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[algogeeks] Re: google question Dumanshu Sun Feb 26 07:00:13 2012

You are assuming is to be a binary tree, its not. Some nodes will
share a common "pour".

On Feb 25, 9:24 pm, atul anand <[EMAIL PROTECTED]> wrote:
> i guess this would work...
> n=number of nodes
> h=height;
> pour=quantity poured;
> capacity = capacity of each cup
>
> n=pow(2,h+1) -1;
> call(capacity,pour,0,n)
>
> node* fillCup(float capacity,float pour,int left,int right)
> {
> node *root;
> int mid;
> if(left > right)
> return NULL;
>
> root=(node *)malloc(sizeof(node));
> if(left==right)
> {
> if(pour >=capacity)
> root->data=capacity;
> else
> root->data=pour;
> root->left=root->right=NULL;}
>
> else
> {
> mid=left+(right-left)/2;
> if(pour >= capacity)
> {
> root->data=capacity;
> pour=pour-capacity;
> pour=pour/2;}
>
> else
> {
> root->data=pour;
> root->left=root->right=NULL;
> return root;
>
> }
>
> root->left=fillCup(capacity,pour,left,mid-1);
> root->right=fillCup(capacity,pour,mid+1,right);
>
> }
>
> return root;
>
> }
>
> On Sat, Feb 25, 2012 at 5:05 PM, Ravi Ranjan <[EMAIL PROTECTED]>wrote:
>
>
>
>
>
>
>
> > |_|
> > |_| |_|
> > |_| |_| |_|
> > |_| |_| |_| |_|
> > |_| |_| |_| |_| |_|
>
> > Each cup has capacity C and once a cup gets full, it drops half extra
> > amount to left child and half extra amount to right child
>
> > for Eg : let' first cups get 2C amount of liquid then extra amount C(2C-C)
> > will be divided equally to left and right child cup of next level
>
> > i.e. C/2 to left child and C/2 to right child
>
> > Write a function which takes input parameter as amount of liquid poured at
> > top (L) and height of particular cup (h) index of that cup (i) and it
> > should return amount of liquid absorbed in that cup.
>
> > source
>
> >http://www.careercup.com/question?id=12770661
>
> > whats exactly the qestion???
>
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