Given a linked list and a target value T, partition it such that all nodes less than T are listed before the nodes larger than or equal to target value T.
I found it is a little confused to me to understand codes.
while(head!=null){
if(head.value<target){
curSmall.next=head;
curSmall=curSmall.next;
} else{
curLarge.next=head;
curLarge=curLarge.next;
}
head=head.next;
}
curSmall.next=large.next;
curLarge.next=null;
return small.next;
I just don't understand these two parts large.next and return small.next?
In other word, when we connect the small and large linked list, the curSmall.next --> large.next, why we put large.next here, I am quite confused. And why we need to return small.next?
For example, after while loop of that code, two linked list are like below.
dummy_head(small)->a->c->d
dummy_head(large)->f->h
And curSmall and curLarge points the node 'd' and 'h' respectively.
Therefore you can get a desired linked list in the following steps.
The small linked list append the large linked list
=> curSmall.next = large.next;
The tail node 'h' of large linked list points to null => curLarge.next = null;
Return the head node 'a' of entire linked list => return small.next;
Related
we are trying to run a cypher query but we are not able to get the results we want.
It is important to note that we cannot make this work with subqueries because we are using Neo4j 3.5 and in this version, they are still not available.
The problem is that we have two parts for a query, the first one will fix the variables for the second part, and the second part consists of multiple matches, and it has to get the results of the previous query and if at least one of the matches return a result for a row, this row is not filtered out, else if none return result it is discarded.
More specifically, the query we are trying to run is similar to the following one:
//First part of the query where we want to fix variables with the match and where
MATCH (u:User)-[:ASSIGNED_TO]->(t:Task)-[:PENDING]->(ob:Object)<-[:HAS_OPEN_OBJECT]-(do:DataObject)<-[:ASOCIATED]-(:Module)-[:CAN_LIST]->(view:WidgetObject)
WHERE u.uid = 'user_uid'
AND view.uid = 'view_uid'
AND view.object_name = do.object_type
with do, t, ob
// In this second part of the query we want to maintain the variables of the previous part and if at least one matches the value should be returned
// we have tried with UNION but we will need pagination, but even with union it's not working
MATCH (ac:Action)<-[:ASOCIATED]-(t)-[rel:COMPLETED|PENDING]->(ob)<-[:HAS_OPEN_OBJECT|HAS_CLOSED_OBJECT]-(do)
WHERE ac.name CONTAINS 'body'
WITH COLLECT({data_object_uid: do.uid}) as act_filter
MATCH (c:Comment)<-[:COMMENTED]-(t)-[rel:COMPLETED|PENDING]->(ob)<-[:HAS_OPEN_OBJECT|HAS_CLOSED_OBJECT]-(do)
WHERE c.body CONTAINS 'body'
WITH act_filter + COLLECT({data_object_uid: do.uid}) as comment_filter
MATCH (at:Attachment)<-[:HAS_ATTACHMENT]-(t)-[rel:COMPLETED|PENDING]->(ob)<-[:HAS_OPEN_OBJECT|HAS_CLOSED_OBJECT]-(do)
WHERE at.name CONTAINS 'body'
WITH comment_filter + COLLECT({data_object_uid: do.uid}) as attachment_filter
UNWIND attachment_filter as row
return row.data_object_uid
We are not sure if in the second part, the second and third matches are maintaining the same subset of results coming from the first part of the query.
A funny behavior we have found is that if we remove the last match we are getting results but if we add it, we are not getting any results. We do not understand this behavior because if the second match is returning results and they are stored in a variable after a collect, appending this to the next collected results should return something.
For example, if the second match returns as comment_filter [{data_object_uid: "343dienmd3-dasd"}] and the third match is not returning anything, after the concatenation in the WITH clause it should return the same thing, but the result is empty.
We need some light here, we don't know if we are close and we are making a stupid mistake or we are getting all wrong and we need to change the approach completely.
Since you do not know which of the three matches in the second part will yield results, I would try something along the lines below:
NOTE I used ASSOCIATED instead of ASOCIATED
MATCH (n)<-[:ASSOCIATED|COMMENTED|HAS_ATTACHMENT]-(t)-[rel:COMPLETED|PENDING]->(ob)<-[:HAS_OPEN_OBJECT|HAS_CLOSED_OBJECT]-(do)
WHERE
(n:Action AND n.name CONTAINS 'body')
OR
(n:Comment AND n.body CONTAINS 'body')
OR
(n:Attachment AND n.name CONTAINS 'body')
RETURN COLLECT(DISTINCT {data_object_uid: do.uid})
Exactly what the title says.
Also: What would list[0] yield (if one were to implement linked list indexing) if the linked list contained just a head pointing to null (or the programming language in question's functional equivalent of null).
Finally: Would the length be 0 or 1 (with length referring to the number of elements in the list) if the linked list contained only a head pointing to null?
void deleteDups(LinkedListNode n)
HashSet<Integer> set = new HashSet<Integer>();
LinkedListNode previous = null;
while(n!=null){
if(set.contains(n.data)){
previous.next = n.next;
}else {
set.add(n.data);
previous = n;
}
n = n.next;
}
}
This code snippet is to remove duplicate elements in a linked List.
I have been going through the linked list concepts in cracking the coding interview. Since only the code snippet is available I am not able to understand the flow and where the LinkedListNode n in the first line is actually from. I can understand that they are passing the entire linked list as a parameter, it would be helpful if anybody can tell me what the code for that LinkedListNode will be. Thank you in advance.
LinkedList traversal: Given the head (starting node) of the list, you could traverse the whole list. Each currentnode will have the information of nextnode which can be accessed by current.next.
In your code LinkedListNode 'n' is the starting node (head) of the list where duplicates needs to be removed
I can query using Cypher in Neo4j from the Panama database the countries of three types of identity holders (I define that term) namely Entities (companies), officers (shareholders) and Intermediaries (middle companies) as three attributes/columns. Each column has single or double entries separated by colon (eg: British Virgin Islands;Russia). We want to concatenate the countries in these columns into a unique set of countries and hence obtain the count of the number of countries as new attribute.
For this, I tried the following code from my understanding of Cypher:
MATCH (BEZ2:Officer)-[:SHAREHOLDER_OF]->(BEZ1:Entity),(BEZ3:Intermediary)-[:INTERMEDIARY_OF]->(BEZ1:Entity)
WHERE BEZ1.address CONTAINS "Belize" AND
NOT ((BEZ1.countries="Belize" AND BEZ2.countries="Belize" AND BEZ3.countries="Belize") OR
(BEZ1.status IN ["Inactivated", "Dissolved shelf company", "Dissolved", "Discontinued", "Struck / Defunct / Deregistered", "Dead"]))
SET BEZ4.countries= (BEZ1.countries+","+BEZ2.countries+","+BEZ3.countries)
RETURN BEZ3.countries AS IntermediaryCountries, BEZ3.name AS
Intermediaryname, BEZ2.countries AS OfficerCountries , BEZ2.name AS
Officername, BEZ1.countries as EntityCountries, BEZ1.name AS Companyname,
BEZ1.address AS CompanyAddress,DISTINCT count(BEZ4.countries) AS NoofConnections
The relevant part is the SET statement in the 7th line and the DISTINCT count in the last line. The code shows error which makes no sense to me: Invalid input 'u': expected 'n/N'. I guess it means to use COLLECT probably but we tried that as well and it shows the error vice-versa'd between 'u' and 'n'. Please help us obtain the output that we want, it makes our job hell lot easy. Thanks in advance!
EDIT: Considering I didn't define variable as suggested by #Cybersam, I tried the command CREATE as following but it shows the error "Invalid input 'R':" for the command RETURN. This is unfathomable for me. Help really needed, thank you.
CODE 2:
MATCH (BEZ2:Officer)-[:SHAREHOLDER_OF]->(BEZ1:Entity),(BEZ3:Intermediary)-
[:INTERMEDIARY_OF]->(BEZ1:Entity)
WHERE BEZ1.address CONTAINS "Belize" AND
NOT ((BEZ1.countries="Belize" AND BEZ2.countries="Belize" AND
BEZ3.countries="Belize") OR
(BEZ1.status IN ["Inactivated", "Dissolved shelf company", "Dissolved",
"Discontinued", "Struck / Defunct / Deregistered", "Dead"]))
CREATE (p:Connections{countries:
split((BEZ1.countries+";"+BEZ2.countries+";"+BEZ3.countries),";")
RETURN BEZ3.countries AS IntermediaryCountries, BEZ3.name AS
Intermediaryname, BEZ2.countries AS OfficerCountries , BEZ2.name AS
Officername, BEZ1.countries as EntityCountries, BEZ1.name AS Companyname,
BEZ1.address AS CompanyAddress, AS TOTAL, collect (DISTINCT
COUNT(p.countries)) AS NumberofConnections
Lines 8 and 9 are the ones new and to be in examination.
First Query
You never defined the identifier BEZ4, so you cannot set a property on it.
Second Query (which should have been posted in a separate question):
You have several typos and a syntax error.
This query should not get an error (but you will have to determine if it does what you want):
MATCH (BEZ2:Officer)-[:SHAREHOLDER_OF]->(BEZ1:Entity),(BEZ3:Intermediary)- [:INTERMEDIARY_OF]->(BEZ1:Entity)
WHERE BEZ1.address CONTAINS "Belize" AND NOT ((BEZ1.countries="Belize" AND BEZ2.countries="Belize" AND BEZ3.countries="Belize") OR (BEZ1.status IN ["Inactivated", "Dissolved shelf company", "Dissolved", "Discontinued", "Struck / Defunct / Deregistered", "Dead"]))
CREATE (p:Connections {countries: split((BEZ1.countries+";"+BEZ2.countries+";"+BEZ3.countries), ";")})
RETURN BEZ3.countries AS IntermediaryCountries,
BEZ3.name AS Intermediaryname,
BEZ2.countries AS OfficerCountries ,
BEZ2.name AS Officername,
BEZ1.countries as EntityCountries,
BEZ1.name AS Companyname,
BEZ1.address AS CompanyAddress,
SIZE(p.countries) AS NumberofConnections;
Problems with the original:
The CREATE clause was missing a closing } and also a closing ).
The RETURN clause had a dangling AS TOTAL term.
collect (DISTINCT COUNT(p.countries)) was attempting to perform nested aggregation, which is not supported. In any case, even if it had worked, it probably would not have returned what you wanted. I suspect that you actually wanted the size of the p.countries collection, so that is what I used in my query.
I am confused about the last "return" statement in the Java implementation of linked list below.
here is the code:
//remove the node with duplicate data and return the linked list
static Node removeDuplicates(Node head) {
if(head == null)
return head;
Node temp = head;
while(null != temp.next) {
if(temp.data == temp.next.data)
temp.next = temp.next.next;
else
temp = temp.next;
}
return head;
}
The last "return" statement is "return head", but shouldn't it be "return temp"?? My explanation would be that temp node is created to copy the head node, and traverse the entire linked list. At the end of this operation, it is the temp that is modified if there is any duplicate data, so the last statement should be "return temp".
The code above that I am confused about is actually correct, can anyone explain it to me?
Thank you!
A handle to a linked list is always the head of the list. The code doesn't reverse the list. It only removes duplicates. More precisely, if temp.data is identical to temp.next.data, then temp.next is being removed.
The head of the list is never changed in this process, therefore it is correct to return the original head, since it is also the head of the modified list.
For instance, suppose the original list was head --> a --> b --> c, where head,a,b,c all denote nodes in the list, and suppose the the value of a is identical to the values of b. To remove duplicates, we effectively change the "next" pointer in a to point to c instead of b. This yields the modified list head --> a --> c. The head node of this list is identical to the head node of the original list, and this is why we return head.