# Find median of stream of integers

Given continuous stream of integers, find **median of all integers** received till a given point of time. Median can be asked at multiple times.

For example:

Input Median

12 12

7 9.5

8 8

11 9.5

And so…

Median of array of integers is middle of those integers if number of elements are odd and average of middle elements if number of elements is even.

Let’s consider array to store integers as they are received. Array will be unsorted. Finding median will take O(N), insertion as O(1).

How about sorted array? Insertion will take O(N) as we need to shift elements once we identify place for new integer. Median can be found O(1)

Linked list will have same complexity as unsorted array.

So let’s consider some advance data structure. Here, we need some kind of order of elements. We can divide integers in two groups: all the elements which are less than median, let it be group 1 and all elements which are greater, be it group 2.

So greatest element on group 1 should be less than group 2. We have to find the greatest on one side and smallest on other side. Which data structure gives us that kind of information in constant time.

Maximum can be found using max heap while minimum using min heap in constant time.

So it is evident that we need two heaps. One max heap and other min heap.

Max heap contain all elements which are less than current median and min heap which are greater than current median.

Now how we can insert an integer and what should be properties they should have?

Since median is middle of all integers if number for elements are odd, we need to keep size difference of two heaps either equal 0 or 1.

If size difference increases, we need to adjust elements in two heaps.

If size of max heap is 2 more than min heap, extract maximum element from max heap and put it in min heap.

If size of min heap is 2 more than min heap, extract minimum element from max heap and put it in max heap.

## Median of stream of numbers

If number of elements in both heaps is equal, take average of max and minimum elements in both heaps and that would be median.

If sizes of two heaps different, then maximum element of max heap will be the median.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAX_SIZE 100
#define true 1
#define false 0
// Data structure to store heap array as well as current count.
typedef struct heap_t {
int count;
int a[MAX_SIZE];
}heap;
void add_integer(heap *min_heap, heap * max_heap, int value);
void print_median(heap min_heap, heap max_heap);
/* To find children of a given element */
int left_child(int i){
return (i*2);
}
int right_child(int i){
return (2*i)+ 1;
}
int swap(int a[], int i, int j){
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
/* This function heapifies max heap after deleting maximum
element from it */
void heapify(int a[], int i, int len){
int largest = i;
int left, right;
left = left_child(i);
right = right_child(i);
if(left <= len && a[i]>a[left]){
largest = left;
}
if(right <= len && a[largest] > a[right]){
largest = right;
}
if(largest != i){
swap(a, i, largest);
heapify(a, largest, len);
}
}
/* This function heapifies min heap after deleting maximum
element from it */
void min_heapify(int a[], int i, int len){
int smallest = i;
int left, right;
left = left_child(i);
right = right_child(i);
if(left <= len && a[i]>a[left]){
smallest = left;
}
if(right <= len && a[smallest] > a[right]){
smallest = right;
}
if(smallest != i){
swap(a, i, smallest);
min_heapify(a, smallest, len);
}
}
/* This function heapifies min heap after inserting an
element from it */
void shift_up_min(int heap[], int i){
int parent = i/2;
if(i > 1){
if(heap[parent] > heap[i]){
swap(heap, parent, i);
shift_up_min(heap, parent);
}
}
}
/* This function inserts an element in min heap */
void insert_min(heap *min_heap, int val){
if((min_heap->count) >= MAX_SIZE){
printf("\n Underlying data structure gone full\n");
return;
}
else{
min_heap->count++;
min_heap->a[min_heap->count] = val;
shift_up_min(min_heap->a, min_heap->count);
}
return;
}
/* This function heapifies max heap after inserting an
element from it */
void shift_up_max(int heap[], int i){
int parent = i/2;
if(i > 1){
if(heap[parent] < heap[i]){
swap(heap, parent, i);
shift_up_max(heap, parent);
}
}
}
/* This function inserts an element in min heap */
void insert_max(heap *max_heap, int val){
if((max_heap->count) >= MAX_SIZE){
printf("\n Underlying data structure gone full\n");
return;
}
else{
max_heap->count++;
max_heap->a[max_heap->count] = val;
shift_up_max(max_heap->a, max_heap->count);
}
return;
}
/* This function deletes an element in max heap */
void delete_max(heap *heap){
if(heap->count == 0 ){
printf("\n Cannot be deleted");
return;
}
heap->a[1] = heap->a[heap->count--];
heapify(heap->a, 1,heap->count);
return;
}
/* This function deletes an element in min heap */
void delete_min(heap *heap){
if(heap->count == 0 ){
printf("\n Cannot be deleted");
return;
}
heap->a[1] = heap->a[heap->count--];
printf("\n Minimum heap size : %d", heap->count);
min_heapify(heap->a,1, heap->count);
return;
}
/*
1. Add to max heap.
2. Check if max_heap.count- min_heap.count >=2
remove top of max heap and add it to min_heap
*/
void add_integer(heap *min_heap, heap * max_heap, int value){
// Insert in max heap
insert_max(max_heap,value);
/* If difference between two heap size is more than 1
or the maximum element in more than least
element in min heap then remove from max heap
and add to min heap*/
if((max_heap->count - min_heap->count) >1 ||
(min_heap->count != 0
&& max_heap->a[1]>min_heap->a[1])){
int top = max_heap->a[1];
delete_max(max_heap);
insert_min(min_heap,top);
}
/* If difference between two heap size is more than 1 */
if((min_heap->count !=0
&& (min_heap->count - max_heap->count) >1)){
int top = min_heap->a[1];
delete_min(min_heap);
insert_max(max_heap,top);
}
}
void print_median(heap min_heap, heap max_heap){
if(max_heap.count == min_heap.count){
printf("\n Median %d",
(max_heap.a[1] + min_heap.a[1])/2 );
}
else{
if(max_heap.count > min_heap.count){
printf("\n Median : %d ", max_heap.a[1]);
}
else{
printf("\n Median : %d ", min_heap.a[1]);
}
}
}
void print_menu(){
printf("\n1. Add integer");
printf("\n2. Median");
printf("\n3. Exit");
}
void main(){
heap min_heap, max_heap;
min_heap.count =0;
max_heap.count =0;
int choice,n,i;
do {
print_menu();
scanf("%d", &choice);
switch(choice){
case 1 :
printf("\n Enter integer : ");
scanf("%d", &n);
add_integer(&min_heap,
&max_heap,n );
printf("\n Max heap : ");
break;
case 2 :
print_median(min_heap,
max_heap);
break;
case 3:
break;
}
}while(choice != 3);
}

Complexity of finding **median of stream of numbers **is O(1) and insertion will be O(log N).

Please comment if you could not understand or something is not correct in post.