Maximum element in an array which is equal to its frequency
Last Updated :
29 Oct, 2023
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: 3
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: 1
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 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 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# 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
<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>
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 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
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))
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;
}
}
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));
Time Complexity : O(N logN)
Space Complexity : O(1)
So far, we have reduced the space complexity from linear to constant.
Similar Reads
Find maximum element among the elements with minimum frequency in given Array
Given an array arr[] consisting of N integers, the task is to find the maximum element with the minimum frequency. Examples: Input: arr[] = {2, 2, 5, 50, 1}Output: 50Explanation:The element with minimum frequency is {1, 5, 50}. The maximum element among these element is 50. Input: arr[] = {3, 2, 5,
13 min read
Minimum cost to make all elements with same frequency equal
Given an array arr[] of integers, the task is to find the minimum cost for making all the integers that have the same frequency equal. By performing one operation you can either increase the current integer by 1 or decrease it by 1. Examples: Input: arr[] = {1, 2, 3, 2, 6, 5, 6}Output: 12Explanation
9 min read
Find element in a sorted array whose frequency is greater than or equal to n/2.
Given a sorted array of length n, find the number in array that appears more than or equal to n/2 times. It is given that such element always exists. Examples: Input : 2 3 3 4 Output : 3 Input : 3 4 5 5 5 Output : 5 Input : 1 1 1 2 3 Output : 1 To find that number, we traverse the array and check th
3 min read
Maximum frequency of any array element possible by at most K increments
Given an array arr[] of size N and an integer K, the task is to find the maximum possible frequency of any array element by at most K increments. Examples: Input: arr[] = {1, 4, 8, 13}, N = 4, K = 5 Output: 2 Explanation: Incrementing arr[0] twice modifies arr[] to {4, 4, 8, 13}. Maximum frequency =
15+ min read
Find the array element having maximum frequency of the digit K
Given an array arr[] of size N and an integer K, the task is to find an array element that contains the digit K a maximum number of times. If more than one solutions exist, then print any one of them. Otherwise, print -1. Examples: Input: arr[] = {3, 77, 343, 456}, K = 3 Output: 343 Explanation: Fre
15 min read
Find element with highest frequency in given nested Array
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
8 min read
Sum of all maximum frequency elements in Matrix
Given a NxM matrix of integers containing duplicate elements. The task is to find the sum of all maximum occurring elements in the given matrix. That is the sum of all such elements whose frequency is even in the matrix. Examples: Input : mat[] = {{1, 1, 1}, {2, 3, 3}, {4, 5, 3}} Output : 12 The max
6 min read
Top K Frequent Elements in an Array
Given an array arr[] and a positive integer k, the task is to find the k most frequently occurring elements from a given array.Note: If more than one element has same frequency then prioritise the larger element over the smaller one.Examples: Input: arr= [3, 1, 4, 4, 5, 2, 6, 1], k = 2Output: [4, 1]
15+ min read
Find element with the maximum set bits in an array
Given an array arr[]. The task is to find an element from arr[] which has the maximum count of set bits.Examples: Input: arr[] = {10, 100, 1000, 10000} Output: 1000 Binary(10) = 1010 (2 set bits) Binary(100) = 1100100 (3 set bits) Binary(1000) = 1111101000 (6 set bits) Binary(10000) = 10011100010000
5 min read
Maximum Frequency by replacing with elements in a given range
Given a 0-indexed array arr[] of N positive integers and two integers K and X. Find the maximum frequency of any element(not necessary to be present in the array) you can make after performing the below operation at most K times- Choose an index i and replace arr[i] with any integer from the range [
9 min read