The code below returns the kth to last element of a linked list. But I don't understand how this is, because doesn't 'for I in range(k)' return values from 0 to k?? As opposed to k to the last element?
Also in general, can someone please explain the traversal of the below, as I don't quite understand it- I do understand the fundamentals of linked list and the syntax but can't quite follow the code below. e.g. why is there a need for a runner (I'm assuming this is some sort of pointer?)
def kth_to_last(l1,k):
runner = current = l1.head
for i in range(k):
if runner is None:
return None
runner = runner.next
while runner:
current = current.next
runner = runner.next
return current
e.g. input is a linked list: a -> b -> c-> d-> None
if k is b then
output: b -> c -> d -> None
in linked list, each node only knows the next node (some times the previous one too). in that case, you cannot just "jump" k nodes forward. you need to move 1 node forward k times.
in this case, runner is the pointer for the 'current' node. its often also named "current", but since this one 'runs' over the list and doesn't really care much for its content, they named it "runner".
let say runner holds the rout, i.e. node #0
runner = runner.next
the node next to node#0 is node#1, so now runner is node#1
runner = runner.next
and now runner is node#2 and so forth. each iteration of runner = runner.next forward you 1 node.
so in order to move from node#0 to node#27, you need to write runner = runner.next 27 times, or use a loop for it.
(note: the "if" part of the loop is mostly to avoid exceptions, since you cant get the "next" of null)
a simpler way to make\understand this function is by finding the length of the list, decreasing it by k and than moving forward that many nodes.
I have a question about coding up a binary tree, and I'm currently stuck on it.
Basically, the algorithm is as follows where at the initial node, we have the following objects: 1,2,3 placed in order as (None,2,3) with object 3 > object 2 > object 1. In each stage of the tree, one object can move to their immediate right or left and be placed with another object if that object that is moving is smaller than the object that is placed at the current position. If the position to the immediate right or left has a None position, then only the smallest number may move over from the right or left position. For example, it is possible to have the events of (None,0,2+3) and (None,1+2,3) after the initial node.
Hence, the tree is as follows:
I am wondering how to code this up in Python. I dont really care about the replicating events so at each point, I am only interested in the unique tuples/events that occurs.
Im not sure but a rough idea I have is to do this:
def recursion():
pos = list(range(1,4))
index = 0
values = list(range(1,4))
tree = []
if pos[index+1] > pos[index]:
pos[index + 1] -= pos[index+1]
pos[index + 2] += pos[index+1]
tree.append(pos)
if pos[index+1] < pos[index]:
pos[index+1] += pos[index]
pos[index] -= pos[index]
tree.append(pos)
else:
recursion()
return tree
Any help would be greatly appreciated.
#Get length of the longest path through recursion
def max_height(node):
if not node:
return 0
left = max_height(node.left) #Base Case based on my understanding
right = max_height(node.right) #Base Case based on my understanding
return max_height(left, right) + 1
I keep calling the max_height to get the length but I'm getting an error. I've thought of three possibilities:
1) I misunderstand the concept of the base case and I don't actually have a base case.
2) I'm not properly spacing Python code.
3) I'm not recursively getting the height of the BST at all but rather the width of the tree, which is affecting later calculations.
I know it is similar to this question, but the main difference is that I'm really trying to use recursion , where the other question used iteration and merely called it recursion.
how to find the height of a node in binary tree recursively
The base case is where the recursion stops and you have one: not node (node == None)
I don't see an issue with the spacing... Make sure you use only tabs or only spaces
This does produce the height: the number of nodes from root to leaf along the longest root-leaf path. At every node level, you add 1, and follow the higher subtree.
def max_height(node):
if not node: # this is the base case:
return 0 # where all recursive calls eventually stop
left = max_height(node.left) # <- these are the recursive calls:
right = max_height(node.right) # <- function x is called inside function x
return max(left, right) + 1 # max here, not max_height
Note that this is merely a more verbose version of this answer to the question you linked.
All answered were right but, I faced little problem while writing inside the class;
So, the code goes like this, I hope this helps.
class Tree(object):
def height(self, root):
if root == None: #root == None
return 0
else:
return 1 + max(self.height(root->left), self.height(root->left))
t = Tree()
t.height(t)
I am trying to write a function to count the repetitive node in a Binary Search Tree (assume that this tree accept repetition). Here is a line of code that I have
def count(treeroot,item,y):
if treeroot==None:
return 0
elif treeroot.getData()==item:
return 1
else:
return count(treeroot.getLeft(),item,y)+count(treeroot.getRight(),item,y)
where y is the starting number of the search (such as search for how many 10 in the tree, we do count(treeroot,10,0)
However, I tried to put 3 number 10 in and I only receive back the count of 1.
Can anyone tell me what is wrong with my code and how to fix it
You stop recursing down the tree as soon as you found the first item. You need to keep recursing:
def count(treeroot,item,y):
if treeroot==None:
return 0
elif treeroot.getData()==item:
return 1 + count(treeroot.getLeft(),item,y)+count(treeroot.getRight(),item,y)
else:
return count(treeroot.getLeft(),item,y)+count(treeroot.getRight(),item,y)
I hope you see the problem with your algorithm though: You will visit each and every node of the tree, even if you know that there will be no 10s to the left of 9s or to the right of 11s.
Okay, sorry if my problem seems a bit rough. I'll try to explain it in a figurative way, I hope this is satisfactory.
10 children. 5 boxes. Each child chooses three boxes. Each box is opened:
- If it contains something, all children selected this box gets 1 point
- Otherwise, nobody gets a point.
My question is about what I put in bold. Because in my code, there are lots of kids and lots of boxes.
Currently, I proceed as follows:
children = {"child_1" : 0, ... , "child_10": 0}
gp1 = ["child_3", "child_7", "child_10"] #children who selected the box 1
...
gp5 = ["child_2", "child_5", "child_8", "child_10"]
boxes = [(0,gp1), (0,gp2), (1,gp3), (1,gp4), (0,gp5)]
for box in boxes:
if box[0] == 1: #something inside
for child in box[1]:
children[child] += 1
I worry mainly about the for loop that assigns each child an extra point. Because in my final code, I have many many children, I fear that doing so would slow the program too.
Is there a more efficient way for all children of the same group may have their point faster?
Represent children as indices into arrays, not as strings:
childrenScores = [0] * 10
gp1 = [2,6,9] # children who selected box 1
...
gp5 = [1,4,7,9]
boxes = [(0,gp1), (0,gp2), (1,gp3), (1,gp4), (0,gp5)]
Then, you can store childrenScores as a NumPy array and use advanced indexing:
childrenScores = np.zeros(10, dtype=int)
...
for box in boxes:
if box[0]:
childrenScores[box[1]] += 1 # NumPy advanced indexing
This still involves a loop somewhere, but the loop is deep inside NumPy instead, which should provide a meaningful speedup.
The only speed up that I can think of is to use numpy arrays and stream the sum operation.
children[child] += np.ones(len(children[child]))
You should benchmark the operation and see if that is too slow for your business case.
What I would do
In the gpX lists don't save the "name of the child" (e.g. "child_10") but save a reference to the child's number of points.
How to do that
Using the fact that lists are objects in python, you can:
Change the children dict to look like: children = {"child_0": [0], "child_1": [0], ...} and so on.
When you assign to group, don't assign the key but assign the value (e.g. gp1.append(children["child_0"])).
The loop should then look like: for child in box[1]: child[0]+=1. This WILL update the children dict.
EDIT:
Why this is faster:
Because you leave out the part where you search for children[child], which might be costly.
This technique works because by storing the totals in a mutable type, and appending those values to the group lists, both the dict value and each box's list value will point to the same list entries, and changing one will change the other.
Two general points:
(1) Based on what you've told us, there's no reason to focus your energy on minor performance optimizations. Your time would be better spent thinking about ways to make your data structures less awkward and more communicative. A bunch of interrelated dicts, lists, and tuples quickly becomes difficult to maintain. For an alternative, see the example below.
(2) As the game designer, you understand that events follow a certain sequence: first the kids select their boxes, and later they discover whether they get points for them. But you don't have to implement it that way. A kid can choose a box and get points (or not) immediately. If there's a need to preserve the child's ignorance about such outcomes, the parts of your algorithm that depend on such ignorance can enforce that veil of secrecy as needed. The upshot: there is no need for a box to loop through its children, awarding points to each one; instead, award the points immediately to kids as boxes are selected.
import random
class Box(object):
def __init__(self, name):
self.name = name
self.prize = random.randint(0,1)
class Child(object):
def __init__(self, name):
self.name = name
self.boxes = []
self.score = 0
self._score = 0
def choose(self, n, boxes):
bs = random.sample(boxes, n)
for b in bs:
self.boxes.append(b)
self._score += b.prize
def reveal_score(self):
self.score = self._score
boxes = [Box(i) for i in range(5)]
kids = [Child(i) for i in range(10)]
for k in kids:
k.choose(3, boxes)
# Later in the game ...
for k in kids:
k.reveal_score()
print (k.name, k.score), '=>', [(b.name, b.prize) for b in k.boxes]
One way or another, you're going to be looping over the children, and your answer appears to avoid looping over children who don't get any points.
It might be slightly faster to use filter or itertools.ifilter to pick the boxes that have something in them:
import itertools
...
for box in itertools.ifilter(lambda x: x[0], boxes):
for child in box[1]
children[child] += 1
If you don't need to immediately print the number of points for every child, you could calculate it on demand, thus saving time. This could help if you only need to query a child every now and then for how many points it has. You can cache each result as you obtain is so you don't go about calculating it again the next time you need it.
Firstly, you'll need to know which groups a child belongs to. We'll store this information as map we'll call childToGroupsMap, which will map each child to an array holding the boxes it belongs to, like so:
childToGroupsMap = {}
for child in children:
childToGroupsMap[child[0]] = []
for box in boxes:
for child in box[1]:
if (box[1] not in childToGroupsMap[child]):
childToGroupsMap[child].append(box[1])
This constructs the reverse map from children to boxes.
It'll also help to have a map from each box to a boolean representing whether it's been opened:
boxToOpenedMap = {}
for box in boxes:
boxToOpenedMap[box[1]] = box[0]
Now, when someone queries how many points a child has, you can go through each of the boxes it belongs to (using childToGroupsMap, of course), and simply count how many of those boxes have been mapped to 1 in the map boxes:
def countBoxesForChild(child):
points = 0
for box in childToGroupsMap[child]
if boxToOpenedMap[box] == 1:
points += 1
return points
To make this better, you can cache the resulting number of points. Make a map like so:
childToPointsCalculated = {}
for child in children:
childToPointsCalculated[child[0]] = -1
Where the -1 denotes that we don't yet know how many points this child has.
Finally, you can modify your countBoxesForChild function to exploit the cache:
def countBoxesForChild(child):
if childToPointsCalculated[child] != -1
return childToPointsCalculated[child]
points = 0
for box in childToGroupsMap[child]
if boxToOpenedMap[box] == 1:
points += 1
childToPointsCalculated[child] = points
return points