Alok :
I have been solving this set of a challenge on Hackerrank, link : Hackerrank in a String
I have come up with my Algo which goes like this :
- To check whether it has atleast two a's, r's and k's according to the hackerrank string
- Check that it contains following chars h, a, c, k, e, r, n, k
- let p[0], p(1), p[2].....,p[9] be the respective indices of h, a, c, k, e, r, r, a, n, k in string . If p[0] < p(1) < p[2] < .... < p[9] is true, then contains hackerrank.
I know is very bad, and the output is also not coming due to this error : Terminated due to timeout
CODE:
I have a method named hackerrankInString(String s)
, it returns "YES" or "NO" string
// Complete the hackerrankInString function below.
static String hackerrankInString(String s) {
int countR = 0, countA = 0, countK = 0;
//for reading multi-line inputs, it will perform operation individually
// on the inputs
while(scanner.hasNextLine()){
for(int i=0; i < s.length(); i++){
if(s.charAt(i) == 'r')
countR++;
if(s.charAt(i) == 'a')
countA++;
if(s.charAt(i) == 'k')
countK++;
}
//It should atleast have two As, two Rs, and two Ks
if(countR >= 2 && countA >= 2 && countK >= 2){
//If the string contains hackerrank
if(s.contains(Character.toString('h')) && s.contains(Character.toString('a')) && s.contains(Character.toString('c')) && s.contains(Character.toString('k')) && s.contains(Character.toString('e')) && s.contains(Character.toString('r')) && s.contains(Character.toString('n'))){
if((s.indexOf('h') > s.indexOf('a')) && (s.indexOf('h') > s.indexOf('c')) && (s.indexOf('h') > s.indexOf('k')) && (s.indexOf('h') > s.indexOf('e')) && (s.indexOf('h') > s.indexOf('r')) && (s.indexOf('h') > s.lastIndexOf('r')) && (s.indexOf('h') > s.lastIndexOf('a')) && (s.indexOf('h') > s.indexOf('n')) && (s.indexOf('h') > s.lastIndexOf('k'))){
if((s.indexOf('a') > s.indexOf('c')) && (s.indexOf('a') > s.indexOf('k')) && (s.indexOf('a') > s.indexOf('e')) && (s.indexOf('a') > s.indexOf('r')) && (s.indexOf('a') > s.lastIndexOf('r')) && (s.indexOf('a') > s.lastIndexOf('a')) && (s.indexOf('a') > s.indexOf('n')) && (s.indexOf('a') > s.lastIndexOf('k'))){
if((s.indexOf('c') > s.indexOf('k')) && (s.indexOf('c') > s.indexOf('e')) && (s.indexOf('c') > s.indexOf('r')) && (s.indexOf('c') > s.lastIndexOf('r')) && (s.indexOf('c') > s.lastIndexOf('a')) && (s.indexOf('c') > s.indexOf('n')) && (s.indexOf('c') > s.lastIndexOf('k'))){
if((s.indexOf('k') > s.indexOf('e')) && (s.indexOf('k') > s.indexOf('r')) && (s.indexOf('k') > s.lastIndexOf('r')) && (s.indexOf('k') > s.lastIndexOf('a')) && (s.indexOf('k') > s.indexOf('n')) && (s.indexOf('k') > s.lastIndexOf('k'))){
if((s.indexOf('e') > s.indexOf('r')) && (s.indexOf('e') > s.lastIndexOf('r')) && (s.indexOf('e') > s.lastIndexOf('a')) && (s.indexOf('e') > s.indexOf('n')) && (s.indexOf('e') > s.lastIndexOf('k'))){
if((s.indexOf('r') > s.lastIndexOf('r')) && (s.indexOf('r') > s.lastIndexOf('a')) && (s.indexOf('r') > s.indexOf('n')) && (s.indexOf('r') > s.lastIndexOf('k'))){
if((s.lastIndexOf('r') > s.lastIndexOf('a')) && (s.lastIndexOf('r') > s.indexOf('n')) && (s.lastIndexOf('r') > s.lastIndexOf('k'))){
if((s.lastIndexOf('a') > s.indexOf('n')) && (s.lastIndexOf('a') > s.lastIndexOf('k'))){
if(s.indexOf('n') > s.lastIndexOf('k')){
return "YES";
}
}
}
}
}
}
}
}
}
}
}
}
return "NO";
}
Please help me with this, I want to learn two thigs,
- How to solve this challenge
- How to solve my third algo statement efficiently
Any help would be appreciated.
Nicholas K :
You can make use of a regex to solve the problem:
static String hackerrankInString(String input) {
final String regexHack = ".*h.*a.*c.*k.*e.*r.*r.*a.*n.*k.*";
return input.matches(regexHack) ? "YES" : "NO";
}
Call it using :
System.out.println(hackerrankInString("hhaacckkekraraannk")); // YES
System.out.println(hackerrankInString("hackerworld")); // NO
You can try out the regex here.