Skip to main content

Median of stream of numbers

Problem: Design a data structure which can perform most efficiently insert and median.


My Solution:
struct node
{
int number;
node *left;
node *right;
int leftTreeNode;
};

struct header
{
int totalNodes;
Int valid;
int median;
node *root;
} myhead ; //myhead ......thats where it will start.

void initialize() { valid = 0, median = 0, root =NULL, totalNodes =0; }
void AddNode ( int n )
{
myhead.totalNodes++;
myhead.valid = 0;
node *prev = NULL;
node *temp = myhead->root;
while ( temp != NULL )
{
if ( n <>number )
{
prev = temp;
temp->leftTreeNodes +=1;
temp = temp->left;
}
else
{
prev = temp;
temp = temp->right;
}
}
if ( prev != NULL )
{
if ( n <>number )
prev->left = createNode(n)
else
prev->right = createNode(n);
}
else
myhead->root = createNode(n);
}

int median ()
{
if (myhead.valid ) return median;
int mid = myhead.totalNodes/2;
node *prev = NULL;
node *temp = myhead.root;
while ( mid )
{
if ( mid > temp->leftTreeNodes )
{
mid = mid - (temp->leftTreeNodes)-1;//1 for root node
prev = temp;
temp = temp->right;
}
else
{
prev = temp;
temp = temp->left;
}
}
myhead.median = prev->number;
myhead.valid = 1;
return myhead.median;
}

Comments

Popular posts from this blog

Car Parking Problem

There is n parking slots and n-1 car already parked. Lets say car parked with initial arrangement and we want to make the car to be parked to some other arrangement. Lets say n = 5, inital = free, 3, 4, 1, 2 desired = 1, free, 2, 4 ,3 Give an algorithm with minimum steps needed to get desired arrangement. Told by one of my friend and after a lot of search i really got a nice solution. I will post solution in comment part

DEShaw Interview Questions

ther are N numbers frm 1 to N and starting from index 1 we will keep deleting every alternate going in cyclic order with array. Only one element will be left at the end. Tell us the index of element in array we started. e.g. there are 5 nums 1 2 3 4 5 then after 1st iteration 1 3 5 will be remained. .. then 1 will be next to be elliminated and then 5 3 will remain alone... give sum efficient algorithm to calculate which numer will remain at the end Answer: 2*(n-2^p)+1 where p=floor(log2 n)

Permutations Sum(xi)

You have given "k" dice. How many way you can get a sum "S" and yes you have to throw all the dice. Write program for this. Its same permutations program...but we have to try with all the six S(1,2,3,4,5,6) possibilities for a dice. Exit condition will be If all the dice run out. SumP(dice,sum) = SumP(dice-1,sum-i)+i (from S).