Find Union and Intersection of two unsorted arrays

2.3

Given two unsorted arrays that represent two sets (elements in every array are distinct), find union and intersection of two arrays.

For example, if the input arrays are:
arr1[] = {7, 1, 5, 2, 3, 6}
arr2[] = {3, 8, 6, 20, 7}
Then your program should print Union as {1, 2, 3, 5, 6, 7, 8, 20} and Intersection as {3, 6}. Note that the elements of union and intersection can be printed in any order.

Method 1 (Naive)
Union:
1) Initialize union U as empty.
2) Copy all elements of first array to U.
3) Do following for every element x of second array:
…..a) If x is not present in first array, then copy x to U.
4) Return U.

Intersection:
1) Initialize intersection I as empty.
2) Do following for every element x of first array
…..a) If x is present in second array, then copy x to I.
4) Return I.

Time complexity of this method is O(mn) for both operations. Here m and n are number of elements in arr1[] and arr2[] respectively.

Method 2 (Use Sorting)
1) Sort arr1[] and arr2[]. This step takes O(mLogm + nLogn) time.
2) Use O(m + n) algorithms to find union and intersection of two sorted arrays.

Overall time complexity of this method is O(mLogm + nLogn).

Method 3 (Use Sorting and Searching)
Union:
1) Initialize union U as empty.
2) Find smaller of m and n and sort the smaller array.
3) Copy the smaller array to U.
4) For every element x of larger array, do following
…….b) Binary Search x in smaller array. If x is not present, then copy it to U.
5) Return U.

Intersection:
1) Initialize intersection I as empty.
2) Find smaller of m and n and sort the smaller array.
3) For every element x of larger array, do following
…….b) Binary Search x in smaller array. If x is present, then copy it to I.
4) Return I.

Time complexity of this method is min(mLogm + nLogm, mLogn + nLogn) which can also be written as O((m+n)Logm, (m+n)Logn). This approach works much better than the previous approach when difference between sizes of two arrays is significant.

Thanks to use_the_force for suggesting this method in a comment here.

Below is the implementation of this method.

C

// A C++ program to print union and intersection 
/// of two unsorted arrays
#include <iostream>
#include <algorithm>
using namespace std;

int binarySearch(int arr[], int l, int r, int x);

// Prints union of arr1[0..m-1] and arr2[0..n-1]
void printUnion(int arr1[], int arr2[], int m, int n)
{
    // Before finding union, make sure arr1[0..m-1] 
    // is smaller
    if (m > n)
    {
        int *tempp = arr1;
        arr1 = arr2;
        arr2 = tempp;

        int temp = m;
        m = n;
        n = temp;
    }

    // Now arr1[] is smaller

    // Sort the first array and print its elements (these two
    // steps can be swapped as order in output is not important)
    sort(arr1, arr1 + m);
    for (int i=0; i<m; i++)
        cout << arr1[i] << " ";

    // Search every element of bigger array in smaller array
    // and print the element if not found
    for (int i=0; i<n; i++)
        if (binarySearch(arr1, 0, m-1, arr2[i]) == -1)
            cout << arr2[i] << " ";
}

// Prints intersection of arr1[0..m-1] and arr2[0..n-1]
void printIntersection(int arr1[], int arr2[], int m, int n)
{
    // Before finding intersection, make sure arr1[0..m-1] 
    // is smaller
    if (m > n)
    {
        int *tempp = arr1;
        arr1 = arr2;
        arr2 = tempp;

        int temp = m;
        m = n;
        n = temp;
    }

    // Now arr1[] is smaller

    // Sort smaller array arr1[0..m-1]
    sort(arr1, arr1 + m);

    // Search every element of bigger array in smaller
    // array and print the element if found
    for (int i=0; i<n; i++)
        if (binarySearch(arr1, 0, m-1, arr2[i]) != -1)
            cout << arr2[i] << " ";
}

// A recursive binary search function. It returns 
// location of x in given array arr[l..r] is present, 
// otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
    if (r >= l)
    {
        int mid = l + (r - l)/2;

        // If the element is present at the middle itself
        if (arr[mid] == x)  return mid;

        // If element is smaller than mid, then it can only
        // be presen in left subarray
        if (arr[mid] > x) 
          return binarySearch(arr, l, mid-1, x);

        // Else the element can only be present in right subarray
        return binarySearch(arr, mid+1, r, x);
    }

    // We reach here when element is not present in array
    return -1;
}

/* Driver program to test above function */
int main()
{
    int arr1[] = {7, 1, 5, 2, 3, 6};
    int arr2[] = {3, 8, 6, 20, 7};
    int m = sizeof(arr1)/sizeof(arr1[0]);
    int n = sizeof(arr2)/sizeof(arr2[0]);
    cout << "Union of two arrays is n";
    printUnion(arr1, arr2, m, n);
    cout << "nIntersection of two arrays is n";
    printIntersection(arr1, arr2, m, n);
    return 0;
}

Java

import java.util.Arrays;

class UnionAndIntersection 
{
    // Prints union of arr1[0..m-1] and arr2[0..n-1]
    void printUnion(int arr1[], int arr2[], int m, int n) 
    {
        // Before finding union, make sure arr1[0..m-1] 
        // is smaller
        if (m > n) 
        {
            int tempp[] = arr1;
            arr1 = arr2;
            arr2 = tempp;

            int temp = m;
            m = n;
            n = temp;
        }

        // Now arr1[] is smaller
        // Sort the first array and print its elements (these two
        // steps can be swapped as order in output is not important)
        Arrays.sort(arr1);
        for (int i = 0; i < m; i++)
            System.out.print(arr1[i] + " ");

        // Search every element of bigger array in smaller array
        // and print the element if not found
        for (int i = 0; i < n; i++) 
        {
            if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1)
                System.out.print(arr2[i] + " ");
        }
    }

    // Prints intersection of arr1[0..m-1] and arr2[0..n-1]
    void printIntersection(int arr1[], int arr2[], int m, int n) 
    {
        // Before finding intersection, make sure arr1[0..m-1] 
        // is smaller
        if (m > n) 
        {
            int tempp[] = arr1;
            arr1 = arr2;
            arr2 = tempp;

            int temp = m;
            m = n;
            n = temp;
        }

        // Now arr1[] is smaller
        // Sort smaller array arr1[0..m-1]
        Arrays.sort(arr1);

        // Search every element of bigger array in smaller array
        // and print the element if found
        for (int i = 0; i < n; i++) 
        {
            if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1) 
                System.out.print(arr2[i] + " ");
        }
    }

    // A recursive binary search function. It returns location of x in
    // given array arr[l..r] is present, otherwise -1
    int binarySearch(int arr[], int l, int r, int x) 
    {
        if (r >= l) 
        {
            int mid = l + (r - l) / 2;

            // If the element is present at the middle itself
            if (arr[mid] == x)
                return mid;

            // If element is smaller than mid, then it can only 
            // be present in left subarray
            if (arr[mid] > x)
                return binarySearch(arr, l, mid - 1, x);

            // Else the element can only be present in right subarray
            return binarySearch(arr, mid + 1, r, x);
        }

        // We reach here when element is not present in array
        return -1;
    }

    // Driver program to test above functions
    public static void main(String[] args) 
    {
        UnionAndIntersection u_i = new UnionAndIntersection();
        int arr1[] = {7, 1, 5, 2, 3, 6};
        int arr2[] = {3, 8, 6, 20, 7};
        int m = arr1.length;
        int n = arr2.length;
        System.out.println("Union of two arrays is ");
        u_i.printUnion(arr1, arr2, m, n);
        System.out.println("");
        System.out.println("Intersection of two arrays is ");
        u_i.printIntersection(arr1, arr2, m, n);
    }
}


Output:
Union of two arrays is
3 6 7 8 20 1 5 2
Intersection of two arrays is
7 3 6

Method 4 (Use Hashing)
Union:
Union
1. Initialize an empty hash set hs.
2. Iterate through the first array and put every element of the first array in the set S.
3. Repeat the process for the second array.
4. Print the set hs.

Intersection
1. Initialize an empty set hs.
2. Iterate through the first array and put every element of the first array in the set S.
3. For every element x of the second array, do the following :
Search x in the set hs. If x is present, then print it.

Time complexity of this method is Θ(m+n) under the assumption that hash table search and insert operations take Θ(1) time.

C++

// CPP program to find union and intersection
// using sets
#include <bits/stdc++.h>
using namespace std;

// Prints union of arr1[0..n1-1] and arr2[0..n2-1]
void printUnion(int arr1[], int arr2[], int n1, int n2)
{
    set<int> hs;

    // Inhsert the elements of arr1[] to set hs
    for (int i = 0; i < n1; i++)
        hs.insert(arr1[i]);

    // Insert the elements of arr2[] to set hs
    for (int i = 0; i < n2; i++)
        hs.insert(arr2[i]);

    // Print the content of set hs
    for (auto it = hs.begin(); it != hs.end(); it++)
        cout << *it << " ";
    cout << endl;
}

// Prints intersection of arr1[0..n1-1] and
// arr2[0..n2-1]
void printIntersection(int arr1[], int arr2[],
                               int n1, int n2)
{
    set<int> hs;

    // Insert the elements of arr1[] to set S
    for (int i = 0; i < n1; i++)
        hs.insert(arr1[i]);

    for (int i = 0; i < n2; i++)

        // If element is present in set then
        // push it to vector V
        if (hs.find(arr2[i]) != hs.end())
            cout << arr2[i] << " ";
}

// Driver Program
int main()
{
    int arr1[] = { 7, 1, 5, 2, 3, 6 };
    int arr2[] = { 3, 8, 6, 20, 7 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    printUnion(arr1, arr2, n1, n2);
    printIntersection(arr1, arr2, n1, n2);

    return 0;
}

Java

// Java program to find union and intersection
// using Hashing
import java.util.HashSet;

class Test
{   
    // Prints union of arr1[0..m-1] and arr2[0..n-1]
    static void printUnion(int arr1[], int arr2[])
    {
        HashSet<Integer> hs = new HashSet<>();
        
        for (int i = 0; i < arr1.length; i++) 
            hs.add(arr1[i]);        
        for (int i = 0; i < arr2.length; i++) 
            hs.add(arr2[i]);
        System.out.println(hs);        
    }
    
    // Prints intersection of arr1[0..m-1] and arr2[0..n-1]
    static void printIntersection(int arr1[], int arr2[])
    {
        HashSet<Integer> hs = new HashSet<>();
        HashSet<Integer> hs1 = new HashSet<>();
        
        for (int i = 0; i < arr1.length; i++) 
            hs.add(arr1[i]);
        
        for (int i = 0; i < arr2.length; i++) 
            if (hs.contains(arr2[i]))
               System.out.print(arr2[i] + " ");
    }
    
    // Driver method to test the above function
    public static void main(String[] args) 
    {
        int arr1[] = {7, 1, 5, 2, 3, 6};
        int arr2[] = {3, 8, 6, 20, 7};

        System.out.println("Union of two arrays is : ");
        printUnion(arr1, arr2);
        
        System.out.println("Intersection of two arrays is : ");
        printIntersection(arr1, arr2);        
    }
}


This method is contributed by Ankur Singh.

Asked in: Rockstand

See following post for sorted arrays.
Find Union and Intersection of two sorted arrays

Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem.

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