wordappend challenge

is it like  return (number of times a found divided by 2 )number of times of a??

Loop over the given array of strings to build a result string like this: when a string appears the 2nd, 4th, 6th, etc. time in the array, append the string to the result. Return the empty string if no string appears a 2nd time.

what it mean by appears at 4th 6th etc??bit confusing description

i got meaning i think

public String wordAppend(String[] strings) {
  
  
 // public Map<String, Integer> wordCount(String[] strings) {
  Map<String, Integer> map=new TreeMap();
  String result="";
  for(int i=0;i<strings.length;i++){
   // String buffer=strings[i]
  String test=strings[i];
  if(map.containsKey(test)){
  int count=map.get(test);
  map.put(test,count+1);
  }
  
  else{
  map.put(test,1);
  }

  result= map.keySet();
  }
  return result;


}

Select all Open in new window

tried as above and getting errors
Error:      result= map.keySet();
              ^^^^^^^^^^^^
Type mismatch: cannot convert from Set<String> to String

[1]

is it like  return (number of times a found divided by 2 )number of times of a??

That's what I said, yes.

[2]

what it mean by appears at 4th 6th etc??bit confusing description

A 2nd, 4th, 6th occurrence of the string "a"

[3]
The code in your last post is badly indented

[4]
The method keySet() returns a Set of Strings (the keys of the map), so you can't assign that to a String variable.

[5]

 Map<String, Integer> map=new TreeMap();

Select all Open in new window

Do you have a good reason for using a TreeMap?

public String wordAppend(String[] strings) {
  
 Map<String, Integer> test    = new HashMap<String, Integer>();
 String               res= "";
  
 for (int i = 0; i < strings.length; i++) {
    
   String key = strings[i];
    
   if (test.containsKey(key)) {
     int val = test.get(key);
     val++;
     if (val % 2 == 0) {
      res = res+key;
     }
     test.put(key, val);
   } else {
     test.put(key, 1);
   }
    
 }
  
 return res;
}

Select all Open in new window

above pass all tests.
Basically if map does not contain key adding 1 as value.otherwise we are getting the value and if it is multiple of 2 adding any improvements or alternate approaches?