Open In App

Maximum element in an array which is equal to its frequency

Improve
Improve
Like Article
Like
Save
Share
Report

Given an array of integers arr[] of size N, the task is to find the maximum element in the array whose frequency equals to it’s value 
Examples: 
 

Input: arr[] = {3, 2, 2, 3, 4, 3} 
Output:
Frequency of element 2 is 2 
Frequency of element 3 is 3 
Frequency of element 4 is 1 
2 and 3 are elements which have same frequency as it’s value and 3 is the maximum.
Input: arr[] = {1, 2, 3, 4, 5, 6} 
Output:
 

 

Approach: Store the frequency of every element of the array using the map, and finally find out the maximum of those element whose frequency is equal to their value.

Algorithm:

Step 1: Start
Step 2: Create a static function with an int return type name “find_maxm” which takes an array and an integer value as input                      parameter.
             a. Create a map “mpp” of integer type to store the frequency of each element in the array arr.
             b. start a for loop and traverse through i = 0 to n-1
             c. For each element in the array, increment the frequency count in the map “mpp”.
Step 3: Create an int variable “ans” and initialize it with 0.
Step 4: Traverse the map “mpp” using a for-each loop.
             a. Set x.first to value and x.second to freq for each key-value pair x in the map.
             b. Verify that the value and frequency are equivalent.
             c. If the answer is yes, see if it exceeds the current value of ans. Update ans with the value if the answer is yes.
             d. Give the ans value back.
Step 5: End
Below is the implementation of the above approach:
 

CPP




// C++ program to find the maximum element
// whose frequency equals to it’s value
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum element
// whose frequency equals to it’s value
int find_maxm(int arr[], int n)
{
    // Hash map for counting frequency
    map<int, int> mpp;
 
    for (int i = 0; i < n; i++) {
        // Counting freq of each element
        mpp[arr[i]] += 1;
    }
 
    int ans = 0;
    for (auto x : mpp)
    {
        int value = x.first;
        int freq = x.second;
         
        // Check if value equals to frequency
        // and it is the maximum element or not
        if (value == freq) {
            ans = max(ans, value);
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 2, 2, 3, 4, 3 };
     
    // Size of array
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Function call
    cout << find_maxm(arr, n);
 
    return 0;
}


Java




// Java program to find the maximum element
// whose frequency equals to it’s value
import java.util.*;
 
class GFG{
  
// Function to find the maximum element
// whose frequency equals to it’s value
static int find_maxm(int arr[], int n)
{
    // Hash map for counting frequency
    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
  
    for (int i = 0; i < n; i++) {
         
        // Counting freq of each element
        if(mp.containsKey(arr[i])){
            mp.put(arr[i], mp.get(arr[i])+1);
        }else{
            mp.put(arr[i], 1);
    }
    }
  
    int ans = 0;
    for (Map.Entry<Integer,Integer> x : mp.entrySet())
    {
        int value = x.getKey();
        int freq = x.getValue();
          
        // Check if value equals to frequency
        // and it is the maximum element or not
        if (value == freq) {
            ans = Math.max(ans, value);
        }
    }
  
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 2, 3, 4, 3 };
      
    // Size of array
    int n = arr.length;
      
    // Function call
    System.out.print(find_maxm(arr, n));
  
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python3 program to find the maximum element
# whose frequency equals to it’s value
 
# Function to find the maximum element
# whose frequency equals to it’s value
def find_maxm(arr, n) :
 
    # Hash map for counting frequency
    mpp = {}
 
    for i in range(0,n):
        # Counting freq of each element
        if(arr[i] in mpp):
            mpp.update( {arr[i] : mpp[arr[i]] + 1} )
        else:
            mpp[arr[i]] = 1
 
    ans = 0
    for value,freq in mpp.items():
        # Check if value equals to frequency
        # and it is the maximum element or not
        if (value == freq):
            ans = max(ans, value)
 
    return ans
 
# Driver code
arr = [ 3, 2, 2, 3, 4, 3 ]
     
# Size of array
n = len(arr)
 
# Function call
print(find_maxm(arr, n))
 
# This code is contributed by Sanjit_Prasad


C#




// C# program to find the maximum element
// whose frequency equals to it’s value
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to find the maximum element
// whose frequency equals to it’s value
static int find_maxm(int []arr, int n)
{
    // Hash map for counting frequency
    Dictionary<int,int> mp = new Dictionary<int,int>();
   
    for (int i = 0; i < n; i++) {
          
        // Counting freq of each element
        if(mp.ContainsKey(arr[i])){
            mp[arr[i]] = mp[arr[i]]+1;
        }else{
            mp.Add(arr[i], 1);
    }
    }
   
    int ans = 0;
    foreach (KeyValuePair<int,int> x in mp)
    {
        int value = x.Key;
        int freq = x.Value;
           
        // Check if value equals to frequency
        // and it is the maximum element or not
        if (value == freq) {
            ans = Math.Max(ans, value);
        }
    }
   
    return ans;
}
   
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 2, 3, 4, 3 };
       
    // Size of array
    int n = arr.Length;
       
    // Function call
    Console.Write(find_maxm(arr, n));
}
}
 
// This code is contributed by Rajput-Ji


Javascript




<script>
 
// Javascript program to find the maximum element
// whose frequency equals to it’s value
 
// Function to find the maximum element
// whose frequency equals to it’s value
function find_maxm(arr, n)
{
    // Hash map for counting frequency
    var mpp = new Map();
 
    for (var i = 0; i < n; i++)
    {
     
        // Counting freq of each element
        if(mpp.has(arr[i]))
            mpp.set(arr[i], mpp.get(arr[i])+1)
        else
            mpp.set(arr[i], 1)
    }
 
    var ans = 0;
    mpp.forEach((value, key) => {
         
        var value = value;
        var freq = key;
         
        // Check if value equals to frequency
        // and it is the maximum element or not
        if (value == freq) {
            ans = Math.max(ans, value);
        }
    });
 
    return ans;
}
 
// Driver code
var arr = [3, 2, 2, 3, 4, 3 ];
 
// Size of array
var n = arr.length;
 
// Function call
document.write( find_maxm(arr, n));
 
// This code is contributed by famously.
</script>


Output

3







Time Complexity : O(N) 

Space Complexity : O(N) for Hash Map

where N is length of array

Approach : Using Binary Search

The idea of using Binary Search is that we can use the indices values of elements after sorting the array.

If the difference of indices is equal to its value then we can say that the no of appearances of the element is equal to its value.

Below is the implementation of the above idea.

C++




#include <iostream>
#include <algorithm>
using namespace std;
 
int binarySearch(int arr[], int target, int left, int right) {
    int index = -1;
 
    while (left <= right) {
        int middle = left + (right - left) / 2;
 
        // if found, keep searching forward to the last occurrence,
        // otherwise, search to the left
        if (arr[middle] == target) {
            index = middle;
            left = middle + 1;
        } else {
            right = middle - 1;
        }
    }
 
    return index;
}
 
int find_max(int arr[], int n) {
    sort(arr, arr + n);
 
    int max_num = -1;
    int i = 0;
 
    while (i < n) {
        int current = arr[i];
 
        // search the last occurrence from the current index
        int j = binarySearch(arr, current, i, n - 1);
 
        // if the number of occurrences of the current element equals the
        // current element itself, update the lucky integer
        if ((j - i) + 1 == current) {
            max_num = current;
        }
 
        // move index to the next different element
        i = j + 1;
    }
 
    return max_num;
}
 
// Driver code
int main() {
    int arr[] = { 3, 2, 2, 3, 4, 3 };
 
    // Size of array
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << find_max(arr, n);
 
    return 0;
}


Java




// Java program to find the maximum element
// whose frequency equals to it’s value
import java.util.*;
 
class GFG {
 
    // Function to find the maximum element
    // whose frequency equals to it’s value
    public static int find_max(int[] arr,int n) {
        Arrays.sort(arr);
         
        int max_num = -1;
        int i = 0;
         
        while(i < arr.length){
            int current = arr[i];
             
            // search the last occurence from the current index
            int j = binarySearch(arr, current, i, arr.length - 1);
             
            // if the number of occurences of current element equal the
            //current element itself, update the lucky integer
            if((j - i) + 1 == current){
                max_num = current;
            }
 
            // move index to the next different element
            i = j + 1;
        }
         
        return max_num;
    }
     
    public static int binarySearch(int[] arr, int target, int left, int right){
        int index = -1;
         
        while(left <= right){
            int middle = left + (right - left) / 2;
 
            // if found, keep searching fwd to the last occurence,
            //otherwise, search to the left
            if(arr[middle] == target){
                index = middle;
                left = middle + 1;
            } else {
                right = middle - 1;
            }
        }
         
        return index;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 3, 2, 2, 3, 4, 3 };
 
        // Size of array
        int n = arr.length;
 
        // Function call
        System.out.print(find_max(arr, n));
    }
}
 
// This code is contributed by aeroabrar_31


Python3




def binarySearch(arr, target, left, right):
    index = -1
 
    while left <= right:
        middle = left + (right - left) // 2
 
        # if found, keep searching forward to the last occurrence,
        # otherwise, search to the left
        if arr[middle] == target:
            index = middle
            left = middle + 1
        else:
            right = middle - 1
 
    return index
 
 
def find_max(arr, n):
    arr.sort()
 
    max_num = -1
    i = 0
 
    while i < n:
        current = arr[i]
 
        # search the last occurrence from the current index
        j = binarySearch(arr, current, i, n - 1)
 
        # if the number of occurrences of the current element equals the
        # current element itself, update the lucky integer
        if (j - i) + 1 == current:
            max_num = current
 
        # move index to the next different element
        i = j + 1
 
    return max_num
 
 
# Driver code
if __name__ == "__main__":
    arr = [3, 2, 2, 3, 4, 3]
 
    # Size of array
    n = len(arr)
 
    # Function call
    print(find_max(arr, n))


C#




using System;
 
class Program
{
    static void Main(string[] args)
    {
        int[] arr = { 3, 2, 2, 3, 4, 3 };
 
        // Function call
        Console.WriteLine(FindMax(arr));
    }
 
    static int FindMax(int[] arr)
    {
        // Sort the array in ascending order
        Array.Sort(arr);
 
        int maxNum = -1;
        int i = 0;
 
        while (i < arr.Length)
        {
            int current = arr[i];
 
            // Search for the last occurrence of the current element
            int j = BinarySearch(arr, current, i, arr.Length - 1);
 
            // If the number of occurrences of the current element equals
            // the current element itself, update the lucky integer
            if ((j - i) + 1 == current)
            {
                maxNum = current;
            }
 
            // Move the index to the next different element
            i = j + 1;
        }
 
        return maxNum;
    }
 
    static int BinarySearch(int[] arr, int target, int left, int right)
    {
        int index = -1;
 
        while (left <= right)
        {
            int middle = left + (right - left) / 2;
 
            // If found, keep searching forward to the last occurrence,
            // otherwise, search to the left
            if (arr[middle] == target)
            {
                index = middle;
                left = middle + 1;
            }
            else
            {
                right = middle - 1;
            }
        }
 
        return index;
    }
}


Javascript




function findMax(arr) {
    arr.sort((a, b) => a - b);
     
    let maxNum = -1;
    let i = 0;
     
    while (i < arr.length) {
        const current = arr[i];
         
        // Search the last occurrence from the current index
        const j = binarySearch(arr, current, i, arr.length - 1);
         
        // If the number of occurrences of the current element equals the
        // current element itself, update the lucky integer
        if ((j - i) + 1 === current) {
            maxNum = current;
        }
         
        // Move index to the next different element
        i = j + 1;
    }
     
    return maxNum;
}
 
function binarySearch(arr, target, left, right) {
    let index = -1;
     
    while (left <= right) {
        const middle = left + Math.floor((right - left) / 2);
         
        // If found, keep searching forward to the last occurrence,
        // otherwise, search to the left
        if (arr[middle] === target) {
            index = middle;
            left = middle + 1;
        } else {
            right = middle - 1;
        }
    }
     
    return index;
}
 
// Driver code
const arr = [3, 2, 2, 3, 4, 3];
 
// Function call
console.log(findMax(arr));


Output

3







Time Complexity : O(N logN)

Space Complexity : O(1)

So far, we have reduced the space complexity from linear to constant.



Last Updated : 29 Oct, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads