Mixing
Generating all valid parenthesis
I want to create all valid parenthesis strings given an input number n. For example, if n=3, output should be as follows:
["((()))","(()())","(())()","()(())","()()()"]
My code for this problem is as follows:
private void allParenthesis(List<String> result, int n){
if(n == 1){
result.add("()");
return;
}
allParenthesis(result, n-1);
List<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
System.out.println(newResult+" for n:"+n);
result = new ArrayList<String>(newResult);
}
And I use this function in the following function,
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<String>();
allParenthesis(result,n);
return result;
}
But when I input n = 3, I get the following output,
["()"]
Where am I going wrong? Am I missing anything very simple?
java recursion
asked Nov 14 '18 at 3:38
mettleapmettleap
1,065316
add a comment |
I want to create all valid parenthesis strings given an input number n. For example, if n=3, output should be as follows:
["((()))","(()())","(())()","()(())","()()()"]
My code for this problem is as follows:
private void allParenthesis(List<String> result, int n){
if(n == 1){
result.add("()");
return;
}
allParenthesis(result, n-1);
List<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
System.out.println(newResult+" for n:"+n);
result = new ArrayList<String>(newResult);
}
And I use this function in the following function,
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<String>();
allParenthesis(result,n);
return result;
}
But when I input n = 3, I get the following output,
["()"]
Where am I going wrong? Am I missing anything very simple?
java recursion
asked Nov 14 '18 at 3:38
mettleapmettleap
1,065316
Did you try debugging?
– shmosel
Nov 14 '18 at 3:39
Yes, i did ... the print statements print [(()), ()(), ()()] for n:2 [(()), ()(), ()()] for n:3 which suggests that the result list is not being updated properly ... but it should because I modify it inside the function. I am not understanding why it does not get modified after I get the result list for n=2
– mettleap
Nov 14 '18 at 3:41
add a comment |
I want to create all valid parenthesis strings given an input number n. For example, if n=3, output should be as follows:
["((()))","(()())","(())()","()(())","()()()"]
My code for this problem is as follows:
private void allParenthesis(List<String> result, int n){
if(n == 1){
result.add("()");
return;
}
allParenthesis(result, n-1);
List<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
System.out.println(newResult+" for n:"+n);
result = new ArrayList<String>(newResult);
}
And I use this function in the following function,
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<String>();
allParenthesis(result,n);
return result;
}
But when I input n = 3, I get the following output,
["()"]
Where am I going wrong? Am I missing anything very simple?
java recursion
asked Nov 14 '18 at 3:38
mettleapmettleap
1,065316
I want to create all valid parenthesis strings given an input number n. For example, if n=3, output should be as follows:
["((()))","(()())","(())()","()(())","()()()"]
My code for this problem is as follows:
private void allParenthesis(List<String> result, int n){
if(n == 1){
result.add("()");
return;
}
allParenthesis(result, n-1);
List<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
System.out.println(newResult+" for n:"+n);
result = new ArrayList<String>(newResult);
}
And I use this function in the following function,
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<String>();
allParenthesis(result,n);
return result;
}
But when I input n = 3, I get the following output,
["()"]
Where am I going wrong? Am I missing anything very simple?
java recursion
java recursion
asked Nov 14 '18 at 3:38
mettleapmettleap
1,065316
asked Nov 14 '18 at 3:38
mettleapmettleap
1,065316
asked Nov 14 '18 at 3:38
mettleapmettleap
1,065316
asked Nov 14 '18 at 3:38
mettleapmettleap
1,065316
asked Nov 14 '18 at 3:38
mettleapmettleap
1,065316
1,065316
Did you try debugging?
– shmosel
Nov 14 '18 at 3:39
Yes, i did ... the print statements print [(()), ()(), ()()] for n:2 [(()), ()(), ()()] for n:3 which suggests that the result list is not being updated properly ... but it should because I modify it inside the function. I am not understanding why it does not get modified after I get the result list for n=2
– mettleap
Nov 14 '18 at 3:41
add a comment |
Did you try debugging?
– shmosel
Nov 14 '18 at 3:39
Yes, i did ... the print statements print [(()), ()(), ()()] for n:2 [(()), ()(), ()()] for n:3 which suggests that the result list is not being updated properly ... but it should because I modify it inside the function. I am not understanding why it does not get modified after I get the result list for n=2
– mettleap
Nov 14 '18 at 3:41
Did you try debugging?
– shmosel
Nov 14 '18 at 3:39
Did you try debugging?
– shmosel
Nov 14 '18 at 3:39
Yes, i did ... the print statements print [(()), ()(), ()()] for n:2 [(()), ()(), ()()] for n:3 which suggests that the result list is not being updated properly ... but it should because I modify it inside the function. I am not understanding why it does not get modified after I get the result list for n=2
– mettleap
Nov 14 '18 at 3:41
Yes, i did ... the print statements print [(()), ()(), ()()] for n:2 [(()), ()(), ()()] for n:3 which suggests that the result list is not being updated properly ... but it should because I modify it inside the function. I am not understanding why it does not get modified after I get the result list for n=2
– mettleap
Nov 14 '18 at 3:41
add a comment |
3 Answers
3
active
oldest
votes
You are losing the results that you have created and are not passing it through the recursion. I have fixed it and simplified to one function. Also using an array list will mean there are duplicates. For eg.
Str = "()";
newResult.add("()"+str); //this will result in ()()
newResult.add(str+"()"); //this will also result in the same
If you want the above results then keep using arraylist. if not i suggest using LinkedHashSet as set dont have duplicates and a linked once so that the order of insertion is maintained. HashSet can be used if you dont care about the ordering of the results.
I have provided both ArrayList and LinkedHashSet versions.
HashSet - No dups
private LinkedHashSet<String> generateParenthesis(int n){
if(n == 1){
LinkedHashSet<String> result = new LinkedHashSet<String>();
result.add("()");
return result;
}
LinkedHashSet<String> result = generateParenthesis(n-1);
LinkedHashSet<String> newResult = new LinkedHashSet<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
ArrayList - keep dups
private ArrayList<String> generateParenthesis(int n){
if(n == 1){
ArrayList<String> result = new ArrayList<String>();
result.add("()");
return result;
}
ArrayList<String> result = generateParenthesis(n-1);
ArrayList<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
You can use this function like so.
LinkedHashSet<String> result = generateParenthesis(3);
System.out.println(result);
answered Nov 14 '18 at 3:55
Mohammad CMohammad C
1,1951312
Good work, but if you would explain a bit more about your answer, it would be useful for new readers, and this would also improve your answer's credibility.
– PradyumanDixit
Nov 14 '18 at 3:59
@PradyumanDixit I was just testing my solution and make changes. mettleap, if you want it to be two functions like your post let me know and i will adjust and add it to the answer. As you can see it can be done in function.
– Mohammad C
Nov 14 '18 at 4:36
I understand that somethings can just be done in one line or one method, but a bit of explaining makes it better for the new readers of Stack Overflow, and this I think would be great on our part. This answer looks great. Good job!
– PradyumanDixit
Nov 14 '18 at 4:46
Thank you for the answer :)
– mettleap
Nov 14 '18 at 16:45
add a comment |
You are discarding your newResult.
Change
result = new ArrayList<String>(newResult);
to
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
KartikKartik
3,66231435
Thanks a lot :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:47
add a comment |
When you do
result = new ArrayList<String>(newResult);
You are updating the variable defined at allParenthesis,
the one you passed it from generateParenthesis remains unchanged.
Do this instead
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
JuanJuan
525
Thank you :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:46
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are losing the results that you have created and are not passing it through the recursion. I have fixed it and simplified to one function. Also using an array list will mean there are duplicates. For eg.
Str = "()";
newResult.add("()"+str); //this will result in ()()
newResult.add(str+"()"); //this will also result in the same
If you want the above results then keep using arraylist. if not i suggest using LinkedHashSet as set dont have duplicates and a linked once so that the order of insertion is maintained. HashSet can be used if you dont care about the ordering of the results.
I have provided both ArrayList and LinkedHashSet versions.
HashSet - No dups
private LinkedHashSet<String> generateParenthesis(int n){
if(n == 1){
LinkedHashSet<String> result = new LinkedHashSet<String>();
result.add("()");
return result;
}
LinkedHashSet<String> result = generateParenthesis(n-1);
LinkedHashSet<String> newResult = new LinkedHashSet<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
ArrayList - keep dups
private ArrayList<String> generateParenthesis(int n){
if(n == 1){
ArrayList<String> result = new ArrayList<String>();
result.add("()");
return result;
}
ArrayList<String> result = generateParenthesis(n-1);
ArrayList<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
You can use this function like so.
LinkedHashSet<String> result = generateParenthesis(3);
System.out.println(result);
answered Nov 14 '18 at 3:55
Mohammad CMohammad C
1,1951312
Good work, but if you would explain a bit more about your answer, it would be useful for new readers, and this would also improve your answer's credibility.
– PradyumanDixit
Nov 14 '18 at 3:59
@PradyumanDixit I was just testing my solution and make changes. mettleap, if you want it to be two functions like your post let me know and i will adjust and add it to the answer. As you can see it can be done in function.
– Mohammad C
Nov 14 '18 at 4:36
I understand that somethings can just be done in one line or one method, but a bit of explaining makes it better for the new readers of Stack Overflow, and this I think would be great on our part. This answer looks great. Good job!
– PradyumanDixit
Nov 14 '18 at 4:46
Thank you for the answer :)
– mettleap
Nov 14 '18 at 16:45
add a comment |
You are losing the results that you have created and are not passing it through the recursion. I have fixed it and simplified to one function. Also using an array list will mean there are duplicates. For eg.
Str = "()";
newResult.add("()"+str); //this will result in ()()
newResult.add(str+"()"); //this will also result in the same
If you want the above results then keep using arraylist. if not i suggest using LinkedHashSet as set dont have duplicates and a linked once so that the order of insertion is maintained. HashSet can be used if you dont care about the ordering of the results.
I have provided both ArrayList and LinkedHashSet versions.
HashSet - No dups
private LinkedHashSet<String> generateParenthesis(int n){
if(n == 1){
LinkedHashSet<String> result = new LinkedHashSet<String>();
result.add("()");
return result;
}
LinkedHashSet<String> result = generateParenthesis(n-1);
LinkedHashSet<String> newResult = new LinkedHashSet<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
ArrayList - keep dups
private ArrayList<String> generateParenthesis(int n){
if(n == 1){
ArrayList<String> result = new ArrayList<String>();
result.add("()");
return result;
}
ArrayList<String> result = generateParenthesis(n-1);
ArrayList<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
You can use this function like so.
LinkedHashSet<String> result = generateParenthesis(3);
System.out.println(result);
answered Nov 14 '18 at 3:55
Mohammad CMohammad C
1,1951312
Good work, but if you would explain a bit more about your answer, it would be useful for new readers, and this would also improve your answer's credibility.
– PradyumanDixit
Nov 14 '18 at 3:59
@PradyumanDixit I was just testing my solution and make changes. mettleap, if you want it to be two functions like your post let me know and i will adjust and add it to the answer. As you can see it can be done in function.
– Mohammad C
Nov 14 '18 at 4:36
I understand that somethings can just be done in one line or one method, but a bit of explaining makes it better for the new readers of Stack Overflow, and this I think would be great on our part. This answer looks great. Good job!
– PradyumanDixit
Nov 14 '18 at 4:46
Thank you for the answer :)
– mettleap
Nov 14 '18 at 16:45
add a comment |
You are losing the results that you have created and are not passing it through the recursion. I have fixed it and simplified to one function. Also using an array list will mean there are duplicates. For eg.
Str = "()";
newResult.add("()"+str); //this will result in ()()
newResult.add(str+"()"); //this will also result in the same
If you want the above results then keep using arraylist. if not i suggest using LinkedHashSet as set dont have duplicates and a linked once so that the order of insertion is maintained. HashSet can be used if you dont care about the ordering of the results.
I have provided both ArrayList and LinkedHashSet versions.
HashSet - No dups
private LinkedHashSet<String> generateParenthesis(int n){
if(n == 1){
LinkedHashSet<String> result = new LinkedHashSet<String>();
result.add("()");
return result;
}
LinkedHashSet<String> result = generateParenthesis(n-1);
LinkedHashSet<String> newResult = new LinkedHashSet<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
ArrayList - keep dups
private ArrayList<String> generateParenthesis(int n){
if(n == 1){
ArrayList<String> result = new ArrayList<String>();
result.add("()");
return result;
}
ArrayList<String> result = generateParenthesis(n-1);
ArrayList<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
You can use this function like so.
LinkedHashSet<String> result = generateParenthesis(3);
System.out.println(result);
answered Nov 14 '18 at 3:55
Mohammad CMohammad C
1,1951312
You are losing the results that you have created and are not passing it through the recursion. I have fixed it and simplified to one function. Also using an array list will mean there are duplicates. For eg.
Str = "()";
newResult.add("()"+str); //this will result in ()()
newResult.add(str+"()"); //this will also result in the same
If you want the above results then keep using arraylist. if not i suggest using LinkedHashSet as set dont have duplicates and a linked once so that the order of insertion is maintained. HashSet can be used if you dont care about the ordering of the results.
I have provided both ArrayList and LinkedHashSet versions.
HashSet - No dups
private LinkedHashSet<String> generateParenthesis(int n){
if(n == 1){
LinkedHashSet<String> result = new LinkedHashSet<String>();
result.add("()");
return result;
}
LinkedHashSet<String> result = generateParenthesis(n-1);
LinkedHashSet<String> newResult = new LinkedHashSet<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
ArrayList - keep dups
private ArrayList<String> generateParenthesis(int n){
if(n == 1){
ArrayList<String> result = new ArrayList<String>();
result.add("()");
return result;
}
ArrayList<String> result = generateParenthesis(n-1);
ArrayList<String> newResult = new ArrayList<String>();
for(String str : result){
newResult.add("("+str+")");
newResult.add("()"+str);
newResult.add(str+"()");
}
result.addAll(newResult);
return result;
}
You can use this function like so.
LinkedHashSet<String> result = generateParenthesis(3);
System.out.println(result);
answered Nov 14 '18 at 3:55
Mohammad CMohammad C
1,1951312
edited Nov 14 '18 at 4:39
answered Nov 14 '18 at 3:55
Mohammad CMohammad C
1,1951312
answered Nov 14 '18 at 3:55
Mohammad CMohammad C
1,1951312
answered Nov 14 '18 at 3:55
Mohammad CMohammad C
1,1951312
1,1951312
Good work, but if you would explain a bit more about your answer, it would be useful for new readers, and this would also improve your answer's credibility.
– PradyumanDixit
Nov 14 '18 at 3:59
@PradyumanDixit I was just testing my solution and make changes. mettleap, if you want it to be two functions like your post let me know and i will adjust and add it to the answer. As you can see it can be done in function.
– Mohammad C
Nov 14 '18 at 4:36
I understand that somethings can just be done in one line or one method, but a bit of explaining makes it better for the new readers of Stack Overflow, and this I think would be great on our part. This answer looks great. Good job!
– PradyumanDixit
Nov 14 '18 at 4:46
Thank you for the answer :)
– mettleap
Nov 14 '18 at 16:45
add a comment |
Good work, but if you would explain a bit more about your answer, it would be useful for new readers, and this would also improve your answer's credibility.
– PradyumanDixit
Nov 14 '18 at 3:59
@PradyumanDixit I was just testing my solution and make changes. mettleap, if you want it to be two functions like your post let me know and i will adjust and add it to the answer. As you can see it can be done in function.
– Mohammad C
Nov 14 '18 at 4:36
I understand that somethings can just be done in one line or one method, but a bit of explaining makes it better for the new readers of Stack Overflow, and this I think would be great on our part. This answer looks great. Good job!
– PradyumanDixit
Nov 14 '18 at 4:46
Thank you for the answer :)
– mettleap
Nov 14 '18 at 16:45
Good work, but if you would explain a bit more about your answer, it would be useful for new readers, and this would also improve your answer's credibility.
– PradyumanDixit
Nov 14 '18 at 3:59
Good work, but if you would explain a bit more about your answer, it would be useful for new readers, and this would also improve your answer's credibility.
– PradyumanDixit
Nov 14 '18 at 3:59
@PradyumanDixit I was just testing my solution and make changes. mettleap, if you want it to be two functions like your post let me know and i will adjust and add it to the answer. As you can see it can be done in function.
– Mohammad C
Nov 14 '18 at 4:36
@PradyumanDixit I was just testing my solution and make changes. mettleap, if you want it to be two functions like your post let me know and i will adjust and add it to the answer. As you can see it can be done in function.
– Mohammad C
Nov 14 '18 at 4:36
I understand that somethings can just be done in one line or one method, but a bit of explaining makes it better for the new readers of Stack Overflow, and this I think would be great on our part. This answer looks great. Good job!
– PradyumanDixit
Nov 14 '18 at 4:46
I understand that somethings can just be done in one line or one method, but a bit of explaining makes it better for the new readers of Stack Overflow, and this I think would be great on our part. This answer looks great. Good job!
– PradyumanDixit
Nov 14 '18 at 4:46
Thank you for the answer :)
– mettleap
Nov 14 '18 at 16:45
Thank you for the answer :)
– mettleap
Nov 14 '18 at 16:45
add a comment |
You are discarding your newResult.
Change
result = new ArrayList<String>(newResult);
to
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
KartikKartik
3,66231435
Thanks a lot :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:47
add a comment |
You are discarding your newResult.
Change
result = new ArrayList<String>(newResult);
to
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
KartikKartik
3,66231435
Thanks a lot :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:47
add a comment |
You are discarding your newResult.
Change
result = new ArrayList<String>(newResult);
to
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
KartikKartik
3,66231435
You are discarding your newResult.
Change
result = new ArrayList<String>(newResult);
to
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
KartikKartik
3,66231435
answered Nov 14 '18 at 3:55
KartikKartik
3,66231435
answered Nov 14 '18 at 3:55
KartikKartik
3,66231435
answered Nov 14 '18 at 3:55
KartikKartik
3,66231435
3,66231435
Thanks a lot :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:47
add a comment |
Thanks a lot :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:47
Thanks a lot :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:47
Thanks a lot :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:47
add a comment |
When you do
result = new ArrayList<String>(newResult);
You are updating the variable defined at allParenthesis,
the one you passed it from generateParenthesis remains unchanged.
Do this instead
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
JuanJuan
525
Thank you :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:46
add a comment |
When you do
result = new ArrayList<String>(newResult);
You are updating the variable defined at allParenthesis,
the one you passed it from generateParenthesis remains unchanged.
Do this instead
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
JuanJuan
525
Thank you :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:46
add a comment |
When you do
result = new ArrayList<String>(newResult);
You are updating the variable defined at allParenthesis,
the one you passed it from generateParenthesis remains unchanged.
Do this instead
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
JuanJuan
525
When you do
result = new ArrayList<String>(newResult);
You are updating the variable defined at allParenthesis,
the one you passed it from generateParenthesis remains unchanged.
Do this instead
result.clear();
result.addAll(newResult);
answered Nov 14 '18 at 3:55
JuanJuan
525
answered Nov 14 '18 at 3:55
JuanJuan
525
answered Nov 14 '18 at 3:55
JuanJuan
525
answered Nov 14 '18 at 3:55
JuanJuan
525
525
Thank you :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:46
add a comment |
Thank you :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:46
Thank you :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:46
Thank you :) ... appreciate your help
– mettleap
Nov 14 '18 at 16:46
add a comment |
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Did you try debugging?
– shmosel
Nov 14 '18 at 3:39
Yes, i did ... the print statements print [(()), ()(), ()()] for n:2 [(()), ()(), ()()] for n:3 which suggests that the result list is not being updated properly ... but it should because I modify it inside the function. I am not understanding why it does not get modified after I get the result list for n=2
– mettleap
Nov 14 '18 at 3:41