计算第一个数组中存在但不在第二个数组中的元素

给定两个大小分别为 M 和 N 的数组（可以排序也可以不排序）。有这样的数组，它们可能有一些共同的元素。您需要计算第一个数组中出现次数多于第二个的元素的数量。

例子：

Input : arr1[] = {41, 43, 45, 50}, M = 4 
arr2[] = {55, 67, 65, 61, 62}, N = 5 
Output : 4 
Explanation: 
arr1[] contains 41, 43, 45, 50 with frequency all having 1 
arr2[] does not contain any such element which is common in both hence count will be 4

Input : arr1[] = {40, 45, 56, 60}, M = 4 
arr2[] = {40, 45, 56, 58}, N = 4 
Output : 1 
Explanation: 
arr1[] contains 40, 45, 56, 60 with frequency all having 1 
arr2[] contains 3 elements in common with arr1[] which are 40, 45, 56 
but not 60 hence count becomes 1

这个想法是使用哈希映射并保留第一个数组中数字频率的计数。在遍历第二个数组时，减少映射中遇到的每个整数的计数。现在计算频率大于 0 的元素，否则返回 0。

C++

// C++ to count number of elements present in arr1 whose
// occurrence is more than in arr2
#include 
using namespace std;
 
int Largercount(int arr1[], int arr2[], int m, int n)
{
    bool f = false;
    int count = 0;
 
    // map to store frequency of elements present in arr1
    unordered_map mp;
 
    // frequency of elements of arr1 is calculated
    for (int i = 0; i < m; i++)
        mp[arr1[i]]++;
 
    // check if the elements of arr2
    // is present in arr2 or not
    for (int i = 0; i < n; i++)
        if (mp.find(arr2[i]) != mp.end() && mp[arr2[i]] != 0)
            mp[arr2[i]]--;
 
    // count the elements of arr1 whose
    // frequency is more than arr2
    for (int i = 0; i < m; i++) {
        if (mp[arr1[i]] != 0) {
            count++;
            mp[arr1[i]] = 0;
        }
    }
 
    return count;
}
 
// Driver code
int main()
{
    int arr1[] = { 2, 4, 4, 6, 6, 6, 8, 9 };
    int arr2[] = { 2, 2, 4, 6, 6 };
    cout << Largercount(arr1, arr2, 8, 5);
    return 0;
}

Java

// Java to count number of elements present in arr1 whose
// occurrence is more than in arr2
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG {
    public static int Largercount(int arr1[], int arr2[], int m, int n)
    {
        int count = 0;
 
        // map to store frequency of elements present in arr1
        HashMap mp = new HashMap();
 
        // frequency of elements of arr1 is calculated
        for (int i = 0; i < m; i++) {
            int key = arr1[i];
            if (mp.containsKey(key)) {
                int freq = mp.get(key);
                freq++;
                mp.put(key, freq);
            }
            else
                mp.put(key, 1);
        }
 
        // check if the elements of arr2 is present in arr2 or not
        for (int i = 0; i < n; i++) {
            if (mp.containsKey(arr2[i]) && mp.get(arr2[i]) != 0) {
                int freq = mp.get(arr2[i]);
                freq--;
                mp.put(arr2[i], freq);
            }
        }
 
        // count the elements of arr1 whose
        // frequency is more than arr2
        for (int i = 0; i < m; i++) {
            if (mp.get(arr1[i]) != 0) {
                count++;
                mp.put(arr1[i], 0);
            }
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = new int[] { 2, 4, 4, 6, 6, 6, 8, 9};
        int arr2[] = new int[] { 2, 2, 4, 6, 6 };
 
        System.out.print(Largercount(arr1, arr2, 8, 5));
    }
}

Python3

# Python3 to count number of elements
# present in arr1 whose occurrence is
# more than in arr2
def Largercount(arr1, arr2, m, n):
 
    count = 0
 
    # map to store frequency of
    # elements present in arr1
    mp=dict()
 
    # frequency of elements of arr1
    # is calculated
    for i in range(m):
        mp[arr1[i]] = mp.get(arr1[i], 0) + 1
 
    # check if the elements of arr2
    # is present in arr2 or not
    for i in range(n):
        if (arr2[i] in mp.keys() and
                       mp[arr2[i]] != 0):
            mp[arr2[i]] -= 1
 
    # count the elements of arr1 whose
    # frequency is more than arr2
    for i in range(m):
        if (mp[arr1[i]] != 0):
            count += 1
            mp[arr1[i]] = 0
             
    return count
 
# Driver code
arr1 = [2, 4, 4, 6, 6, 6, 8, 9]
arr2 = [2, 2, 4, 6, 6 ]
print(Largercount(arr1, arr2, 8, 5))
 
# This code is contributed by mohit kumar

C#

// C# to count number of elements
// present in arr1 whose occurrence
// is more than in arr2
using System;
using System.Collections.Generic;
 
class GFG
{
public static int Largercount(int[] arr1,
                              int[] arr2,
                              int m, int n)
{
    int count = 0;
 
    // map to store frequency of
    // elements present in arr1
    Dictionary mp = new Dictionary();
 
    // frequency of elements
    // of arr1 is calculated
    for (int i = 0; i < m; i++)
    {
        int key = arr1[i];
        if (mp.ContainsKey(key))
        {
            int freq = mp[key];
            freq++;
            mp[key] = freq;
        }
        else
        {
            mp[key] = 1;
        }
    }
 
    // check if the elements of arr2
    // is present in arr2 or not
    for (int i = 0; i < n; i++)
    {
        if (mp.ContainsKey(arr2[i]) &&
                  mp[arr2[i]] != 0)
        {
            int freq = mp[arr2[i]];
            freq--;
            mp[arr2[i]] = freq;
        }
    }
 
    // count the elements of arr1 whose
    // frequency is more than arr2
    for (int i = 0; i < m; i++)
    {
        if (mp[arr1[i]] != 0)
        {
            count++;
            mp[arr1[i]] = 0;
        }
    }
 
    return count;
}
 
// Driver code
public static void Main(string[] args)
{
    int[] arr1 = new int[] {2, 4, 4, 6,
                            6, 6, 8, 9};
    int[] arr2 = new int[] {2, 2, 4, 6, 6};
 
    Console.Write(Largercount(arr1, arr2, 8, 5));
}
}
 
// This code is contributed by Shrikant13

Javascript

// Javascript to count number of elements present in arr1 whose
// occurrence is more than in arr2
     
    function Largercount(arr1,arr2,m,n)
    {
        let count = 0;
   
        // map to store frequency of elements present in arr1
        let mp = new Map();
   
        // frequency of elements of arr1 is calculated
        for (let i = 0; i < m; i++) {
            let key = arr1[i];
            if (mp.has(key)) {
                let freq = mp.get(key);
                freq++;
                mp.set(key, freq);
            }
            else
                mp.set(key, 1);
        }
   
        // check if the elements of arr2 is present in arr2 or not
        for (let i = 0; i < n; i++) {
            if (mp.has(arr2[i]) && mp.get(arr2[i]) != 0) {
                let freq = mp.get(arr2[i]);
                freq--;
                mp.set(arr2[i], freq);
            }
        }
   
        // count the elements of arr1 whose
        // frequency is more than arr2
        for (let i = 0; i < m; i++) {
            if (mp.get(arr1[i]) != 0) {
                count++;
                mp.set(arr1[i], 0);
            }
        }
   
        return count;
    }
     
    // Driver code
    let arr1=[2, 4, 4, 6, 6, 6, 8, 9];
    let arr2=[2, 2, 4, 6, 6 ];
    document.write(Largercount(arr1, arr2, 8, 5));
     
     
// This code is contributed by unknown2108

输出：

4

时间复杂度： O(m + n)

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