1(70)
/ \
/ \
2(40) 5(10)
/ \ \
/ \ \
3(60) 4(80) 6(20)
/ \
/ \
7(30) 8(50)
This is for an online challenge (not live contest). I don't need someone to solve for me, just to push in right direction. Trying to learn.
Each node has a unique ID, no two people have same salary. Person #1 has salary $70, person #7 has $30 salary, for example. Tree structure denotes who supervises who. Question is who has kth lowest salary on a person's subordinates.
For example I choose person #2. Who is 2nd lowest among subordinates? #2's subordinates are 3, 4, 7, 8. 2nd lowest salary is $50 belonging to person #8.
There are many queries so structure to must be efficient.
I thought about this problem and researched data structures. Binary tree seems like a good idea but I need help.
For example I think ideal structure look like, for person #2:
2(40)
/ \
/ \
7(30) 3(60)
/ \
/ \
8(50) 4(80)
Every child node is subordinate of #2, every left branch has lower salary than on right. If I store how many children at each node I can get kth lowest.
For example: From #2, left branch 1 node, right branch 3 nodes. So 2nd lowest - 1 means I now want 1st lowest in right branch.
Move to #3, 1st lowest points to #8 with $50 which is correct.
My question:
Is this approach as I describe it a good one? Is it a valid approach?
I am having trouble figuring out how to construct this kind of tree. I think I can make them recursively. But hard to figure out how to make all children into new tree sorted by salary. Need some light help.
Here's a solution that uses O(n log^2 n + q log n) time and O(n log^2 n) space (not the best on the latter count, but probably good enough given the limits).
Implement a purely functional sorted list (as an augmented binary search tree) with the following operations and some way to iterate.
EmptyList() -> returns the empty list
Insert(list, key) -> returns the list where |key| has been inserted into |list|
Length(list) -> returns the length of the list
Get(list, k) -> returns the element at index |k| in |list|
On top of these operations, implement an operation
Merge(list1, list2) -> returns the union of |list1| and |list2|
by inserting the elements of the shorter list into the longer.
Now do the obvious thing: traverse the employee hierarchy from leaves to root, setting the ordered list for each employee to the appropriate merge of her subordinate lists, and answer the queries.
Analysis (sketch)
Each query takes O(log n) time. The interesting part of the analysis pertains to the preprocessing.
The cost of preprocessing is dominated by the cost of calling Insert(), specifically from Merge(), since there are n other insertions. Each insertion takes O(log n) time and costs O(log n) space (measuring in words).
What keeps the preprocessing from being quadratic is an implicit heavy path decomposition. Every time we merge two lists, neither list is merged subsequently. Since the shorter list is inserted into the longer, every time a key is inserted into a list, that list is at least twice as long as the list into which that key was previously inserted. It follows that each key is the subject of at most lg n insertions, which suffices to establish a bound of O(n log n) insertions overall and thus the claimed resource bounds.
Here is one possible solution. For each node, we will construct an array of all children's values of that node and keep this in sorted order. The result we are looking for is a dictionary of the form
{ 1 : [10, 20, 30, 40, 60 80],
2 : [30, 50, 60, 80]
...
}
Once we have this, to query for any node for the ith lowest salary, just take the ith element of the array. The total time to do all the queries is O( q ) where q is the number of queries.
How do we construct this? Assuming you have a pointer to the root node, you can recursively construct the sorted salaries for each child. Store those values in the result. Make a copy of each child's array, and insert each child's salary into the child's copied array. Use binary search to find the position, since each array is sorted. Now you have k sorted arrays, you merge them to get a sorted array. If you are merging two arrays, this can be done in linear time. Simply loop, picking the first element of the array that is smaller each time.
For the case where each node has 2 children, merging the two children's arrays is O(n) tine. Inserting the salaries of each node to its corresponding array is O(log(n)) per node since we use binary search. Copying the children array is O(n), and there are n nodes, so we have O(n^2) total time for pre-processing.
Total run time is O(n^2 + q)
What if we can not assume each node has at most 2 children? Then to merge the arrays, use this algorithm. This runs in O(nlog(k)) where k is the number of arrays to merge, since we pop from the heap once per element, and resizing the heap takes O(log(k)) when there are k arrays. k<=n so we can simplify this to O(nlog(n)). So the total running time is unchanged.
The space complexity of this solution is O(n^2).
The question has two parts: first, finding the specified person, and then finding the kth subordinate.
Since the tree is not ordered by id, to find the specified perspn by id requires walking the whole tree until the specified id is found. To speed up this part, we can builda hash map that would allow us to find the person node by id in O(1) time and require O(n) space and set-up time.
Then, to find the subordinate with the kth lowest salary, we need to search the subtree. Since its not ordered by salary, we would have to scan the whole subtree and find the kth lowest salary. This could be done using an array or a heap (putting the subtree nodes into array or heap). This second part would O(m log k) time using the heap to keep the lowest k items, where m is the number of sub-ordinates, and require O(k) space. This should be acceptable if m (number of subordinates of specified person), and k are small.
I'm looking for some help on a specific augmented Red Black Binary Tree. My goal is to make every single operation run in O(log(n)) in the worst case. The nodes of the tree will have an integer as there key. This integer can not be negative, and the tree should be sorted by a simple compare function off of this integer. Additionally, each node will also store another value: its power. (Note that this has nothing to do with mathematical exponents). Power is a floating point value. Both power and key are always non-negative. The tree must be able to provide these operations in O(log(n)) runtime.:
insert(key, power): Insert into the tree. The node in the tree should also store the power, and any other variables needed to augment the tree in such a way that all other operations are also O(log(n)). You can assume that there is no node in the tree which already has the same key.
get(key): Return the power of the node identified by the key.
delete(key): Delete the node with key (assume that the key does exist in the tree prior to the delete.
update(key,power): Update the power at the node given by key.
Here is where it gets interesting:
highestPower(key1, key2): Return the maximum power of all nodes with key k in the range key1 <= k <= key2. That is, all keys from key1 to key2, inclusive on both ends.
powerSum(key1, key2): Return the sum of the powers of all nodes with key k in the ragne key1 <= k <= key2. That is, all keys from key1 to key2, inclusive on both ends.
The main thing I would like to know is what extra variables should I store at each node. Then I need to work out how to use each one of these in each of the above functions so that the tree stays balanced and all operations can run in O(log(n)) My original thought was to store the following:
highestPowerLeft: The highest power of all child nodes to the right of this node.
highestPowerRight: The highest power of all child nodes to the right of this node.
powerSumLeft: The sum of the powers of all child nodes to the left of this node.
powerSumRight: The sum of the powers of all child nodes to the right of this node.
Would just this extra information work? If so, I'm not sure how to deal with it in the functions that are required. Frankly my knowledge of Red Black Tree's isn't great because I feel like every explanation of them gets convoluted really fast, and all the rotations and things confuse the hell out of me. Thanks to anyone willing to attempt helping here, I know what I'm asking is far from simple.
A very interesting problem! For the sum, your proposed method should work (it should be enough to only store the sum of the powers to the left of the current node, though; this technique is called prefix sum). For the max, it doesn't work, since if both max values are equal, that value is outside of your interval, so you have no idea what the max value in your interval is. My only idea is to use a segment tree (in which the leaves are the nodes of your red-black tree), which lets you answer the question "what is the maximal value within the given range?" in logarithmic time, and also lets you update individual values in logarithmic time. However, since you need to insert new values into it, you need to keep it balanced as well.
A company is planning a party for its employees. A fun rating is assigned to every employee.
The employees are organized into a strict hierarchy, i.e. a tree rooted the president. There is one
restriction, though, on the guest list to the party: an employee and his/her immediate supervisor
(parent in the tree) cannot both attend the party. You wish to prepare a guest list for the party that
maximizes the sum of fun ratings of the guests. Show that greedily choosing guests according to fun
rating, will not work. Then, formulate a dynamic programming solution
I could not understand some of the conditions like is the fun rate of the president higher than that of his descendants and how many employees are there for each of his supervisor. Can someone help me in proceeding with this ?
From the phrasing on the problem, the fun rating assigned to someone in the hierarchy tree is not necessarily greater than their descendants in the hierarchy tree.
However, even if this were the case, to see that it is not optimal to pick the best employee, consider a tree of height 2 with a root of fun=10 and 20 children of fun=1. Then the optimal solution is to skip the greedy choice (the root) and choose the 20 children.
In any case, with dynamic programming you can find the best solution even if parents can have lower fun than their descendants. For a node v in the tree, let F(v) be the maximum fun that can be attained within subtree rooted at v. Then either you choose v in which case the children are skipped and you look at all subtrees that are rooted at children of children (taking the sum of max fun over these subtrees, and adding to fun(v)), or you skip v and then you get the maximum fun is the sum of maximum fun over all subtrees rooted at children of v. This gives a linear time dynamic programming algorithm.
For greedy example show a simple counterexample.
1
|
1--3--1
|
1
Choosing greedily i.e. first selecting 3 won't allow us to select any other employee. But if we don't select 3, than max will be 4(all 1's). So greedy approach won't work.
For dynamic programming we can formulate problem as selecting a given root of subtree. If we select a node, than none of its children can be selected. However if we don't select the node, than we can either select the child or not select the child.
for all v initialize
c(v,true) = fun(v)
c(v,false)= 0
Than use the following recursion to solve the problem
weight(v,true) = for all children sum( weight(ci,false) ) + fun(i)
weight(v,false) = for all children sum( max(weight(ci,false), weight(ci,true)))
The answer will be max(weight(v,true),weight(v,false)) for root node.
I'm working on putting together a problem set for an intro-level CS course and came up with a question that, on the surface, seems very simple:
You are given a list of people with the names of their parents, their birth dates, and their death dates. You are interested in finding out who, at some point in their lifetime, was a parent, a grandparent, a great-grandparent, etc. Devise an algorithm to label each person with this information as an integer (0 means the person never had a child, 1 means that the person was a parent, 2 means that the person was a grandparent, etc.)
For simplicity, you can assume that the family graph is a DAG whose undirected version is a tree.
The interesting challenge here is that you can't just look at the shape of the tree to determine this information. For example, I have 8 great-great-grandparents, but since none of them were alive when I was born, in their lifetimes none of them were great-great-grandparents.
The best algorithm I can come up with for this problem runs in time O(n2), where n is the number of people. The idea is simple - start a DFS from each person, finding the furthest descendant down in the family tree that was born before that person's death date. However, I'm pretty sure that this is not the optimal solution to the problem. For example, if the graph is just two parents and their n children, then the problem can be solved trivially in O(n). What I'm hoping for is some algorithm that is either beats O(n2) or whose runtime is parameterized over the shape of the graph that makes it fast for wide graphs with a graceful degradation to O(n2) in the worst-case.
Update: This is not the best solution I have come up with, but I've left it because there are so many comments relating to it.
You have a set of events (birth/death), parental state (no descendants, parent, grandparent, etc) and life state (alive, dead).
I would store my data in structures with the following fields:
mother
father
generations
is_alive
may_have_living_ancestor
Sort your events by date, and then for each event take one of the following two courses of logic:
Birth:
Create new person with a mother, father, 0 generations, who is alive and may
have a living ancestor.
For each parent:
If generations increased, then recursively increase generations for
all living ancestors whose generations increased. While doing that,
set the may_have_living_ancestor flag to false for anyone for whom it is
discovered that they have no living ancestors. (You only iterate into
a person's ancestors if you increased their generations, and if they
still could have living ancestors.)
Death:
Emit the person's name and generations.
Set their is_alive flag to false.
The worst case is O(n*n) if everyone has a lot of living ancestors. However in general you've got the sorting preprocessing step which is O(n log(n)) and then you're O(n * avg no of living ancestors) which means that the total time tends to be O(n log(n)) in most populations. (I hadn't counted the sorting prestep properly, thanks to #Alexey Kukanov for the correction.)
I thought of this this morning, then found that #Alexey Kukanov had similar thoughts. But mine is more fleshed out and has some more optimization, so I'll post it anyways.
This algorithm is O(n * (1 + generations)), and will work for any dataset. For realistic data this is O(n).
Run through all records and generate objects representing people which include date of birth, links to parents, and links to children, and several more uninitialized fields. (Time of last death between self and ancestors, and an array of dates that they had 0, 1, 2, ... surviving generations.)
Go through all people and recursively find and store the time of last death. If you call the person again, return the memoized record. For each person you can encounter the person (needing to calculate it), and can generate 2 more calls to each parent the first time you calculate it. This gives a total of O(n) work to initialize this data.
Go through all people and recursively generate a record of when they first added a generation. These records only need go to the maximum of when the person or their last ancestor died. It is O(1) to calculate when you had 0 generations. Then for each recursive call to a child you need to do O(generations) work to merge that child's data in to yours. Each person gets called when you encounter them in the data structure, and can be called once from each parent for O(n) calls and total expense O(n * (generations + 1)).
Go through all people and figure out how many generations were alive at their death. This is again O(n * (generations + 1)) if implemented with a linear scan.
The sum total of all of these operations is O(n * (generations + 1)).
For realistic data sets, this will be O(n) with a fairly small constant.
My suggestion:
additionally to the values described in the problem statement, each personal record will have two fields: child counter and a dynamically growing vector (in C++/STL sense) which will keep the earliest birthday in each generation of a person's descendants.
use a hash table to store the data, with the person name being the key. The time to build it is linear (assuming a good hash function, the map has amortized constant time for inserts and finds).
for each person, detect and save the number of children. It's also done in linear time: for each personal record, find the record for its parents and increment their counters. This step can be combined with the previous one: if a record for a parent is not found, it is created and added, while details (dates etc) will be added when found in the input.
traverse the map, and put references to all personal records with no children into a queue. Still O(N).
for each element taken out of the queue:
add the birthday of this person into descendant_birthday[0] for both parents (grow that vector if necessary). If this field is already set, change it only if the new date is earlier.
For all descendant_birthday[i] dates available in the vector of the current record, follow the same rule as above to update descendant_birthday[i+1] in parents' records.
decrement parents' child counters; if it reaches 0, add the corresponding parent's record into the queue.
the cost of this step is O(C*N), with C being the biggest value of "family depth" for the given input (i.e. the size of the longest descendant_birthday vector). For realistic data it can be capped by some reasonable constant without correctness loss (as others already pointed out), and so does not depend on N.
traverse the map one more time, and "label each person" with the biggest i for which descendant_birthday[i] is still earlier than the death date; also O(C*N).
Thus for realistic data the solution for the problem can be found in linear time. Though for contrived data like suggested in #btilly's comment, C can be big, and even of the order of N in degenerate cases. It can be resolved either by putting a cap on the vector size or by extending the algorithm with step 2 of #btilly's solution.
A hash table is key part of the solution in case if parent-child relations in the input data are provided through names (as written in the problem statement). Without hashes, it would require O(N log N) to build a relation graph. Most other suggested solutions seem to assume that the relationship graph already exists.
Create a list of people, sorted by birth_date. Create another list of people, sorted by death_date. You can travel logically through time, popping people from these lists, in order to get a list of the events as they happened.
For each Person, define an is_alive field. This'll be FALSE for everyone at first. As people are born and die, update this record accordingly.
Define another field for each person, called has_a_living_ancestor, initialized to FALSE for everyone at first. At birth, x.has_a_living_ancestor will be set to x.mother.is_alive || x.mother.has_a_living_ancestor || x.father.is_alive || x.father.has_a_living_ancestor. So, for most people (but not everyone), this will be set to TRUE at birth.
The challenge is to identify occasions when has_a_living_ancestor can be set to FALSE. Each time a person is born, we do a DFS up through the ancestors, but only those ancestors for which ancestor.has_a_living_ancestor || ancestor.is_alive is true.
During that DFS, if we find an ancestor that has no living ancestors, and is now dead, then we can set has_a_living_ancestor to FALSE. This does mean, I think, that sometimes has_a_living_ancestor will be out of date, but it will hopefully be caught quickly.
The following is an O(n log n) algorithm that work for graphs in which each child has at most one parent (EDIT: this algorithm does not extend to the two-parent case with O(n log n) performance). It is worth noting that I believe the performance can be improved to O(n log(max level label)) with extra work.
One parent case:
For each node x, in reverse topological order, create a binary search tree T_x that is strictly increasing both in date of birth and in number of generations removed from x. (T_x contains the first born child c1 in the subgraph of the ancestry graph rooted at x, along with the next earliest born child c2 in this subgraph such that c2's 'great grandparent level' is a strictly greater than that of c1, along with the next earliest born child c3 in this subgraph such that c3's level is strictly greater than that of c2, etc.) To create T_x, we merge the previously-constructed trees T_w where w is a child of x (they are previously-constructed because we are iterating in reverse topological order).
If we are careful with how we perform the merges, we can show that the total cost of such merges is O(n log n) for the entire ancestry graph. The key idea is to note that after each merge, at most one node of each level survives in the merged tree. We associate with each tree T_w a potential of h(w) log n, where h(w) is equal to the length of the longest path from w to a leaf.
When we merge the child trees T_w to create T_x, we 'destroy' all of the trees T_w, releasing all of the potential that they store for use in building the tree T_x; and we create a new tree T_x with (log n)(h(x)) potential. Thus, our goal is to spend at most O((log n)(sum_w(h(w)) - h(x) + constant)) time to create T_x from the trees T_w so that the amortized cost of the merge will be only O(log n). This can be achieved by choosing the tree T_w such that h(w) is maximal as a starting point for T_x and then modifying T_w to create T_x. After such a choice is made for T_x, we merge each of the other trees, one by one, into T_x with an algorithm that is similar to the standard algorithm for merging two binary search trees.
Essentially, the merging is accomplished by iterating over each node y in T_w, searching for y's predecessor z by birth date, and then inserting y into T_x if it is more levels removed from x than z; then, if z was inserted into T_x, we search for the node in T_x of the lowest level that is strictly greater than z's level, and splice out the intervening nodes to maintain the invariant that T_x is ordered strictly both by birth date and level. This costs O(log n) for each node in T_w, and there are at most O(h(w)) nodes in T_w, so the total cost of merging all trees is O((log n)(sum_w(h(w))), summing over all children w except for the child w' such that h(w') is maximal.
We store the level associated with each element of T_x in an auxiliary field of each node in the tree. We need this value so that we can figure out the actual level of x once we've constructed T_x. (As a technical detail, we actually store the difference of each node's level with that of its parent in T_x so that we can quickly increment the values for all nodes in the tree. This is a standard BST trick.)
That's it. We simply note that the initial potential is 0 and the final potential is positive so the sum of the amortized bounds is an upper bound on the total cost of all merges across the entire tree. We find the label of each node x once we create the BST T_x by binary searching for the latest element in T_x that was born before x died at cost O(log n).
To improve the bound to O(n log(max level label)), you can lazily merge the trees, only merging the first few elements of the tree as necessary to provide the solution for the current node. If you use a BST that exploits locality of reference, such as a splay tree, then you can achieve the above bound.
Hopefully, the above algorithm and analysis is at least clear enough to follow. Just comment if you need any clarification.
I have a hunch that obtaining for each person a mapping (generation -> date the first descendant in that generation is born) would help.
Since the dates must be strictly increasing, we would be able to use use binary search (or a neat datastructure) to find the most distant living descendant in O(log n) time.
The problem is that merging these lists (at least naively) is O(number of generations) so this could get to be O(n^2) in the worst case (consider A and B are parents of C and D, who are parents of E and F...).
I still have to work out how the best case works and try to identify the worst cases better (and see if there is a workaround for them)
We recently implemented relationship module in one of our project in which we had everything in database and yes I think algorithm was best 2nO(m) (m is max branch factor). I multiplied operations twice to N because in first round we create relationship graph and in second round we visit every Person. We have stored bidirectional relationship between every two nodes. While navigating, we only use one direction to travel. But we have two set of operations, one traverse only children, other traverse only parent.
Person{
String Name;
// all relations where
// this is FromPerson
Relation[] FromRelations;
// all relations where
// this is ToPerson
Relation[] ToRelations;
DateTime birthDate;
DateTime? deathDate;
}
Relation
{
Person FromPerson;
Person ToPerson;
RelationType Type;
}
enum RelationType
{
Father,
Son,
Daughter,
Mother
}
This kind of looks like bidirectional graph. But in this case, first you build list of all Person, and then you can build list relations and setup FromRelations and ToRelations between each node. Then all you have to do is, for every Person, you have to only navigate ToRelations of type (Son,Daughter) only. And since you have date, you can calculate everything.
I dont have time to check correctness of the code, but this will give you idea of how to do it.
void LabelPerson(Person p){
int n = GetLevelOfChildren(p, p.birthDate, p.deathDate);
// label based on n...
}
int GetLevelOfChildren(Person p, DateTime bd, DateTime? ed){
List<int> depths = new List<int>();
foreach(Relation r in p.ToRelations.Where(
x=>x.Type == Son || x.Type == Daughter))
{
Person child = r.ToPerson;
if(ed!=null && child.birthDate <= ed.Value){
depths.Add( 1 + GetLevelOfChildren( child, bd, ed));
}else
{
depths.Add( 1 + GetLevelOfChildren( child, bd, ed));
}
}
if(depths.Count==0)
return 0;
return depths.Max();
}
Here's my stab:
class Person
{
Person [] Parents;
string Name;
DateTime DOB;
DateTime DOD;
int Generations = 0;
void Increase(Datetime dob, int generations)
{
// current person is alive when caller was born
if (dob < DOD)
Generations = Math.Max(Generations, generations)
foreach (Person p in Parents)
p.Increase(dob, generations + 1);
}
void Calculate()
{
foreach (Person p in Parents)
p.Increase(DOB, 1);
}
}
// run for everyone
Person [] people = InitializeList(); // create objects from information
foreach (Person p in people)
p.Calculate();
There's a relatively straightforward O(n log n) algorithm that sweeps the events chronologically with the help of a suitable top tree.
You really shouldn't assign homework that you can't solve yourself.