Exercise #31: Counting Intersection of Two Unsorted Arrays by Zoran Horvat @zoranh75
Problem Statement
Given two unsorted arrays of integer numbers, write a function which returns
number of elements that exist in both arrays. Elements may be repeated in any
of the arrays, and all common appearances should be counted.
Example: Suppose that the first array is 3, 1, 4, 1, 2 and the second array is
5, 1, 2, 4, 2, 1. Common elements of the two arrays are 1, 1 and 2, and
therefore the function should return result 3.
Problem Analysis
In this problem, we have to traverse the two arrays and identify elements that
are in common. This exercise is closely connected with Finding Intersection of Two Unsorted Arrays, in which we were actually constructing the array containing the common
elements.
In that exercise, we have discussed several approaches:
 Indexing solution – Pass through one array and put all its elements into a hash
table. Then iterate through the other array and see if any element can be found
in the hash table. If so, remove it from the hash table (to avoid future hits)
and push it to the output.
 Brute force solution – Compare any two elements of both arrays, while ensuring
that elements that take part in any matching are not considered again.
 Sorting solution – Sort both arrays and then progress through them and extract
common elements.
 Partial sorting solution – Divide and conquer both arrays around the same pivot
values and then push common pivot occurrences to the output.
For more details about each of these methods, please refer to Finding Intersection of Two Unsorted Arrays exercise. In this article, we will provide solution which is based on partial
sorting.
Partial Sorting Solution
As already indicated, this solution is based on divideandconquer idea. We
would find a pivot in the first array and then partition the array around that
pivot value. Then we would partition the second array around the same pivot
value. In this way, we would split both arrays around the same value. The whole
idea boils down to this sequence of steps:
1. Partition both arrays around the same pivot value
2. Count how many occurrences of the pivot value they have in common
3. Repeat recursively on left and on right partitions of both arrays, if
nonempty
Once recursion unfolds, all calls will report how many pivot values they have
in common. Overall result is then the sum of all partial results.
Below is the pseudocode which implements this idea.
function CountIntersection(a, b, n, m)
 a  array
 b  array
 n  length of a
 m  length of b
begin
return CountIntersectionRecursive(a, b, 1, n, 1, m)
end
function CountIntersectionRecursive(a, b, aStart, aEnd, bStart, bEnd)
 a  array
 b  array
 aStart  first position in a to intersect
 aEnd  last position in a to intersect
 bStart  first position in b to intersect
 bEnd  last position in b to intersect
begin
pivot = MedianOfMedians(a, aStart, aEnd);
Partition(a, aStart, aEnd, pivot, out aPivotStart, out aPivotLen)
Partition(b, bStart, bEnd, pivot, out bPivotStart, out bPivotLen)
count = min(aPivotLen, bPivotLen)
 Recursively solve left partitions
if aPivotStart > aStart AND bPivotStart > bStart then
count = count +
CountIntersectionRecursive(a, b, aStart, aPivotStart  1,
bStart, bPivotStart  1)
 Recursively solve right partitions
if aPivotStrat + aPivotLen < aEnd AND bPivotStart + bPivotLen < bEnd then
count = count +
CountIntersectionRecursive(a, b, aPivotStart + aPivotLen, aEnd,
bPivotStart + bPivotLen, bEnd)
return count
end
Partition is a utility function which partitions the array around specified pivot value. MedianOfMedians is another utility function which calculates convenient pivot value using
median of medians algorithm. You can read more about this algorithm in exercise
Finding Kth Smallest Element in an Unsorted Array.
Implementation
Below is the complete Console application in C# which implements the
intersection counting function.
using System;
namespace IntersectionCount
{
public class Program
{
static int IntersectionCount(int[] a, int[] b)
{
return IntersectionCountRecursive(a, b,
0, a.Length  1,
0, b.Length  1);
}
static int IntersectionCountRecursive(int[] a, int[] b,
int aStart, int aEnd,
int bStart, int bEnd)
{
int pivot = MedianOfMedians(a, aStart, aEnd);
int aPivotStart;
int aPivotLen;
int bPivotStart;
int bPivotLen;
Partition(a, aStart, aEnd, pivot,
out aPivotStart, out aPivotLen);
Partition(b, bStart, bEnd, pivot,
out bPivotStart, out bPivotLen);
int count = aPivotLen;
if (bPivotLen < count)
count = bPivotLen;
if (aPivotStart > aStart && bPivotStart > bStart)
count += IntersectionCountRecursive(a, b,
aStart, aPivotStart  1,
bStart, bPivotStart  1);
if (aPivotStart + aPivotLen  1 < aEnd &&
bPivotStart + bPivotLen  1 < bEnd)
count += IntersectionCountRecursive(a, b,
aPivotStart + aPivotLen, aEnd,
bPivotStart + bPivotLen, bEnd);
return count;
}
static void Partition(int[] a, int start, int end, int pivot,
out int pivotStart, out int pivotLen)
{
int i = start;
while (i <= end)
{
if (a[i] < pivot)
{
a[start++] = a[i++];
}
else if (a[i] == pivot)
{
i++;
}
else
{
int tmp = a[end];
a[end] = a[i];
a[i] = tmp;
end;
}
}
for (i = start; i <= end; i++)
a[i] = pivot;
pivotStart = start;
pivotLen = end  start + 1;
}
static int MedianOfMedians(int[] a, int start, int end)
{
while (end > start + 1)
{
int pos = start;
for (int segment = start; segment < end; segment += 5)
{
int tmp;
int segmentEnd = segment + 5;
if (segmentEnd > end)
segmentEnd = end;
for (int i = segment; i < segmentEnd  1; i++)
for (int j = i + 1; j < segmentEnd; j++)
if (a[i] > a[j])
{
tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
int middlePos = (segment + segmentEnd  1) 2;
tmp = a[pos];
a[pos] = a[middlePos];
a[middlePos] = tmp;
pos++;
}
end = pos;
}
return a[start];
}
static int[] ReadArray()
{
Console.Write("Enter array elements (ENTER to exit): ");
string[] elements =
Console
.ReadLine()
.Split(new char[] { ' ' },
StringSplitOptions.RemoveEmptyEntries);
int[] a = null;
if (elements.Length > 0)
{
a = new int[elements.Length];
for (int i = 0; i < elements.Length; i++)
a[i] = int.Parse(elements[i]);
}
return a;
}
static void Main(string[] args)
{
while (true)
{
int[] a = ReadArray();
if (a == null)
break;
int[] b = ReadArray();
if (b == null)
break;
int intersectionCount = IntersectionCount(a, b);
Console.WriteLine("Arrays share {0} element(s).\n",
intersectionCount);
}
}
}
}
Demonstration
When Console application shown above is run and tried on a few arrays, it
produces the following output:
Enter array elements (ENTER to exit): 3 1 4 1 2
Enter array elements (ENTER to exit): 5 1 2 4 2
Arrays share 3 element(s).
Enter array elements (ENTER to exit): 2 5 1 2 7 3
Enter array elements (ENTER to exit): 9 8 7 6 5 4
Arrays share 2 element(s).
Enter array elements (ENTER to exit): 5 5 5 5 5
Enter array elements (ENTER to exit): 4 4 4 4 4
Arrays share 0 element(s).
Enter array elements (ENTER to exit):
Discussion
Good thing about solution presented in this article is that analysis of one
partition of an array terminates if the corresponding partition in the other
array is empty. Searching for new common elements between the two arrays
progresses only as long as there is hope to find more of them. This measure
makes positive effect on the efficiency of the solution.
See also:
Published: Sep 17, 2015; Modified: Sep 16, 2015
ZORAN HORVAT Zoran is software architect dedicated to clean design and CTO in a growing software company. Since 2014 Zoran is an author at Pluralsight where he is preparing a series of courses on objectoriented and functional design, design patterns, writing unit and integration tests and applying methods to improve code design and longterm maintainability. Follow him on Twitter @zoranh75 to receive updates and links to new articles. Watch Zoran's video courses at pluralsight.com (requires registration): Making Your C# Code More ObjectOriented This course will help leverage your conceptual understanding to produce proper objectoriented code, where objects will completely replace procedural code for the sake of flexibility and maintainability. More... Advanced Defensive Programming Techniques This course will lead you step by step through the process of developing defensive design practices, which can substitute common defensive coding, for the better of software design and implementation. More... Tactical Design Patterns in .NET: Creating Objects This course sheds light on issues that arise when implementing creational design patterns and then provides practical solutions that will make our code easier to write and more stable when running. More... Tactical Design Patterns in .NET: Managing Responsibilities Applying a design pattern to a realworld problem is not as straightforward as literature implicitly tells us. It is a more engaged process. This course gives an insight to tactical decisions we need to make when applying design patterns that have to do with separating and implementing class responsibilities. More... Tactical Design Patterns in .NET: Control Flow Improve your skills in writing simpler and safer code by applying coding practices and design patterns that are affecting control flow. More... Writing Highly Maintainable Unit Tests This course will teach you how to develop maintainable and sustainable tests as your production code grows and develops. More... Improving Testability Through Design This course tackles the issues of designing a complex application so that it can be covered with high quality tests. More... Share this article
