Find Union and Intersection of two unsorted arrays

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.

public class IntersectUnionArray {

	private int binarySearch(int[] A, int target) {
		int start = 0, end = A.length-1;
		while(start <= end) {
			int m = (start+end)/2;
			if(A[m] == target) {
				return m;
			} else if(A[m] < target) {
				start = m+1;
			} else {
				end = m-1;
			}
		}
		return -1;
	}
	
	public List<Integer> unionArray(int[] A, int[] B) {
		if(A.length > B.length) {
			return unionArray(B, A);
		}
		Arrays.sort(A);
		List<Integer> list = new ArrayList<>();
		for(int num:A) {
			list.add(num);
		}
		for(int num:B) {
			if(binarySearch(A, num) == -1) {
				list.add(num);
			}
		}
		return list;
	}
	
	public List<Integer> intersectArray(int[] A, int[] B) {
		if(A.length > B.length) {
			return intersectArray(B, A);
		}
		Arrays.sort(A);
		List<Integer> list = new ArrayList<>();
		for(int num:B) {
			if(binarySearch(A, num) != -1) {
				list.add(num);
			}
		}
		return list;
	}
	
	public static void main(String[] args) {
		int[] A = {7, 1, 5, 2, 3, 6};
		int[] B = {3, 8, 6, 20, 7};
		IntersectUnionArray iu = new IntersectUnionArray();
		System.out.println(iu.unionArray(A, B));
		System.out.println(iu.intersectArray(A, B));
	}
}

Method 4 (Use Hashing)
Union:
1) Initialize union U as empty.
1) Initialize an empty hash table.
2) Iterate through first array and put every element of first array in the hash table, and in U.
4) For every element x of second array, do following
…….a) Search x in the hash table. If x is not present, then copy it to U.
5) Return U.

Intersection:
1) Initialize intersection I as empty.
2) In initialize an empty hash table.
3) Iterate through first array and put every element of first array in the hash table.
4) For every element x of second array, do following
…….a) Search x in the hash table. If x is present, then copy it to I.
5) Return I.

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

 

Reference:

http://www.geeksforgeeks.org/find-union-and-intersection-of-two-unsorted-arrays/