Anil Kumar
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AnswerGiven two arrays. One is for tasks (Processes) and each element depicts the amount of cores required to run the task. 2nd array is an array of CPU where each element depicts the no of cores in it.We have to tell how many maximum number of tasks can be allocated. Example: Task: [3, 5, 7], Cores: [1, 3, 5] . Here only task 0 and 1 can be allocated to CPU 1 and 2 . So, answer=2.
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#include <stdio.h>
- Anil Kumar March 07, 2014#include <string.h>
struct tree *newnode(int data);
void push(int **data);
struct list
{
int data;
struct list *next;
}*top, *currentlevel=NULL,*nextlevel=NULL;
struct tree
{
int data;
struct tree *left;
struct tree *right;
};
main()
{
struct tree *tree1=newnode(1);
tree1->left=newnode(2);
tree1->right=newnode(3);
tree1->left->left=newnode(4);
tree1->left->right=newnode(5);
tree1->right->left=newnode(6);
tree1->right->right=newnode(7);
tree1->left->left->left=newnode(8);
tree1->left->left->right=newnode(9);
tree1->left->right->left=newnode(10);
tree1->left->right->right=newnode(11);
tree1->right->left->left=newnode(12);
tree1->right->left->right=newnode(13);
tree1->right->right->left=newnode(14);
tree1->right->right->right=newnode(15);
zigzag(tree1);
}
int zigzag(struct list *tree1)
{
int lefttoright=1;
push(tree1);
while(!isEmpty())
{
struct tree *tree3=pop();
printf(" %d",tree3->data);
if(lefttoright)
{
if(tree3->left!=0)
{
push1(tree3->left);
}
if(tree3->right!=0)
{
push1(tree3->right);
}
}
else
{
if(tree3->right!=0)
{
push1(tree3->right);
}
if(tree3->left!=0)
{
push1(tree3->left);
}
}
while(isEmpty())
{
lefttoright=1-lefttoright;
swap();
}
}
}
void swap()
{
currentlevel=nextlevel;
nextlevel=NULL;
}
int pop()
{
struct list *ptr;
ptr=currentlevel->data;
if(currentlevel==NULL)
{
printf(" overflow ");
}
currentlevel=currentlevel->next;
return ptr;
}
int isEmpty()
{
if(currentlevel==NULL)
{
return 1;
}
else
{
return 0;
}
}
struct tree *newnode(int data)
{
struct tree *tree3;
tree3=(struct tree *)malloc(sizeof(struct tree));
tree3->data=data;
tree3->left=NULL;
tree3->right=NULL;
return tree3;
}
void push(int **data)
{
struct list *p1=(struct list *)malloc(sizeof(struct list));
p1->data=data;
if(currentlevel==NULL)
{
p1->next=NULL;
currentlevel=p1;
}
else
{
p1->next=currentlevel;
currentlevel=p1;
}
}
void push1(int **data)
{
struct list *p11=(struct list *)malloc(sizeof(struct list));
p11->data=data;
if(nextlevel==NULL)
{
p11->next=NULL;
nextlevel=p11;
}
else
{
p11->next=nextlevel;
nextlevel=p11;
}
}