Find maximum element among the elements with minimum frequency in given Array
Last Updated :
23 Jul, 2025
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: 50
Explanation:
The element with minimum frequency is {1, 5, 50}. The maximum element among these element is 50.
Input: arr[] = {3, 2, 5, 6, 1}
Output: 6
Approach: Sort the array in non-decreasing order, which brings elements with the same value together. This allows us to easily count and identify when the frequency of an element changes. When the frequency changes ('currFreq' <= 'minFreq'), we update the minimum frequency and as the array is sorted, the current element (arr[i-1]) becomes the maximum element, representing the highest value within its frequency group.
Follow the steps below to solve the given problem:
- Sort the input array in non-decreasing order.
- Initialize minFreq to array size+1, maxElement to the first element of the sorted array, and currFreq to 1.
- Iterate over the sorted array from the second element to the last element.
- If the current element is the same as the previous element, increment currFreq.
- Otherwise, If currFreq is less than or equal to minFreq, update minFreq with currFreq and update maxElement with the previous element.
- Reset currFreq to 1 for the new element.
- After the loop, check the frequency of the last element and update minFreq and maxElement if necessary.
- Return the maxElement, which represents the maximum element with the minimum frequency in the input array.
C++14
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
int maxElementWithMinFreq(int *arr, int N)
{
sort(arr, arr + N); // Sort the array in non-decreasing order
int minFreq = N + 1; // Initialize minFreq to a value greater than the array size
int maxElement = arr[0]; // Initialize maxElement to the first element of the sorted array
int currFreq = 1; // Keep track of the frequency of the current element
// Iterate over the sorted array
for (int i = 1; i < N; i++)
{
if (arr[i] == arr[i - 1])
{
currFreq++; // Increment the frequency if the current element is the same as the previous element
}
else
{
// If the frequency of the current element is less than minFreq, update minFreq and maxElement
if (currFreq <= minFreq)
{
minFreq = currFreq;
maxElement = arr[i - 1];
}
currFreq = 1; // Reset the frequency for the new element
}
}
// Check the frequency of the last element
if (currFreq <= minFreq)
{
minFreq = currFreq;
maxElement = arr[N - 1];
}
return maxElement; // Return the maximum element with the minimum frequency
}
int main()
{
int arr[] = {2, 2, 5, 50, 1};
int N = sizeof(arr) / sizeof(arr[0]);
cout << maxElementWithMinFreq(arr, N);
return 0;
}
//This code is contributed by Abhishek Kumar
// Java code for the above approach
import java.util.Arrays;
public class gfg {
public static int maxElementWithMinFreq(int[] arr, int N) {
Arrays.sort(arr); // Sort the array in non-decreasing order
int minFreq = N + 1; // Initialize minFreq to a value greater than the array size
int maxElement = arr[0]; // Initialize maxElement to the first element of the sorted array
int currFreq = 1; // Keep track of the frequency of the current element
// Iterate over the sorted array
for (int i = 1; i < N; i++) {
if (arr[i] == arr[i - 1]) {
currFreq++; // Increment the frequency if the current element is the same as the previous element
} else {
// If the frequency of the current element is less than minFreq, update minFreq and maxElement
if (currFreq <= minFreq) {
minFreq = currFreq;
maxElement = arr[i - 1];
}
currFreq = 1; // Reset the frequency for the new element
}
}
// Check the frequency of the last element
if (currFreq <= minFreq) {
minFreq = currFreq;
maxElement = arr[N - 1];
}
return maxElement; // Return the maximum element with the minimum frequency
}
public static void main(String[] args) {
int[] arr = {2, 2, 5, 50, 1};
int N = arr.length;
System.out.println(maxElementWithMinFreq(arr, N));
}
}
# Python code for the above approach
def max_element_with_min_freq(arr):
arr.sort() # Sort the array in non-decreasing order
# Initialize min_freq to a value greater than the array size
min_freq = len(arr) + 1
# Initialize max_element to the first element of the sorted array
max_element = arr[0]
curr_freq = 1 # Keep track of the frequency of the current element
# Iterate over the sorted array
for i in range(1, len(arr)):
if arr[i] == arr[i - 1]:
curr_freq += 1 # Increment the frequency if the current element is the same as the previous element
else:
# If the frequency of the current element is less than min_freq, update min_freq and max_element
if curr_freq <= min_freq:
min_freq = curr_freq
max_element = arr[i - 1]
curr_freq = 1 # Reset the frequency for the new element
# Check the frequency of the last element
if curr_freq <= min_freq:
min_freq = curr_freq
max_element = arr[-1]
return max_element # Return the maximum element with the minimum frequency
arr = [2, 2, 5, 50, 1]
print(max_element_with_min_freq(arr))
using System;
class GFG
{
static int MaxElementWithMinFreq(int[] arr, int N)
{
Array.Sort(arr); // Sort the array in non-decreasing order
int minFreq = N + 1; // Initialize minFreq to a value greater than the array size
int maxElement = arr[0]; // Initialize maxElement to the first element of the sorted array
int currFreq = 1; // Keep track of the frequency of the current element
// Iterate over the sorted array
for (int i = 1; i < N; i++)
{
if (arr[i] == arr[i - 1])
{
currFreq++; // Increment the frequency if the current element is the same as the previous element
}
else
{
// If the frequency of the current element is less than minFreq, update minFreq and maxElement
if (currFreq <= minFreq)
{
minFreq = currFreq;
maxElement = arr[i - 1];
}
currFreq = 1; // Reset the frequency for the new element
}
}
// Check the frequency of the last element
if (currFreq <= minFreq)
{
minFreq = currFreq;
maxElement = arr[N - 1];
}
return maxElement; // Return the maximum element with the minimum frequency
}
static void Main()
{
int[] arr = { 2, 2, 5, 50, 1 };
int N = arr.Length;
Console.WriteLine(MaxElementWithMinFreq(arr, N));
}
}
function maxElementWithMinFreq(arr) {
arr.sort(); // Sort the array in non-decreasing order
let minFreq = arr.length + 1; // Initialize minFreq to a value greater than the array size
let maxElement = arr[0]; // Initialize maxElement to the first element of the sorted array
let currFreq = 1; // Keep track of the frequency of the current element
// Iterate over the sorted array
for (let i = 1; i < arr.length; i++) {
if (arr[i] == arr[i - 1]) {
currFreq++; // Increment the frequency if the current element is the same as the previous element
} else {
// If the frequency of the current element is less than minFreq, update minFreq and maxElement
if (currFreq <= minFreq) {
minFreq = currFreq;
maxElement = arr[i - 1];
}
currFreq = 1; // Reset the frequency for the new element
}
}
// Check the frequency of the last element
if (currFreq <= minFreq) {
minFreq = currFreq;
maxElement = arr[arr.length - 1];
}
return maxElement; // Return the maximum element with the minimum frequency
}
const arr = [2, 2, 5, 50, 1];
console.log(maxElementWithMinFreq(arr));
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
Approach: The given problem can be solved by storing the frequency of the array element in a HashMap and then finding the maximum value having a minimum frequency. Follow the steps below to solve the given problem:
- Store the frequency of each element in a HashMap, say M.
- Initialize two variables, say maxValue as INT_MIN and minFreq as INT_MAX that store the resultant maximum element and stores the minimum frequency among all the frequencies.
- Iterate over the map M and perform the following steps:
- If the frequency of the current element is less than minFreq then update the value of minFreq to the current frequency and the value of maxValue to the current element.
- If the frequency of the current element is equal to the minFreq and the value of maxValue is less than the current value then update the value of maxValue to the current element.
- After completing the above steps, print the value of maxValue as the resultant element.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum element
// with the minimum frequency
int maxElementWithMinFreq(int* arr, int N)
{
// Stores the frequency of array
// elements
unordered_map<int, int> mp;
// Find the frequency and store
// in the map
for (int i = 0; i < N; i++) {
mp[arr[i]]++;
}
// Initialize minFreq to the maximum
// value and minValue to the minimum
int minFreq = INT_MAX;
int maxValue = INT_MIN;
// Traverse the map mp
for (auto x : mp) {
int num = x.first;
int freq = x.second;
// If freq < minFreq, then update
// minFreq to freq and maxValue
// to the current element
if (freq < minFreq) {
minFreq = freq;
maxValue = num;
}
// If freq is equal to the minFreq
// and current element > maxValue
// then update maxValue to the
// current element
else if (freq == minFreq
&& maxValue < num) {
maxValue = num;
}
}
// Return the resultant maximum value
return maxValue;
}
// Driver Code
int main()
{
int arr[] = { 2, 2, 5, 50, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << maxElementWithMinFreq(arr, N);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the maximum element
// with the minimum frequency
static int maxElementWithMinFreq(int[] arr, int N)
{
// Stores the frequency of array
// elements
HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
// Find the frequency and store
// in the map
for (int i = 0; i < N; i++) {
if(mp.containsKey(arr[i])){
mp.put(arr[i], mp.get(arr[i])+1);
}else{
mp.put(arr[i], 1);
}
}
// Initialize minFreq to the maximum
// value and minValue to the minimum
int minFreq = Integer.MAX_VALUE;
int maxValue = Integer.MIN_VALUE;
// Traverse the map mp
for (Map.Entry<Integer,Integer> x : mp.entrySet()){
int num = x.getKey();
int freq = x.getValue();
// If freq < minFreq, then update
// minFreq to freq and maxValue
// to the current element
if (freq < minFreq) {
minFreq = freq;
maxValue = num;
}
// If freq is equal to the minFreq
// and current element > maxValue
// then update maxValue to the
// current element
else if (freq == minFreq
&& maxValue < num) {
maxValue = num;
}
}
// Return the resultant maximum value
return maxValue;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 2, 5, 50, 1 };
int N = arr.length;
System.out.print(maxElementWithMinFreq(arr, N));
}
}
// This code is contributed by shikhasingrajput
# Python 3 program for the above approach
import sys
from collections import defaultdict
# Function to find the maximum element
# with the minimum frequency
def maxElementWithMinFreq(arr, N):
# Stores the frequency of array
# elements
mp = defaultdict(int)
# Find the frequency and store
# in the map
for i in range(N):
mp[arr[i]] += 1
# Initialize minFreq to the maximum
# value and minValue to the minimum
minFreq = sys.maxsize
maxValue = -sys.maxsize-1
# Traverse the map mp
for x in mp:
num = x
freq = mp[x]
# If freq < minFreq, then update
# minFreq to freq and maxValue
# to the current element
if (freq < minFreq):
minFreq = freq
maxValue = num
# If freq is equal to the minFreq
# and current element > maxValue
# then update maxValue to the
# current element
elif (freq == minFreq
and maxValue < num):
maxValue = num
# Return the resultant maximum value
return maxValue
# Driver Code
if __name__ == "__main__":
arr = [2, 2, 5, 50, 1]
N = len(arr)
print(maxElementWithMinFreq(arr, N))
# This code is contributed by ukasp.
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum element
// with the minimum frequency
static int maxElementWithMinFreq(int []arr, int N)
{
// Stores the frequency of array
// elements
Dictionary<int, int> mp = new Dictionary<int,int>();
// Find the frequency and store
// in the map
for (int i = 0; i < N; i++) {
if(mp.ContainsKey(arr[i]))
mp[arr[i]]++;
else
mp.Add(arr[i],1);
}
// Initialize minFreq to the maximum
// value and minValue to the minimum
int minFreq = Int32.MaxValue;
int maxValue = Int32.MinValue;
// Traverse the map mp
foreach(KeyValuePair<int,int> x in mp) {
int num = x.Key;
int freq = x.Value;
// If freq < minFreq, then update
// minFreq to freq and maxValue
// to the current element
if (freq < minFreq) {
minFreq = freq;
maxValue = num;
}
// If freq is equal to the minFreq
// and current element > maxValue
// then update maxValue to the
// current element
else if (freq == minFreq
&& maxValue < num) {
maxValue = num;
}
}
// Return the resultant maximum value
return maxValue;
}
// Driver Code
public static void Main()
{
int []arr = { 2, 2, 5, 50, 1 };
int N = arr.Length;
Console.Write(maxElementWithMinFreq(arr, N));
}
}
// This code is contributed by SURENDRA_GANGWAR.
<script>
// JavaScript Program to implement
// the above approach
// Function to find the maximum element
// with the minimum frequency
function maxElementWithMinFreq(arr, N) {
// Stores the frequency of array
// elements
let mp = new Map();
// Find the frequency and store
// in the map
for (let i = 0; i < N; i++) {
if (mp.has(arr[i])) {
mp.set(arr[i], mp.get(arr[i]) + 1);
}
else {
mp.set(arr[i], 1);
}
}
// Initialize minFreq to the maximum
// value and minValue to the minimum
let minFreq = Number.MAX_VALUE
let maxValue = Number.MIN_VALUE;
// Traverse the map mp
for (let [key, value] of mp) {
let num = key;
let freq = value;
// If freq < minFreq, then update
// minFreq to freq and maxValue
// to the current element
if (freq < minFreq) {
minFreq = freq;
maxValue = num;
}
// If freq is equal to the minFreq
// and current element > maxValue
// then update maxValue to the
// current element
else if (freq == minFreq
&& maxValue < num) {
maxValue = num;
}
}
// Return the resultant maximum value
return maxValue;
}
// Driver Code
let arr = [2, 2, 5, 50, 1];
let N = arr.length;
document.write(maxElementWithMinFreq(arr, N));
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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