# 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 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 from B[]

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

## C++

 `// CPP program to find minimum element ` `// to remove so no common element ` `// exist in both array ` `#include ` `using` `namespace` `std; ` ` `  `// To find no elements to remove ` `// so no common element exist ` `int` `minRemove(``int` `a[], ``int` `b[], ``int` `n, ``int` `m) ` `{ ` `    ``// To store count of array element ` `    ``unordered_map<``int``, ``int``> countA, countB; ` ` `  `    ``// Count elements of a ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``countA[a[i]]++; ` ` `  `    ``// Count elements of b ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``countB[b[i]]++; ` ` `  `    ``// Traverse through all common element, and ` `    ``// pick minimum occurrence from two arrays ` `    ``int` `res = 0; ` `    ``for` `(``auto` `x : countA) ` `        ``if` `(countB.find(x.first) != countB.end()) ` `            ``res += min(x.second, countB[x.first]); ` ` `  `    ``// To return count of minimum elements ` `    ``return` `res; ` `} ` ` `  `// Driver program to test minRemove() ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 2, 3, 4 }; ` `    ``int` `b[] = { 2, 3, 4, 5, 8 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` `    ``int` `m = ``sizeof``(b) / ``sizeof``(b[0]); ` ` `  `    ``cout << minRemove(a, b, n, m); ` ` `  `    ``return` `0; ` `} `

## 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 countA = ``new` `HashMap< ` `                                          ``Integer, Integer>(); ` `        ``HashMap 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 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

## C#

 `// C# Code to Remove minimum number of elements ` `// such that no common element exist in both array ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `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 ` `        ``Dictionary<``int``,``int``> countA = ``new` `Dictionary<``int``,``int``>(); ` `        ``Dictionary<``int``,``int``>countB = ``new` `Dictionary<``int``,``int``>(); ` `     `  `        ``// Count elements of a ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `(countA.ContainsKey(a[i])) ` `            ``{ ` `                ``var` `v = countA[a[i]]; ` `                ``countA.Remove(countA[a[i]]); ` `                ``countA.Add(a[i], v + 1); ` `            ``} ` `            ``else` `countA.Add(a[i], 1); ` `                 `  `        ``}    ` `         `  `        ``// Count elements of b ` `        ``for` `(``int` `i = 0; i < m; i++) ` `        ``{ ` `            ``if` `(countB.ContainsKey(b[i])) ` `            ``{ ` `                ``var` `v = countB[b[i]]; ` `                ``countB.Remove(countB[b[i]]); ` `                ``countB.Add(b[i], v + 1); ` `            ``}  ` `            ``else` `countB.Add(b[i], 1); ` `        ``} ` `         `  `        ``// Traverse through all common element, and ` `        ``// pick minimum occurrence from two arrays ` `        ``int` `res = 0; ` ` `  `        ``foreach` `(``int` `x ``in` `countA.Keys) ` `            ``if``(countB.ContainsKey(x)) ` `                ``res += Math.Min(countB[x],  ` `                            ``countA[x]); ` `     `  `        ``// To return count of minimum elements ` `        ``return` `res; ` `    ``} ` `     `  `    ``/* Driver code */` `    ``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; ` `         `  `            ``Console.WriteLine(minRemove(a, b, n, m)); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```3
```