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wordappend challenge
Hi,
I am working on below chalenge
http://codingbat.com/prob/p103593
wordAppend(["a", "b", "a"]) → "a"
i expected aa for above as a is at index 2.
please advise
I am working on below chalenge
http://codingbat.com/prob/p103593
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.
wordAppend(["a", "b", "a"]) → "a"
wordAppend(["a", "b", "a", "c", "a", "d", "a"]) → "aa"
wordAppend(["a", "", "a"]) → "a"
wordAppend(["a", "b", "a"]) → "a"
i expected aa for above as a is at index 2.
please advise
Asked:
-
4
2
- +1
3 Solutions
In fact you could count the number of occurrences of "a".
In the string to return you need to have (total number of "a" / 2) times "a".
No "a" found ==> ""
Once "a" found ==> ""
Twice "a" found ==> return "a"
3x "a" found ==> return "a"
4x "a" found ==> return "aa"
5x "a" found ==> return "aa"
6x "a" found ==> return "aaa"
...
In the string to return you need to have (total number of "a" / 2) times "a".
No "a" found ==> ""
Once "a" found ==> ""
Twice "a" found ==> return "a"
3x "a" found ==> return "a"
4x "a" found ==> return "aa"
5x "a" found ==> return "aa"
6x "a" found ==> return "aaa"
...
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
tried as above and getting errors
Error: result= map.keySet();
^^^^^^^^^^^^
Type mismatch: cannot convert from Set<String> to String
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;
}
tried as above and getting errors
Error: result= map.keySet();
^^^^^^^^^^^^
Type mismatch: cannot convert from Set<String> to String
String result="";
Error: result= map.keySet();
^^^^^^^^^^^^
Type mismatch: cannot convert from Set<String> to String
Error: result= map.keySet();
^^^^^^^^^^^^
Type mismatch: cannot convert from Set<String> to String
"Set<K> keySet() Returns a Set view of the keys contained in this map."
[1]
[2]
[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]
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 descriptionA 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();
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;
}
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?
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Examples:
["a", "", "a"] = 2 (as "a" is appeared 2 times) - final result "a".
["a", "a", ""] = 2 (as "a" is appeared 2 times) - final result "a".
["a", "a", "", "a"] = 3 (as "a" is appeared 3 times) - final result "a".
["a", "a", "a", "a", ""] = 4 (as "a" is appeared 4 times) - final result "aa".
Hope it explains.