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Topic: Simple Question -Number of distinct BSTs (Read 4769 times) |
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mad
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Simple Question -Number of distinct BSTs
« on: Mar 7th, 2008, 10:13pm » |
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Given n nodes, derive the number of distinct binary search trees that can be constructed from it.
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towr
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Re: Simple Question -Number of distinct BSTs
« Reply #1 on: Mar 8th, 2008, 2:27am » |
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Do the n nodes have distinct values?
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mad
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Re: Simple Question -Number of distinct BSTs
« Reply #2 on: Mar 8th, 2008, 2:47am » |
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I think it would be n! if thay are distinct, right?
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towr
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Re: Simple Question -Number of distinct BSTs
« Reply #3 on: Mar 8th, 2008, 3:02am » |
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on Mar 8th, 2008, 2:47am, mad wrote:| I think it would be n! if thay are distinct, right? |
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I think it would, generally, be less. Because different orders of insertion might still yield the same tree.
e.g. 31245 or 34512 should give the same tree (1,2 goes in the left subtree in both case; and 4,5 in the right)
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mad
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Re: Simple Question -Number of distinct BSTs
« Reply #4 on: Mar 8th, 2008, 5:07pm » |
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Q)Given the root of a binary tree and any two nodes in it, device an algorithm to find out the nearest ancestor to the given two nodes. Assume the tree has only left and right child pointers and no parent pointers.
Maybe we could mark the depth of each node and the do a dfs traversal visiting all nodes and say we get the string
1012, we take as ancestor the least value lying between the two required node values.
But, can anybody tell me how to mark depth at each node and what the complexity would be?
Is there a better method?
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mad
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Re: Simple Question -Number of distinct BSTs
« Reply #5 on: Mar 8th, 2008, 5:24pm » |
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Another Question-
The structure of node of a binary tree is defined as
struct node
{
int info;
struct node *parent;
}
You are given pointers to two nodes of a tree and you have to find whether these two nodes lie on the same side of the root or not.
Achieve:
time - O(n)
space - O(1)
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towr
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Re: Simple Question -Number of distinct BSTs
« Reply #6 on: Mar 9th, 2008, 3:46am » |
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on Mar 8th, 2008, 5:24pm, mad wrote:The structure of node of a binary tree is defined as
struct node
{
int info;
struct node *parent;
} |
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The only thing known is the parent? Not even a left and right subtree?
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Grimbal
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Re: Simple Question -Number of distinct BSTs
« Reply #7 on: Mar 9th, 2008, 6:31am » |
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It is an ancestor's tree for people with a long tradition of being monoparental.
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Colonel Panic
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Re: Simple Question -Number of distinct BSTs
« Reply #8 on: Jul 10th, 2008, 6:09am » |
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on Mar 8th, 2008, 5:07pm, mad wrote:Q)Given the root of a binary tree and any two nodes in it, device an algorithm to find out the nearest ancestor to the given two nodes. Assume the tree has only left and right child pointers and no parent pointers.
Maybe we could mark the depth of each node and the do a dfs traversal visiting all nodes and say we get the string
1012, we take as ancestor the least value lying between the two required node values.
But, can anybody tell me how to mark depth at each node and what the complexity would be?
Is there a better method? |
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1. Mark each node with corresponding depth (recursive DFS lets you mark depth easily, each level of recursion would increase depth by 1)
2. Do a Euler tour (DFS with repetition) of the tree and list the nodes. In this list, the node with least depth between the 2 nodes in question is the least common ancestor.
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Colonel Panic
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Re: Simple Question -Number of distinct BSTs
« Reply #9 on: Jul 10th, 2008, 6:31am » |
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on Mar 8th, 2008, 5:24pm, mad wrote:Another Question-
The structure of node of a binary tree is defined as
struct node
{
int info;
struct node *parent;
}
You are given pointers to two nodes of a tree and you have to find whether these two nodes lie on the same side of the root or not.
Achieve:
time - O(n)
space - O(1) |
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Since we have access to parent , we could traverse back to the root and make note of root's child that is encountered on the way. Complexity would be O(log n ) in time and O(1) in space, did I miss something ?
// returns true if nodes 'a' and 'b'
// are on the same side of root
bool
func(node *p, node *q){
node *a = p;
node *b = q;
// when the loop terminates, ((a->parent)->parent == NULL)
// we point to the child of root
while ((a->parent)->parent)
a=a->parent;
}
while ((b->parent)->parent)
b=b->parent;
}
// on our way back to root if we meet the same child of root
// then we must be on the same side of the root.
return (a == b)
}
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towr
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Re: Simple Question -Number of distinct BSTs
« Reply #10 on: Aug 7th, 2008, 12:38am » |
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on Aug 6th, 2008, 9:23pm, acceleracer wrote:Given n nodes, derive the number of distinct binary search trees that can be constructed from it.
it is 2^(n-1) |
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Try it for n=3. I get 5 distinct BSTs, not 4
(. 1 ( . 2 (3))
(. 1 ( (2) 3 .)
( (1) 2 (3) )
( (1) 2 . ) 3 .)
( (. 1 (2) ) 3 .)
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acceleracer
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Re: Simple Question -Number of distinct BSTs
« Reply #11 on: Aug 7th, 2008, 8:10am » |
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sorry it is ((2^n)-n)
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SMQ
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Re: Simple Question -Number of distinct BSTs
« Reply #12 on: Aug 7th, 2008, 8:33am » |
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on Aug 7th, 2008, 8:10am, acceleracer wrote:
Sorry, still no.
(. 1 (. 2 (. 3 (4) )
(. 1 (. 2 ( (3) 4 .)
(. 1 ( (2) 3 (4) )
(. 1 ( (2) 3 .) 4 .)
(. 1 ( (. 1 (3) ) 4 .)
( ( (1) 2 .) 3 (4) )
( (. 1 (2) ) 3 (4) )
( (1) 2 ( (3) 4 .) )
( (1) 2 (. 3 (4) ) )
( (. 1 (. 2 (3) ) 4 .)
( (. 1 ( (2) 3 .) 4 .)
( ( (1) 2 (3) ) 4 .)
( ( (1) 2 .) 3 .) 4 .)
( ( (. 1 (2) ) 3 .) 4 .)
--SMQ
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| « Last Edit: Aug 7th, 2008, 8:35am by SMQ » |
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--SMQ
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ankura
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Re: Simple Question -Number of distinct BSTs
« Reply #13 on: Aug 31st, 2008, 1:09pm » |
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i think answer is 2n C n .
catelan number...
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javeed.shariff
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Re: Simple Question -Number of distinct BSTs
« Reply #14 on: Sep 16th, 2008, 10:27pm » |
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Hint: how many x digits number can be formed using y numbers w/o repitations
Answer: yPx
Similarly considering at all levels we derive the following:
r=n
-------
> nP(r+1)
-------
r = 0
ex: for n=3,
we can have 3P1 + 3P3 = 3+6 = 9 BST
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towr
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Re: Simple Question -Number of distinct BSTs
« Reply #15 on: Sep 17th, 2008, 2:05am » |
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on Sep 16th, 2008, 10:27pm, javeed.shariff wrote:Hint: how many x digits number can be formed using y numbers w/o repitations
Answer: yPx |
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What is P?
Quote:ex: for n=3,
we can have 3P1 + 3P3 = 3+6 = 9 BST |
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We only get 5 distinct BSTs though.
I gave them all a few posts up.
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SMQ
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Re: Simple Question -Number of distinct BSTs
« Reply #16 on: Sep 17th, 2008, 5:10am » |
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on Sep 17th, 2008, 2:05am, towr wrote:
At least as I've encountered it (and Mathworld agrees), nPk = k! nCk = n!/(n-k)! is the number of distinct permutations of k elements chosen from a set of n distinct elements.
--SMQ
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kumariitr
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Re: Simple Question -Number of distinct BSTs
« Reply #17 on: Nov 27th, 2008, 10:08am » |
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let x(n) denote the number of distinct BSTs for n elements (assumed that elements are distinct)
then for even n, x(n) = 2 [x(n-1)*x(0) + x(n-2)*x(1) +x(n-3)*x(2)+...upto (n-1)/2 terms] + [x((n-1)/2)]^2
and for odd n, x(n) = 2 [x(n-1)*x(0) + x(n-2)*x(1) +x(n-3)*x(2)+...upto n/2 terms]
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towr
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Re: Simple Question -Number of distinct BSTs
« Reply #18 on: Nov 27th, 2008, 10:14am » |
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But x(3)=5, which isn't even. And according to your formula it should be (considering the factor of 2 in the front)
You seem to have switched which is odd and which is even 
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| « Last Edit: Nov 27th, 2008, 10:22am by towr » |
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kumariitr
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Re: Simple Question -Number of distinct BSTs
« Reply #19 on: Nov 28th, 2008, 8:41am » |
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correct series representing the solution is:
let x(n) denote the number of distinct BSTs for n elements (assumed that elements are distinct)
then for odd n>1, x(n) = 2 [x(n-1)*x(0) + x(n-2)*x(1) +x(n-3)*x(2)+...upto (n-1)/2 terms] + [x((n-1)/2)]^2
and for even n, x(n) = 2 [x(n-1)*x(0) + x(n-2)*x(1) +x(n-3)*x(2)+...upto n/2 terms]
and x(1) = 1 and taking x(0) = 1.
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