[24] find median in a stream of running integers.
find median in a stream of running integers.
Median in a stream of integers (running integers)
Given that integers are read from a data stream. Find median of elements read so for in efficient way. For simplicity assume there are no duplicates. For example, let us consider the stream 5, 15, 1, 3 …
After reading 1st element of stream - 5 -> median - 5
After reading 2nd element of stream - 5, 15 -> median - 10
After reading 3rd element of stream - 5, 15, 1 -> median - 5
After reading 4th element of stream - 5, 15, 1, 3 -> median - 4, so on...
Making it clear, when the input size is odd, we take the middle element of sorted data. If the input size is even, we pick average of middle two elements in sorted stream.
Note that output is effective median of integers read from the stream so far. Such an algorithm is called online algorithm. Any algorithm that can guarantee output of i-elements after processing i-th element, is said to be online algorithm. Let us discuss three solutions for the above problem.
Method 1: Insertion Sort
If we can sort the data as it appears, we can easily locate median element. Insertion Sort is one such online algorithm that sorts the data appeared so far. At any instance of sorting, say after sorting i-th element, the first i elements of array are sorted. The insertion sort doesn’t depend on future data to sort data input till that point. In other words, insertion sort considers data sorted so far while inserting next element. This is the key part of insertion sort that makes it an online algorithm.
However, insertion sort takes O(n2) time to sort n elements. Perhaps we can use binary search on insertion sort to find location of next element in O(log n) time. Yet, we can’t do data movement in O(log n) time. No matter how efficient the implementation is, it takes polynomial time in case of insertion sort.
Interested reader can try implementation of Method 1.
Method 2: Augmented self balanced binary search tree (AVL, RB, etc…)
At every node of BST, maintain number of elements in the subtree rooted at that node. We can use a node as root of simple binary tree, whose left child is self balancing BST with elements less than root and right child is self balancing BST with elements greater than root. The root element always holds effective median.
If left and right subtrees contain same number of elements, root node holds average of left and right subtree root data. Otherwise, root contains same data as the root of subtree which is having more elements. After processing an incoming element, the left and right subtrees (BST) are differed utmost by 1.
Self balancing BST is costly in managing balancing factor of BST. However, they provide sorted data which we don’t need. We need median only. The next method make use of Heaps to trace median.
Method 3: Heaps
Similar to balancing BST in Method 2 above, we can use a max heap on left side to represent elements that are less than effective median, and a min heap on right side to represent elements that are greater than effective median.
After processing an incoming element, the number of elements in heaps differ utmost by 1 element. When both heaps contain same number of elements, we pick average of heaps root data as effective median. When the heaps are not balanced, we select effective median from the root of heap containing more elements.
Given below is implementation of above method. For algorithm to build these heaps, please read the highlighted code.
Example:
Input: 5 10 15
Output: 5
7.5
10
Explanation: Given the input stream as an array of integers [5,10,15]. We will now read integers one by one and print the median correspondingly. So, after reading first element 5,median is 5. After reading 10,median is 7.5 After reading 15 ,median is 10.
The idea is to use max heap and min heap to store the elements of higher half and lower half. Max heap and min heap can be implemented using priority_queue in C++ STL. Below is the step by step algorithm to solve this problem.
Algorithm:
Create two heaps. One max heap to maintain elements of lower half and one min heap to maintain elements of higher half at any point of time..
Take initial value of median as 0.
For every newly read element, insert it into either max heap or min heap and calculate the median based on the following conditions:
If the size of max heap is greater than size of min heap and the element is less than previous median then pop the top element from max heap and insert into min heap and insert the new element to max heap else insert the new element to min heap. Calculate the new median as average of top of elements of both max and min heap.
If the size of max heap is less than size of min heap and the element is greater than previous median then pop the top element from min heap and insert into max heap and insert the new element to min heap else insert the new element to max heap. Calculate the new median as average of top of elements of both max and min heap.
If the size of both heaps are same. Then check if current is less than previous median or not. If the current element is less than previous median then insert it to max heap and new median will be equal to top element of max heap. If the current element is greater than previous median then insert it to min heap and new median will be equal to top element of min heap.
Below is the implementation of above approach:
filter_none
// C++ program to find med in
// stream of running integers
#include<bits/stdc++.h>
using namespace std;
// function to calculate med of stream
void printMedians(double arr[], int n)
{
// max heap to store the smaller half elements
priority_queue<double> s;
// min heap to store the greater half elements
priority_queue<double,vector<double>,greater<double> > g;
double med = arr[0];
s.push(arr[0]);
cout << med << endl;
// reading elements of stream one by one
/* At any time we try to make heaps balanced and
their sizes differ by at-most 1. If heaps are
balanced,then we declare median as average of
min_heap_right.top() and max_heap_left.top()
If heaps are unbalanced,then median is defined
as the top element of heap of larger size */
for (int i=1; i < n; i++)
{
double x = arr[i];
// case1(left side heap has more elements)
if (s.size() > g.size())
{
if (x < med)
{
g.push(s.top());
s.pop();
s.push(x);
}
else
g.push(x);
med = (s.top() + g.top())/2.0;
}
// case2(both heaps are balanced)
else if (s.size()==g.size())
{
if (x < med)
{
s.push(x);
med = (double)s.top();
}
else
{
g.push(x);
med = (double)g.top();
}
}
// case3(right side heap has more elements)
else
{
if (x > med)
{
s.push(g.top());
g.pop();
g.push(x);
}
else
s.push(x);
med = (s.top() + g.top())/2.0;
}
cout << med << endl;
}
}
// Driver program to test above functions
int main()
{
// stream of integers
double arr[] = {5, 15, 10, 20, 3};
int n = sizeof(arr)/sizeof(arr[0]);
printMedians(arr, n);
return 0;
}
Output:
5
10
10
12.5
10
Time Complexity: O(n Log n)
Auxiliary Space : O(n)
// Java program to find med in
// stream of running integers
import java.util.Collections;
import java.util.PriorityQueue;
public class MedianMaintain
{
// method to calculate med of stream
public static void printMedian(int[] a)
{
double med = a[0];
// max heap to store the smaller half elements
PriorityQueue<Integer> smaller = new PriorityQueue<>
(Collections.reverseOrder());
// min heap to store the greater half elements
PriorityQueue<Integer> greater = new PriorityQueue<>();
smaller.add(a[0]);
System.out.println(med);
// reading elements of stream one by one
/* At any time we try to make heaps balanced and
their sizes differ by at-most 1. If heaps are
balanced,then we declare median as average of
min_heap_right.top() and max_heap_left.top()
If heaps are unbalanced,then median is defined
as the top element of heap of larger size */
for(int i = 1; i < a.length; i++)
{
int x = a[i];
// case1(left side heap has more elements)
if(smaller.size() > greater.size())
{
if(x < med)
{
greater.add(smaller.remove());
smaller.add(x);
}
else
greater.add(x);
med = (double)(smaller.peek() + greater.peek())/2;
}
// case2(both heaps are balanced)
else if(smaller.size() == greater.size())
{
if(x < med)
{
smaller.add(x);
med = (double)smaller.peek();
}
else
{
greater.add(x);
med = (double)greater.peek();
}
}
// case3(right side heap has more elements)
else
{
if(x > med)
{
smaller.add(greater.remove());
greater.add(x);
}
else
smaller.add(x);
med = (double)(smaller.peek() + greater.peek())/2;
}
System.out.println(med);
}
}
// Driver code
public static void main(String []args)
{
// stream of integers
int[] arr = new int[]{5, 15, 10, 20, 3};
printMedian(arr);
}
}
// This code is contributed by Kaustav kumar Chanda.
Output:
5
10
10
12.5
10
Time Complexity: O(n Log n)
Auxiliary Space : O(n)
Median in a stream of integers (running integers)
Given that integers are read from a data stream. Find median of elements read so for in efficient way. For simplicity assume there are no duplicates. For example, let us consider the stream 5, 15, 1, 3 …
After reading 1st element of stream - 5 -> median - 5
After reading 2nd element of stream - 5, 15 -> median - 10
After reading 3rd element of stream - 5, 15, 1 -> median - 5
After reading 4th element of stream - 5, 15, 1, 3 -> median - 4, so on...
Making it clear, when the input size is odd, we take the middle element of sorted data. If the input size is even, we pick average of middle two elements in sorted stream.
Note that output is effective median of integers read from the stream so far. Such an algorithm is called online algorithm. Any algorithm that can guarantee output of i-elements after processing i-th element, is said to be online algorithm. Let us discuss three solutions for the above problem.
Method 1: Insertion Sort
If we can sort the data as it appears, we can easily locate median element. Insertion Sort is one such online algorithm that sorts the data appeared so far. At any instance of sorting, say after sorting i-th element, the first i elements of array are sorted. The insertion sort doesn’t depend on future data to sort data input till that point. In other words, insertion sort considers data sorted so far while inserting next element. This is the key part of insertion sort that makes it an online algorithm.
However, insertion sort takes O(n2) time to sort n elements. Perhaps we can use binary search on insertion sort to find location of next element in O(log n) time. Yet, we can’t do data movement in O(log n) time. No matter how efficient the implementation is, it takes polynomial time in case of insertion sort.
Interested reader can try implementation of Method 1.
Method 2: Augmented self balanced binary search tree (AVL, RB, etc…)
At every node of BST, maintain number of elements in the subtree rooted at that node. We can use a node as root of simple binary tree, whose left child is self balancing BST with elements less than root and right child is self balancing BST with elements greater than root. The root element always holds effective median.
If left and right subtrees contain same number of elements, root node holds average of left and right subtree root data. Otherwise, root contains same data as the root of subtree which is having more elements. After processing an incoming element, the left and right subtrees (BST) are differed utmost by 1.
Self balancing BST is costly in managing balancing factor of BST. However, they provide sorted data which we don’t need. We need median only. The next method make use of Heaps to trace median.
Method 3: Heaps
Similar to balancing BST in Method 2 above, we can use a max heap on left side to represent elements that are less than effective median, and a min heap on right side to represent elements that are greater than effective median.
After processing an incoming element, the number of elements in heaps differ utmost by 1 element. When both heaps contain same number of elements, we pick average of heaps root data as effective median. When the heaps are not balanced, we select effective median from the root of heap containing more elements.
Given below is implementation of above method. For algorithm to build these heaps, please read the highlighted code.
Example:
Input: 5 10 15
Output: 5
7.5
10
Explanation: Given the input stream as an array of integers [5,10,15]. We will now read integers one by one and print the median correspondingly. So, after reading first element 5,median is 5. After reading 10,median is 7.5 After reading 15 ,median is 10.
The idea is to use max heap and min heap to store the elements of higher half and lower half. Max heap and min heap can be implemented using priority_queue in C++ STL. Below is the step by step algorithm to solve this problem.
Algorithm:
Create two heaps. One max heap to maintain elements of lower half and one min heap to maintain elements of higher half at any point of time..
Take initial value of median as 0.
For every newly read element, insert it into either max heap or min heap and calculate the median based on the following conditions:
If the size of max heap is greater than size of min heap and the element is less than previous median then pop the top element from max heap and insert into min heap and insert the new element to max heap else insert the new element to min heap. Calculate the new median as average of top of elements of both max and min heap.
If the size of max heap is less than size of min heap and the element is greater than previous median then pop the top element from min heap and insert into max heap and insert the new element to min heap else insert the new element to max heap. Calculate the new median as average of top of elements of both max and min heap.
If the size of both heaps are same. Then check if current is less than previous median or not. If the current element is less than previous median then insert it to max heap and new median will be equal to top element of max heap. If the current element is greater than previous median then insert it to min heap and new median will be equal to top element of min heap.
Below is the implementation of above approach:
filter_none
// C++ program to find med in
// stream of running integers
#include<bits/stdc++.h>
using namespace std;
// function to calculate med of stream
void printMedians(double arr[], int n)
{
// max heap to store the smaller half elements
priority_queue<double> s;
// min heap to store the greater half elements
priority_queue<double,vector<double>,greater<double> > g;
double med = arr[0];
s.push(arr[0]);
cout << med << endl;
// reading elements of stream one by one
/* At any time we try to make heaps balanced and
their sizes differ by at-most 1. If heaps are
balanced,then we declare median as average of
min_heap_right.top() and max_heap_left.top()
If heaps are unbalanced,then median is defined
as the top element of heap of larger size */
for (int i=1; i < n; i++)
{
double x = arr[i];
// case1(left side heap has more elements)
if (s.size() > g.size())
{
if (x < med)
{
g.push(s.top());
s.pop();
s.push(x);
}
else
g.push(x);
med = (s.top() + g.top())/2.0;
}
// case2(both heaps are balanced)
else if (s.size()==g.size())
{
if (x < med)
{
s.push(x);
med = (double)s.top();
}
else
{
g.push(x);
med = (double)g.top();
}
}
// case3(right side heap has more elements)
else
{
if (x > med)
{
s.push(g.top());
g.pop();
g.push(x);
}
else
s.push(x);
med = (s.top() + g.top())/2.0;
}
cout << med << endl;
}
}
// Driver program to test above functions
int main()
{
// stream of integers
double arr[] = {5, 15, 10, 20, 3};
int n = sizeof(arr)/sizeof(arr[0]);
printMedians(arr, n);
return 0;
}
Output:
5
10
10
12.5
10
Time Complexity: O(n Log n)
Auxiliary Space : O(n)
// Java program to find med in
// stream of running integers
import java.util.Collections;
import java.util.PriorityQueue;
public class MedianMaintain
{
// method to calculate med of stream
public static void printMedian(int[] a)
{
double med = a[0];
// max heap to store the smaller half elements
PriorityQueue<Integer> smaller = new PriorityQueue<>
(Collections.reverseOrder());
// min heap to store the greater half elements
PriorityQueue<Integer> greater = new PriorityQueue<>();
smaller.add(a[0]);
System.out.println(med);
// reading elements of stream one by one
/* At any time we try to make heaps balanced and
their sizes differ by at-most 1. If heaps are
balanced,then we declare median as average of
min_heap_right.top() and max_heap_left.top()
If heaps are unbalanced,then median is defined
as the top element of heap of larger size */
for(int i = 1; i < a.length; i++)
{
int x = a[i];
// case1(left side heap has more elements)
if(smaller.size() > greater.size())
{
if(x < med)
{
greater.add(smaller.remove());
smaller.add(x);
}
else
greater.add(x);
med = (double)(smaller.peek() + greater.peek())/2;
}
// case2(both heaps are balanced)
else if(smaller.size() == greater.size())
{
if(x < med)
{
smaller.add(x);
med = (double)smaller.peek();
}
else
{
greater.add(x);
med = (double)greater.peek();
}
}
// case3(right side heap has more elements)
else
{
if(x > med)
{
smaller.add(greater.remove());
greater.add(x);
}
else
smaller.add(x);
med = (double)(smaller.peek() + greater.peek())/2;
}
System.out.println(med);
}
}
// Driver code
public static void main(String []args)
{
// stream of integers
int[] arr = new int[]{5, 15, 10, 20, 3};
printMedian(arr);
}
}
// This code is contributed by Kaustav kumar Chanda.
Output:
5
10
10
12.5
10
Time Complexity: O(n Log n)
Auxiliary Space : O(n)
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