# 移除最小数量的元素，使两个数组中不存在公共元素

2021年3月28日16:15:31 发表评论 466 次浏览

## 本文概述

``````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.``````

## C ++

``````// CPP program to find minimum element
// to remove so no common element
// exist in both array
#include <bits/stdc++.h>
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);
int m = sizeof (b) / sizeof (b);

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<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``````

## 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]]);
}

}

// 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]]);
}
}

// 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``````

``3`` 