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Elements of first array that have more frequencies

  • Difficulty Level : Easy
  • Last Updated : 24 May, 2021
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Given two arrays (which may or may not be sorted). These arrays are such that they might have some common elements in them. We need to find elements whose counts of occurrences are more in first array than second. 
Examples: 
 

Input : ar1[] = {1, 2, 2, 2, 3, 3, 4, 5}
        ar2[] = {2, 2, 3, 3, 3, 4}
Output : 1 2 5
1 occurs one times in first and zero times in second
2 occurs three times in first and two times in second
............................

Input : ar1[] = {1, 3, 4, 2, 3}
        ar2[] = {3, 4, 5}
Output : 3

 

The idea is to use hashing. We traverse first array and insert all elements and their frequencies in a hash table. Now we traverse through the second array and reduce frequencies in hash table for the common elements. Now we traverse through first array again and print those elements whose frequencies are still more than 0. To avoid repeated printing of same elements, we set frequency as 0. 
 

C++




// C++ program to print all those elements of
// first array that have more frequencies than
// second array.
#include <bits/stdc++.h>
using namespace std;
 
// Compares two intervals according to staring times.
void moreFreq(int ar1[], int ar2[], int m, int n)
{
    // Traverse first array and store frequencies
    // of all elements
    unordered_map<int, int> mp;
    for (int i = 0; i < m; i++)
        mp[ar1[i]]++;
 
    // Traverse second array and reduce frequencies
    // of common elements.
    for (int i = 0; i < n; i++)
        if (mp.find(ar2[i]) != mp.end())
            mp[ar2[i]]--;
 
    // Now traverse first array again and print
    // all those elements whose frequencies are
    // more than 0. To avoid repeated printing,
    // we set frequency as 0 after printing.
    for (int i = 0; i < m; i++) {
        if (mp[ar1[i]] > 0) {
            cout << ar1[i] << " ";
            mp[ar1[i]] = 0;
        }
    }
}
 
// Driver code
int main()
{
    int ar1[] = { 1, 2, 2, 2, 3, 3, 4, 5 };
    int ar2[] = { 2, 2, 3, 3, 3, 4 };
    int m = sizeof(ar1) / sizeof(ar1[0]);
    int n = sizeof(ar2) / sizeof(ar2[0]);
    moreFreq(ar1, ar2, m, n);
    return 0;
}

Java




// Java program to print all those elements of
// first array that have more frequencies than
// second array.
import java.util.*;
 
class GFG
{
 
    // Compares two intervals according to staring times.
    static void moreFreq(int ar1[], int ar2[], int m, int n)
    {
        // Traverse first array and store frequencies
        // of all elements
        Map<Integer,Integer> mp = new HashMap<>();
        for (int i = 0 ; i < m; i++)
        {
            if(mp.containsKey(ar1[i]))
            {
                mp.put(ar1[i], mp.get(ar1[i])+1);
            }
            else
            {
                mp.put(ar1[i], 1);
            }
        }
         
        // Traverse second array and reduce frequencies
        // of common elements.
        for (int i = 0; i < n; i++)
            if (mp.containsKey(ar2[i]))
                mp.put(ar2[i], mp.get(ar2[i])-1);
     
        // Now traverse first array again and print
        // all those elements whose frequencies are
        // more than 0. To avoid repeated printing,
        // we set frequency as 0 after printing.
        for (int i = 0; i < m; i++)
        {
            if (mp.get(ar1[i]) > 0)
            {
                System.out.print(ar1[i] + " ");
                mp.put(ar1[i], 0);
            }
        }
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int ar1[] = { 1, 2, 2, 2, 3, 3, 4, 5 };
        int ar2[] = { 2, 2, 3, 3, 3, 4 };
        int m = ar1.length;
        int n = ar2.length;
        moreFreq(ar1, ar2, m, n);
    }
}
 
// This code has been contributed by 29AjayKumar

Python3




# Python3 program to print all those elements of
# first array that have more frequencies than
# second array.
import math as mt
 
# Compares two intervals according to
# staring times.
def moreFreq(ar1, ar2, m, n):
 
    # Traverse first array and store
    # frequencies of all elements
    mp = dict()
    for i in range(m):
        if ar1[i] in mp.keys():
            mp[ar1[i]] += 1
        else:
            mp[ar1[i]] = 1
                 
    # Traverse second array and reduce
    # frequencies of common elements.
    for i in range(n):
        if ar2[i] in mp.keys():
            mp[ar2[i]] -= 1
 
    # Now traverse first array again and print
    # all those elements whose frequencies are
    # more than 0. To avoid repeated printing,
    # we set frequency as 0 after printing.
    for i in range(m):
        if (mp[ar1[i]] > 0):
            print(ar1[i], end = " ")
            mp[ar1[i]] = 0
                                 
# Driver code
ar1 = [ 1, 2, 2, 2, 3, 3, 4, 5 ]
ar2 = [ 2, 2, 3, 3, 3, 4 ]
m = len(ar1)
n = len(ar2)
moreFreq(ar1, ar2, m, n)
 
# This code is contributed
# by mohit kumar 29

C#




// C# pprogram to print all those elements of
// first array that have more frequencies than
// second array.
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Compares two intervals according to
    // staring times.
    static void moreFreq(int []ar1, int []ar2,
                         int m, int n)
    {
        // Traverse first array and store frequencies
        // of all elements
        Dictionary<int,
                   int> mp = new Dictionary<int,
                                            int>();
        for (int i = 0 ; i < m; i++)
        {
            if(mp.ContainsKey(ar1[i]))
            {
                mp[ar1[i]] = mp[ar1[i]] + 1;
            }
            else
            {
                mp.Add(ar1[i], 1);
            }
        }
         
        // Traverse second array and reduce frequencies
        // of common elements.
        for (int i = 0; i < n; i++)
            if (mp.ContainsKey(ar2[i]))
                mp[ar2[i]] = mp[ar2[i]] - 1;
     
        // Now traverse first array again and print
        // all those elements whose frequencies are
        // more than 0. To avoid repeated printing,
        // we set frequency as 0 after printing.
        for (int i = 0; i < m; i++)
        {
            if (mp[ar1[i]] > 0)
            {
                Console.Write(ar1[i] + " ");
                mp[ar1[i]] = 0;
            }
        }
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int []ar1 = { 1, 2, 2, 2, 3, 3, 4, 5 };
        int []ar2 = { 2, 2, 3, 3, 3, 4 };
        int m = ar1.Length;
        int n = ar2.Length;
        moreFreq(ar1, ar2, m, n);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program to print all those elements of
// first array that have more frequencies than
// second array.
 
// Compares two intervals according to staring times.
function moreFreq(ar1, ar2, m, n)
{
    // Traverse first array and store frequencies
    // of all elements
    var mp = new Map();
    for (var i = 0; i < m; i++)
    {
        if(mp.has(ar1[i]))
            mp.set(ar1[i], mp.get(ar1[i])+1)
        else 
            mp.set(ar1[i], 1)
    }
 
    // Traverse second array and reduce frequencies
    // of common elements.
    for (var i = 0; i < n; i++)
        if (mp.has(ar2[i]))
        {
            mp.set(ar2[i], mp.get(ar2[i])-1)
        }
 
    // Now traverse first array again and print
    // all those elements whose frequencies are
    // more than 0. To avoid repeated printing,
    // we set frequency as 0 after printing.
    for (var i = 0; i < m; i++) {
        if (mp.get(ar1[i]) > 0) {
            document.write( ar1[i] + " ");
            mp.set(ar1[i], 0);
        }
    }
}
 
// Driver code
var ar1 = [1, 2, 2, 2, 3, 3, 4, 5];
var ar2 = [2, 2, 3, 3, 3, 4];
var m = ar1.length;
var n = ar2.length;
moreFreq(ar1, ar2, m, n);
 
</script>
Output: 
1 2 5

 

Time Complexity : O(m + n) under the assumption that unordered_map find() and insert() work in O(1) time.
 

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