Find element with highest frequency in given nested Array (original) (raw)

Last Updated : 28 Jun, 2021

Given an array arr[] of N integers. The task is to create a frequency array freq[] of the given array arr[] and find the maximum element of the frequency array. If two elements have the same frequency in the array freq[], then return the element which has a smaller value.

Examples:

Input: arr[] = {1, 1, 1, 2, 3, 2, 2, 3, 5, 5, 5, 5, 4, 4, 4, 4, 4};
Output: 3
Explanation:
frequency of elements is given by:
val -> freq[]
1 -> 3
2 -> 3
3 -> 2
4 -> 5
5 -> 4
Element 3 has the maximum frequency in the frequency array.

Input: arr[] = { 3, 5, 15, 51, 15, 14, 14, 14, 14, 4};
Output: 1
Explanation:
frequency of elements is given by:
val -> freq[]
3 -> 1
4 -> 1
5 -> 1
14 -> 4
15 -> 2
51 -> 1
Element 1 has the maximum frequency in the frequency array.

Approach:

  1. Store the frequency of the elements of arr[] in a map say map1, with elements of arr[] as key and their frequency as value.
  2. Now, store the frequency of elements of map1 in some other map say map2.
  3. Traverse map2 to get the highest element.
  4. If there is multiple highest element than the element which has lower value is print.

Below is the implementation of the above approach:

C++14 `

// C++14 program for the above approach #include <bits/stdc++.h> using namespace std;

// Function to get the highest // frequency of frequency array int findElement(int a[], int n) { // To find the maximum frequency // initialize it with INT_MIN int mx = INT_MIN; int ans = 0;

// Initialize maps to store the
// count of element in array
map<int, int> map1, map2;

// Store the frequency of
// element of array in map1
for (int i = 0; i < n; i++) {
    map1[a[i]]++;
}

// Storing the frequency i.e.,
// (x.second) which is count of
// element in array
for (auto x : map1) {
    map2[x.second]++;
}

for (auto x : map2) {

    // Check if the frequency of
    // element is greater than mx
    if (x.second > mx) {
        mx = x.second;

        // Store the value to check
        // when frequency is same
        ans = x.first;
    }

    // If frequency of 2 element is
    // same than storing minimum value
    else if (x.second == mx) {
        ans = min(ans, x.first);
    }
}

// Return the highest frequency
return ans;

}

// Driver Code int main() { // Given array arr[] int arr[] = { 1, 1, 1, 2, 3, 2, 2, 3, 5, 5, 5, 5, 4, 4, 4, 4, 4 };

// Size of the array
int n = sizeof(arr) / sizeof(arr[0]);

// Function Call
cout << findElement(arr, n) << endl;

}

Java

// Java program for the above approach import java.util.*;

class GFG{

// Function to get the highest // frequency of frequency array static int findElement(int a[], int n) {

// To find the maximum frequency
// initialize it with INT_MIN
int mx = Integer.MIN_VALUE;
int ans = 0;

// Initialize maps to store the
// count of element in array
Map<Integer, Integer> map1 = new HashMap<>(),
                      map2 = new HashMap<>();

// Store the frequency of
// element of array in map1
for(int i = 0; i < n; i++)
{
    map1.put(a[i], map1.getOrDefault(a[i], 0) + 1);
}

// Storing the frequency i.e.,
// (x.second) which is count of
// element in array
for(Integer x : map1.values()) 
{
    map2.put(x, map2.getOrDefault(x, 0) + 1);
}

for(Map.Entry<Integer, Integer> x : map2.entrySet())
{
    
    // Check if the frequency of
    // element is greater than mx
    if (x.getValue() > mx)
    {
        mx = x.getValue();

        // Store the value to check
        // when frequency is same
        ans = x.getKey();
    }

    // If frequency of 2 element is
    // same than storing minimum value
    else if (x.getValue() == mx)
    {
        ans = Math.min(ans, x.getKey());
    }
}

// Return the highest frequency
return ans;

}

// Driver code public static void main (String[] args) {

// Given array arr[]
int arr[] = { 1, 1, 1, 2, 3, 2, 2, 3, 5,
              5, 5, 5, 4, 4, 4, 4, 4 };

// Size of the array
int n = arr.length;

// Function call
System.out.println(findElement(arr, n));

} }

// This code is contributed by offbeat

Python3

Python3 program for the above approach

import sys

Function to get the highest

frequency of frequency array

def findElement(a, n):

# To find the maximum frequency
# initialize it with INT_MIN
mx = -sys.maxsize - 1
ans = 0

# Initialize maps to store the
# count of element in array
map1 = {}
map2 = {}

# Store the frequency of
# element of array in map1
for i in a:
    map1[i] = map1.get(i, 0) + 1

# Storing the frequency i.e.,
# (x.second) which is count of
# element in array
for x in map1:
    map2[map1[x]] = map2.get(map1[x], 0) + 1

for x in map2:

    # Check if the frequency of
    # element is greater than mx
    if (map2[x] > mx):
        mx = map2[x]

        # Store the value to check
        # when frequency is same
        ans = x

    # If frequency of 2 element is
    # same than storing minimum value
    elif (map2[x] == mx):
        ans = min(ans, x)

# Return the highest frequency
return ans

Driver Code

if name == 'main':

# Given array arr[]
arr = [ 1, 1, 1, 2, 3, 2, 2, 3, 
        5, 5, 5, 5, 4, 4, 4, 4, 4]

# Size of the array
n = len(arr)

# Function call
print(findElement(arr, n))

This code is contributed by mohit kumar 29

C#

// C# program for the above approach using System; using System.Collections.Generic;

class GFG{

// Function to get the highest // frequency of frequency array static int findElement(int[] a, int n) {

// To find the maximum frequency
// initialize it with INT_MIN
int mx = Int32.MinValue;
int ans = 0;

// Initialize maps to store the
// count of element in array
Dictionary<int,
           int> map1 = new Dictionary<int, 
                                      int>(),
                map2 = new Dictionary<int, 
                                      int>();

// Store the frequency of
// element of array in map1
for(int i = 0; i < n; i++)
{  
    if (map1.ContainsKey(a[i]))
        map1[a[i]] = map1[a[i]] + 1;
    else
        map1[a[i]] = 1;
}

// Storing the frequency i.e.,
// (x.second) which is count of
// element in array
foreach(KeyValuePair<int, int> xx in map1)
{
    int x = xx.Value;
    if (map2.ContainsKey(x))
        map2[x] = map2[x] + 1;
    else
        map2[x] = 1;
}

foreach(KeyValuePair<int, int> x in map2)
{
    
    // Check if the frequency of
    // element is greater than mx
    if (x.Value > mx)
    {
        mx = x.Value;
        
        // Store the value to check
        // when frequency is same
        ans = x.Key;
    }
    
    // If frequency of 2 element is
    // same than storing minimum value
    else if (x.Value == mx)
    {
        ans = Math.Min(ans, x.Key);
    }
}

// Return the highest frequency
return ans;

}

// Driver code static public void Main () {

// Given array arr[]
int[] arr = { 1, 1, 1, 2, 3, 2, 2, 3, 5,
              5, 5, 5, 4, 4, 4, 4, 4 };

// Size of the array
int n = arr.Length;

// Function call
Console.WriteLine(findElement(arr, n));

} }

// This code is contributed by offbeat

JavaScript

`

Time Complexity: O(N)
Auxiliary Space: O(N)

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