Count elements present in first array but not in second

Given two arrays (which may or may not be sorted) of sizes M and N respectively. There arrays are such that they might have some common elements in them. You need to count the number of elements whose occurrences are more in first array than second.

Examples:

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

The idea is to use a hash map and keep the count of frequency of numbers in the first array. While traversing the second array reduce the count in the map for every integer encountered. Now count the elements whose frequency is more than 0 otherwise return 0.

C++

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// C++ to count number of elements present in arr1 whose
// occurrence is more than in arr2
#include <bits/stdc++.h>
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<int, int> mp;
  
    // frequency of elements of arr1 is calulated
    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;
}

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Java

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// 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<Integer, Integer> mp = new HashMap<Integer, Integer>();
  
        // frequency of elements of arr1 is calulated
        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));
    }
}

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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 calulated
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#

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// 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<int
               int> mp = new Dictionary<int
                                        int>();
  
    // frequency of elements 
    // of arr1 is calulated 
    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

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Output:

4

Time complexity: O(m + n)



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