Answer / hitesh viradiya

m = 1 2 3 4 5 6 7 8
n=1 1
2 1 1
3 2 2 1
4 5 5 3 1
5 14 14 9 4 1
6 42 42 28 14 5 1
7 132 132 90 48 20 6 1
8 429 429 297 165 75 27 7 1

(2n)!/[(n+1)!n!]

Answer / kathir

There are 2 pointers available for each node.
So we can have 2*n pointers totally.

Total no. of edges = n-1

So, Null pointers = n+1.

We need to choose (n-1) pointers from 2n pointers.

So the combination results in (2n)C(n-1).

We can have n distinct roots possible.

So, answer will be (2n) C (n-1) / n.

which leads to,

{2n C n}/{n+1}. ( Unlabelled )


Labelled Structured tree will be,

{2n C n}/{n+1} * {n!}

Answer / durga prasad sahoo

No. of labeled binary tree :

n^(n-2)

No. unlabeled binary tree :

(2n)!/[(n+1)!.n!] (this is known as catlon number)

Answer / atul barve

No. of labeled binary tree :

{(2n)!/[(n+1)!.n!]}*n!

No. unlabeled binary tree :

(2n)!/[(n+1)!.n!] (this is known as catlon number)

Answer / neha shah

(2n)!/(n+1)!*n!

Answer / fawad ghafoor

The number of different trees with 8 nodes is 248
2^n - n

Answer / ajeet

int countTrees(int num)
{
if(num<=1)
return 1;
else
{
int root,left,right,sum=0;
for(root=1;root<=num;root++)
{
left=countTrees(root-1);
right=countTrees(num-root);
sum+=left*right;
}
return sum;
}
}

Answer / munyazikwiye pierre celestin

(2n)!/[(n+1)!*n!]

Answer / sayandip ghosh

The max number of binary trees that can be formed from n
nodes is given by the Catlan Number C(n).

C(n) = (2n)! / (n+1)!*n! for n>=0.

int findNoTree(int low ,int high)
{
int sum=0;
if(low<=high)
{
for(int k=low;k<=high;k++)
{
if(k==low)
sum+=findNoTree(low+1,high);
else
{
if(k==high)
sum+=findNoTree(low,high-1);
else

sum=sum+findNoTree(low,k-1)*findNoTree(k+1,high);
}
}

return sum;
}
return 1;
}

Answer / sachin

(2n)!/(n+1)!

Given an array of size N in which every number is between 1 and N, determine if there are any duplicates in it. You are allowed to destroy the array if you like.

21 Answers   ABC, eBay, Goldman Sachs, Google, HUP, Microsoft, TATA,

{ int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }

4 Answers  

Given an array of characters which form a sentence of words, give an efficient algorithm to reverse the order of the words (not characters) in it.

2 Answers   Wipro,

Program to find the largest sum of contiguous integers in the array. O(n)

11 Answers  

print numbers till we want without using loops or condition statements like specifically(for,do while, while swiches, if etc)!

11 Answers   Wipro,

void main() { int i=i++,j=j++,k=k++; printf(“%d%d%d”,i,j,k); }

1 Answers  

Derive expression for converting RGB color parameters to HSV values

1 Answers  

main() { char *p; p="Hello"; printf("%c\n",*&*p); }

1 Answers  

what is the output of following program ? void main() { int i=5; printf("%d %d %d %d %d ",i++,i--,++i,--i,i); }

10 Answers  

main() { float f=5,g=10; enum{i=10,j=20,k=50}; printf("%d\n",++k); printf("%f\n",f<<2); printf("%lf\n",f%g); printf("%lf\n",fmod(f,g)); }

1 Answers  

void main() { int x,y=2,z; z=(z*=2)+(x=y=z); printf("%d",z); }

4 Answers  

main() { int c[ ]={2.8,3.4,4,6.7,5}; int j,*p=c,*q=c; for(j=0;j<5;j++) { printf(" %d ",*c); ++q; } for(j=0;j<5;j++){ printf(" %d ",*p); ++p; } }

1 Answers