Remove minimum number of elements such that no common element exist in both array

Given two arrays A[] and B[] consisting of n and m elements respectively. Find the minimum number of elements to remove from each array such that no common element exist in both.

Examples:

Input : A[] = { 1, 2, 3, 4}
        B[] = { 2, 3, 4, 5, 8 }
Output : 3
We need to remove 2, 3 and 4 from any array.

Input : A[] = { 4, 2, 4, 4}
        B[] = { 4, 3 }
Output : 1
We need to remove 4 from B[]

Input : A[] = { 1, 2, 3, 4 }
        B[] = { 5, 6, 7 }
Output : 0
There is no common element in both.

Count occurrence of each number in both arrays. If there is a number in both array remove number from array in which it appears less number of times add it to the result. 

Implementation:

Java

// JAVA Code to Remove minimum number of elements
// such that no common element exist in both array
import java.util.*;

class GFG {

    // To find no elements to remove
    // so no common element exist
    public static int minRemove(int a[], int b[], int n,
                                                 int m)
    {
        // To store count of array element
        HashMap<Integer, Integer> countA = new HashMap<
                                          Integer, Integer>();
        HashMap<Integer, Integer> countB = new HashMap<
                                          Integer, Integer>();

        // Count elements of a
        for (int i = 0; i < n; i++){
           if (countA.containsKey(a[i]))
                countA.put(a[i], countA.get(a[i]) + 1);

           else countA.put(a[i], 1);

        }

        // Count elements of b
        for (int i = 0; i < m; i++){
             if (countB.containsKey(b[i]))
                    countB.put(b[i], countB.get(b[i]) + 1);

               else countB.put(b[i], 1);
        }

        // Traverse through all common element, and
        // pick minimum occurrence from two arrays
        int res = 0;

        Set<Integer> s = countA.keySet();

        for (int x : s)
            if(countB.containsKey(x))
                res += Math.min(countB.get(x),
                               countA.get(x));

        // To return count of minimum elements
        return res;
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {

            int a[] = { 1, 2, 3, 4 };
            int b[] = { 2, 3, 4, 5, 8 };
            int n = a.length;
            int m = b.length;

            System.out.println(minRemove(a, b, n, m));

    }
}

// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to find minimum
# element to remove so no common
# element exist in both array

# To find no elements to remove
# so no common element exist
def minRemove(a, b, n, m):

    # To store count of array element
    countA = dict()
    countB = dict()

    # Count elements of a
    for i in range(n):
        countA[a[i]] = countA.get(a[i], 0) + 1

    # Count elements of b
    for i in range(n):
        countB[b[i]] = countB.get(b[i], 0) + 1

    # Traverse through all common
    # element, and pick minimum
    # occurrence from two arrays
    res = 0
    for x in countA:
        if x in countB.keys():
            res += min(countA[x],countB[x])

    # To return count of
    # minimum elements
    return res

# Driver Code
a = [ 1, 2, 3, 4 ]
b = [2, 3, 4, 5, 8 ]
n = len(a)
m = len(b)
print(minRemove(a, b, n, m))

# This code is contributed
# by mohit kumar

Javascript

<script>

// JavaScript Code to Remove
// minimum number of elements
// such that no common element
// exist in both array

    // To find no elements to remove
    // so no common element exist
    function minRemove(a,b,n,m)
    {
        // To store count of array element
        let countA = new Map();
        let countB = new Map();

        // Count elements of a
        for (let i = 0; i < n; i++){
           if (countA.has(a[i]))
                countA.set(a[i],
                countA.get(a[i]) + 1);

           else
               countA.set(a[i], 1);

        }

        // Count elements of b
        for (let i = 0; i < m; i++){
             if (countB.has(b[i]))
                    countB.set(b[i],
                    countB.get(b[i]) + 1);

             else
                   countB.set(b[i], 1);
        }


        // Traverse through all
        // common element, and
        // pick minimum occurrence
        // from two arrays
        let res = 0;


        for (let x of countA.keys())
            if(countB.has(x))
                res += Math.min(countB.get(x),
                               countA.get(x));

        // To return count of minimum elements
        return res;
    }

    /* Driver program to test above function */
    let a=[1, 2, 3, 4 ];
    let b=[2, 3, 4, 5, 8];
    let n = a.length;
    let m = b.length;
    document.write(minRemove(a, b, n, m));



// This code is contributed by unknown2108

</script>

Output

3

Time Complexity: O(n+m)
Auxiliary Space: O(n+m)