mapShare challenge

Hi,

 I am working on below challenge

http://codingbat.com/prob/p148813


Map-1 > mapShare
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Modify and return the given map as follows: if the key "a" has a value, set the key "b" to have that same value. In all cases remove the key "c", leaving the rest of the map unchanged.

mapShare({"b": "bbb", "c": "ccc", "a": "aaa"}) → {"b": "aaa", "a": "aaa"}
mapShare({"b": "xyz", "c": "ccc"}) → {"b": "xyz"}
mapShare({"d": "hi", "c": "meh", "a": "aaa"}) → {"d": "hi", "b": "aaa", "a": "aaa"}


does the  'a' key corresponding value remains same in output map as in below case "aaa"?

mapShare({"b": "bbb", "c": "ccc", "a": "aaa"}) → {"b": "aaa", "a": "aaa"}

please advise
LVL 7
gudii9Asked:
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zzynxSr. Software engineerCommented:
>> does the  'a' key corresponding value remains same in output map as in below case "aaa"?
Yes.

This is really a so basic challenge that you should be able to solve this on your own.
awking00Information Technology SpecialistCommented:
Think of the characteristics and methods for different types of maps -
Modify and return the given map as follows: if the key "a" has a value [contains], set the key [put] "b" to have that same value [get]. In all cases remove the key "c" [remove], leaving the rest of the map unchanged [need a map that preserves order].

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gudii9Author Commented: 
public Map<String, String> mapShare(Map<String, String> map) {
  
  String val=map.get("a");
  String backup=val;
  map.put("b",backup);
 // map.put("a","");
 map.remove("c");
  
  return map;

}

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Expected      Run            
mapShare({"b": "bbb", "c": "ccc", "a": "aaa"}) → {"b": "aaa", "a": "aaa"}      {"b": "aaa", "a": "aaa"}      OK      
mapShare({"b": "xyz", "c": "ccc"}) → {"b": "xyz"}      {"b": null}      X      
mapShare({"d": "hi", "c": "meh", "a": "aaa"}) → {"d": "hi", "b": "aaa", "a": "aaa"}      {"d": "hi", "b": "aaa", "a": "aaa"}      OK      
mapShare({"b": "1234", "c": "yo", "a": "xyz", "z": "zzz"}) → {"b": "xyz", "a": "xyz", "z": "zzz"}      {"b": "xyz", "a": "xyz", "z": "zzz"}      OK      
mapShare({"d": "ddd", "e": "everything", "b": "1234", "c": "yo", "a": "xyz"}) → {"d": "ddd", "e": "everything", "b": "xyz", "a": "xyz"}      {"d": "ddd", "e": "everything", "b": "xyz", "a": "xyz"}      OK      
other tests
X      
Correct for more than half the tests

Your progress graph for this problem

above fails one test.
please advise
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gudii9Author Commented: 
public Map<String, String> mapShare(Map<String, String> map) {
  if(map.containsKey("a")){
  String val=map.get("a");
  String backup=val;
  map.put("b",backup);
 // map.put("a","");
 map.remove("c");
  }
  else{
     map.remove("c");
  }
  return map;

}

Open in new window


above passes all tests. any improvements or alternate approaches?
awking00Information Technology SpecialistCommented:
You might consolidate some of the steps -
map.remove"c";  ==> in all cases
if (map.containsKey("a") {
  map.put("b",map.get("a")); ==>no need for intermediate variable
  map.put("a","");
}
awking00Information Technology SpecialistCommented:
Typo - map.remove("c"); ==> needs parentheses
gudii9Author Commented: 
public Map<String, String> mapShare(Map<String, String> map) {
  if(map.containsKey("a")){
 // String val=map.get("a");
 // String backup=val;
  map.put("b",map.get("a"));
 // map.put("a","");
 map.remove("c");
  }
 // else{
 //    map.remove("c");
 // }
  return map;

}

Open in new window


above failse one test if i try to remove c in all cases

Expected      Run            
mapShare({"b": "bbb", "c": "ccc", "a": "aaa"}) → {"b": "aaa", "a": "aaa"}      {"b": "aaa", "a": "aaa"}      OK      
mapShare({"b": "xyz", "c": "ccc"}) → {"b": "xyz"}      {"b": "xyz", "c": "ccc"}      X      
mapShare({"d": "hi", "c": "meh", "a": "aaa"}) → {"d": "hi", "b": "aaa", "a": "aaa"}      {"d": "hi", "b": "aaa", "a": "aaa"}      OK      
mapShare({"b": "1234", "c": "yo", "a": "xyz", "z": "zzz"}) → {"b": "xyz", "a": "xyz", "z": "zzz"}      {"b": "xyz", "a": "xyz", "z": "zzz"}      OK      
mapShare({"d": "ddd", "e": "everything", "b": "1234", "c": "yo", "a": "xyz"}) → {"d": "ddd", "e": "everything", "b": "xyz", "a": "xyz"}      {"d": "ddd", "e": "everything", "b": "xyz", "a": "xyz"}      OK      
other tests
OK      


please advise
zzynxSr. Software engineerCommented:
>> above failse one test if i try to remove c in all cases
You DON'T remove "c" in all cases.
If you would remove instead of comment out old code ànd indent your code well, you would have seen that your 
map.remove("c");

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statement is INSIDE the IF statement.
A perfect illustration of WHY I always keep telling you that you should indent your code well, to make it good readable.

This works:
public Map<String, String> mapShare(Map<String, String> map) {
    map.remove("c");
    if (map.containsKey("a")) {
        map.put("b", map.get("a"));
    }
    return map;
}

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You had this:
public Map<String, String> mapShare(Map<String, String> map) {
   if (map.containsKey("a")) {
       map.put("b",map.get("a"));
       map.remove("c");
   }
   return map;
}

Open in new window


You see the difference?
awking00Information Technology SpecialistCommented:
Without comments, your code is -
if (map.containsKey("a")) {
    map.put("b",map.get("a"));
    map.remove("c");
}
return map;

You need to remove "c" whether "a" exists or not -
    map.remove("c"); ==> note - if "c" doesn't exist, nothing will happen, if it does, the record is removed from the map
if (map.containsKey("a")) {
    map.put("b",map.get("a"));
    map.put("a",""); ==> this is still needed
}
gudii9Author Commented: 
public Map < String, String > mapShare(Map < String, String > map) {
    if (map.containsKey("a")) {
        String val = map.get("a");
        String backup = val;
        map.put("b", backup);
        // map.put("a","");
        map.remove("c");
    } else {
        map.remove("c");
    }
    return map;

}

Open in new window


i see it is in if loop
gudii9Author Commented: 
public Map < String, String > mapShare(Map < String, String > map) {
    map.remove("c");
    if (map.containsKey("a")) {
        // String val=map.get("a");
        // String backup=val;
        map.put("b", map.get("a"));
        // map.put("a","");
        //map.remove("c");
    }
    // else{
    //    map.remove("c");
    // }
    return map;

}

Open in new window


above passes all tests
gudii9Author Commented:
http://www.tutorialspoint.com/online_java_formatter.htm
above is good formatter tool i found
zzynxSr. Software engineerCommented:
>> above is good formatter tool i found
All right. Try to use it :)
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