);
    String others=str.substring(n,str.length());
    if (others.contains(checkString)) {
    return true;
    } else {
    return false;
    }
    }

    ReplyDelete
  14. UnknownAugust 7, 2020 at 1:40 PM

    public boolean prefixAgain(String str, int n) {
    if (n > str.length() / 2)
    {
    return false;
    }
    else
    {
    for (int i = str.length() - n; i > n - 1; i--)
    {
    if (str.substring(i, i + n).equals(str.substring(0, n)))
    {
    return true;
    }
    }
    return false;
    }
    }

    ReplyDelete
  15. AnarAghaSeptember 17, 2020 at 12:55 AM

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

    ReplyDelete
  16. bkNovember 5, 2020 at 6:29 AM

    single line code:
    public boolean prefixAgain(String str, int n) {
    return str.substring(str.indexOf(str.substring(0,n))+1)
    .contains(str.substring(0,n));
    }

    ReplyDelete
  17. alexanderOctober 23, 2021 at 7:00 PM

    public boolean xyzMiddle(String str) {
    int countLeft = 0;
    int countRight = 0;
    int size = str.length();
    String tar = "xyz";
    String acc = "";

    if(size < 3){
    return false;
    }
    for(int i = 0; i < size - 2; i++){
    acc = "";
    // Traverse 3chars at a time, compare to xyz each step
    acc = acc + str.charAt(i) + str.charAt(i + 1) + str.charAt(i + 2);
    if(!acc.equals(tar)){
    countLeft += 1;
    }
    else{
    // Capture index where char z of xyz is found
    int z = i + 2;
    // Store count of remaining letters after xyz, size - 1 - found index z
    countRight = size - 1 - z;
    }
    }
    countLeft = countLeft - countRight;
    int diff = countLeft - countRight;
    if(diff == 1 || diff == 0 || diff == -1){
    return true;
    }
    return false;
    }

    ReplyDelete
  18. alexanderOctober 23, 2021 at 7:01 PM

    public boolean prefixAgain(String str, int n) {
    String acc = "";
    String comp = "";

    for(int i = 0; i < n; i++){
    acc = acc + str.charAt(i);
    }
    for(int i = n; i <= str.length() - n; i++){
    comp = "";
    for(int j = i; j < i + n; j++){
    comp = comp + str.charAt(j);
    }
    if(acc.equals(comp)){
    return true;
    }
    }
    return false;
    }

    ReplyDelete
  19. Cema2009June 19, 2022 at 3:00 PM

    Another way

    public boolean prefixAgain(String str, int n) {
    return (str.split(str.substring(0, n), - 1).length - 1) > 1;
    }

    ReplyDelete