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Solved
mapShare challenge
Posted on 2016-09-27
Hi,
I am working on below challenge
http://codingbat.com/prob/p148813
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
I am working on below challenge
http://codingbat.com/prob/p148813
Map-1 > mapShare
prev | next | chance
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
[X]
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13 Comments
>> 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.
Yes.
This is really a so basic challenge that you should be able to solve this on your own.
Active 2 days ago
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].
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].
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;
}
Expected Runabove fails one test.
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
please advise
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;
}
above passes all tests. any improvements or alternate approaches?
Active 2 days ago
You might consolidate some of the steps -
map.remove"c"; ==> in all cases
if (map.containsKey("a") {
map.put("b",map.get("a"));
map.put("a","");
}
map.remove"c"; ==> in all cases
if (map.containsKey("a") {
map.put("b",map.get("a"));
map.put("a","");
}
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;
}
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
>> 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
A perfect illustration of WHY I always keep telling you that you should indent your code well, to make it good readable.
This works:
You had this:
You see the difference?
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");
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;
}
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;
}
You see the difference?
Active 2 days ago
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
}
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
}
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;
}
i see it is in if loop
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;
}
above passes all tests
http://www.tutorialspoint.com/online_java_formatter.htm
above is good formatter tool i found
above is good formatter tool i found
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