Check if frequency of each element in given array is unique or not
Last Updated :
20 Sep, 2023
Given an array arr[] of N positive integers where the integers are in the range from 1 to N, the task is to check whether the frequency of the elements in the array is unique or not. If all the frequency is unique then print "Yes", else print "No".
Examples:
Input: N = 5, arr[] = {1, 1, 2, 5, 5}
Output: No
Explanation:
The array contains 2 (1's), 1 (2's) and 2 (5's), since the number of frequency of 1 and 5 are the same i.e. 2 times. Therefore, this array does not satisfy the condition.
Input: N = 10, arr[] = {2, 2, 5, 10, 1, 2, 10, 5, 10, 2}
Output: Yes
Explanation:
Number of 1's -> 1
Number of 2's -> 4
Number of 5's -> 2
Number of 10's -> 3.
Since, the number of occurrences of elements present in the array is unique. Therefore, this array satisfy the condition.
Naive Approach: The idea is to check for every number from 1 to N whether it is present in the array or not. If yes, then count the frequency of that element in the array, and store the frequency in an array. At last, just check for any duplicate element in the array and print the output accordingly.
- Iterate over every number in the range from 1 to N
- Counting the frequency of each element in frequency[] array
- Iterate over the frequency array
- Checking if the frequency[] array contains any duplicates or not
- If any duplicate frequency is found then return false.
- If no duplicate frequency is found, then return true
Below is the implementation of the above approach:
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check whether the
// frequency of elements in array
// is unique or not.
bool checkUniqueFrequency(int arr[], int n)
{
vector<int> frequency(n + 1);
// For counting the frequency of each element
for (int i = 1; i <= n; i++) {
for (int j = 0; j < n; j++) {
if (arr[j] == i) {
frequency[i - 1]++;
}
}
}
// Checking if frequency array contains any duplicate
// or not
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == j || frequency[i] == 0)
continue;
if (frequency[i] == frequency[j]) {
// If any duplicate frequency then return
// false
return false;
}
}
}
// If no duplicate frequency found, then return true
return true;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
int n = sizeof arr / sizeof arr[0];
// Function Call
bool res = checkUniqueFrequency(arr, n);
// Print the result
if (res)
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
// Function to check whether the
// frequency of elements in array
// is unique or not.
static boolean checkUniqueFrequency(int arr[], int n)
{
int[] frequency = new int[n + 1];
// For counting the frequency of each element
for (int i = 1; i <= n; i++) {
for (int j = 0; j < n; j++) {
if (arr[j] == i) {
frequency[i - 1]++;
}
}
}
// Checking if frequency array contains any duplicate
// or not
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == j || frequency[i] == 0)
continue;
if (frequency[i] == frequency[j]) {
// If any duplicate frequency then return
// false
return false;
}
}
}
// If no duplicate frequency found, then return true
return true;
}
public static void main (String[] args) {
// Given array arr[]
int arr[] = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
int n = arr.length;
// Function Call
boolean res = checkUniqueFrequency(arr, n);
// Print the result
if (res)
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by aadityaburujwale.
# Python code for the above approach
# Function to check whether the
# frequency of elements in array
# is unique or not.
def checkUniqueFrequency(arr, n):
frequency=[0]*(n + 1);
# For counting the frequency of each element
for i in range(1,n+1):
for j in range(0,n):
if (arr[j] == i):
frequency[i - 1]+=1;
# Checking if frequency array contains any duplicate
# or not
for i in range(0, n):
for j in range(0, n):
if (i == j or frequency[i] == 0):
continue;
if (frequency[i] == frequency[j]):
# If any duplicate frequency then return
# false
return False;
# If no duplicate frequency found, then return true
return True;
# Driver Code
# Given array arr[]
arr = [ 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 ];
n = len(arr);
# Function Call
res = checkUniqueFrequency(arr, n);
# Print the result
if (res):
print("Yes");
else:
print("No");
using System;
public class GFG {
static bool CheckUniqueFrequency(int[] arr, int n)
{
int[] frequency = new int[n + 1];
// For counting the frequency of each element
for (int i = 1; i <= n; i++) {
for (int j = 0; j < n; j++) {
if (arr[j] == i) {
frequency[i - 1]++;
}
}
}
// Checking if frequency array contains any
// duplicate or not
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == j || frequency[i] == 0)
continue;
if (frequency[i] == frequency[j]) {
// If any duplicate frequency then
// return false
return false;
}
}
}
// If no duplicate frequency found, then return true
return true;
}
static void Main(string[] args)
{
// Given array arr[]
int[] arr = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
int n = arr.Length;
// Function Call
bool res = CheckUniqueFrequency(arr, n);
// Print the result
if (res)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
<script>
// Function to check whether the
// frequency of elements in array
// is unique or not.
function checkUniqueFrequency(arr, n)
{
var frequency = Array(n + 1).fill(0);
// For counting the frequency of each element
for (var i = 1; i <= n; i++)
{
for (var j = 0; j < n; j++)
{
if (arr[j] == i)
{
frequency[i - 1]++;
}
}
}
// Checking if frequency array contains any duplicate
// or not
for (var i = 0; i < n; i++)
{
for (var j = 0; j < n; j++)
{
if (i == j || frequency[i] == 0)
{
continue;
}
if (frequency[i] == frequency[j])
{
// If any duplicate frequency then return
// false
return false;
}
}
}
// If no duplicate frequency found, then return true
return true;
}
// Given array arr[]
let arr = [ 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 ];
let n = arr.length;
// Function call
let res = checkUniqueFrequency(arr, n);
// Print the result
if (res)
document.write("Yes");
else
document.write("No");
</script>
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: The idea is to use Hashing. Below are the steps:
- Traverse the given array arr[] and store the frequency of each element in a Map.
- Now traverse the map and check if the count of any element occurred more than once.
- If the count of any element in the above steps is more than one then print “No”, else print “Yes”.
Below is the implementation of the above approach:
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check whether the
// frequency of elements in array
// is unique or not.
bool checkUniqueFrequency(int arr[],
int n)
{
// Freq map will store the frequency
// of each element of the array
unordered_map<int, int> freq;
// Store the frequency of each
// element from the array
for (int i = 0; i < n; i++) {
freq[arr[i]]++;
}
unordered_set<int> uniqueFreq;
// Check whether frequency of any
// two or more elements are same
// or not. If yes, return false
for (auto& i : freq) {
if (uniqueFreq.count(i.second))
return false;
else
uniqueFreq.insert(i.second);
}
// Return true if each
// frequency is unique
return true;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 1, 2, 5, 5 };
int n = sizeof arr / sizeof arr[0];
// Function Call
bool res = checkUniqueFrequency(arr, n);
// Print the result
if (res)
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
// Java code for the above approach
import java.util.*;
class GFG{
// Function to check whether the
// frequency of elements in array
// is unique or not.
static boolean checkUniqueFrequency(int arr[],
int n)
{
// Freq map will store the frequency
// of each element of the array
HashMap<Integer,
Integer> freq = new HashMap<Integer,
Integer>();
// Store the frequency of each
// element from the array
for(int i = 0; i < n; i++)
{
if(freq.containsKey(arr[i]))
{
freq.put(arr[i], freq.get(arr[i]) + 1);
}else
{
freq.put(arr[i], 1);
}
}
HashSet<Integer> uniqueFreq = new HashSet<Integer>();
// Check whether frequency of any
// two or more elements are same
// or not. If yes, return false
for(Map.Entry<Integer,
Integer> i : freq.entrySet())
{
if (uniqueFreq.contains(i.getValue()))
return false;
else
uniqueFreq.add(i.getValue());
}
// Return true if each
// frequency is unique
return true;
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 1, 1, 2, 5, 5 };
int n = arr.length;
// Function call
boolean res = checkUniqueFrequency(arr, n);
// Print the result
if (res)
System.out.print("Yes" + "\n");
else
System.out.print("No" + "\n");
}
}
// This code is contributed by PrinciRaj1992
# Python3 code for
# the above approach
from collections import defaultdict
# Function to check whether the
# frequency of elements in array
# is unique or not.
def checkUniqueFrequency(arr, n):
# Freq map will store the frequency
# of each element of the array
freq = defaultdict (int)
# Store the frequency of each
# element from the array
for i in range (n):
freq[arr[i]] += 1
uniqueFreq = set([])
# Check whether frequency of any
# two or more elements are same
# or not. If yes, return false
for i in freq:
if (freq[i] in uniqueFreq):
return False
else:
uniqueFreq.add(freq[i])
# Return true if each
# frequency is unique
return True
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [1, 1, 2, 5, 5]
n = len(arr)
# Function Call
res = checkUniqueFrequency(arr, n)
# Print the result
if (res):
print ("Yes")
else:
print ("No")
# This code is contributed by Chitranayal
// C# code for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check whether the
// frequency of elements in array
// is unique or not.
static bool checkUniqueFrequency(int []arr,
int n)
{
// Freq map will store the frequency
// of each element of the array
Dictionary<int,
int> freq = new Dictionary<int,
int>();
// Store the frequency of each
// element from the array
for(int i = 0; i < n; i++)
{
if(freq.ContainsKey(arr[i]))
{
freq[arr[i]] = freq[arr[i]] + 1;
}else
{
freq.Add(arr[i], 1);
}
}
HashSet<int> uniqueFreq = new HashSet<int>();
// Check whether frequency of any
// two or more elements are same
// or not. If yes, return false
foreach(KeyValuePair<int,
int> i in freq)
{
if (uniqueFreq.Contains(i.Value))
return false;
else
uniqueFreq.Add(i.Value);
}
// Return true if each
// frequency is unique
return true;
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 1, 1, 2, 5, 5 };
int n = arr.Length;
// Function call
bool res = checkUniqueFrequency(arr, n);
// Print the result
if (res)
Console.Write("Yes" + "\n");
else
Console.Write("No" + "\n");
}
}
// This code is contributed by sapnasingh4991
<script>
// Javascript code for the above approach
// Function to check whether the
// frequency of elements in array
// is unique or not.
function checkUniqueFrequency(arr, n)
{
// Freq map will store the frequency
// of each element of the array
let freq = new Map();
// Store the frequency of each
// element from the array
for(let i = 0; i < n; i++)
{
if (freq.has(arr[i]))
{
freq.set(arr[i],
freq.get(arr[i]) + 1);
}
else
{
freq.set(arr[i], 1);
}
}
let uniqueFreq = new Set();
// Check whether frequency of any
// two or more elements are same
// or not. If yes, return false
for(let [key, value] of freq.entries())
{
if (uniqueFreq.has(value))
return false;
else
uniqueFreq.add(value);
}
// Return true if each
// frequency is unique
return true;
}
// Driver Code
// Given array arr[]
let arr = [ 1, 1, 2, 5, 5 ];
let n = arr.length;
// Function call
let res = checkUniqueFrequency(arr, n);
// Print the result
if (res)
document.write("Yes" + "<br>");
else
document.write("No" + "<br>");
// This code is contributed by avanitrachhadiya2155
</script>
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N)
Another Approach (set):
We can traverse the array and count the frequency of each element using a hash map. Then, we can insert the frequencies into a set and check if the size of the set is equal to the number of distinct frequencies. If yes, then all the frequencies are unique, otherwise not.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool checkUniqueFrequency(int arr[], int n)
{
unordered_map<int, int> freq;
set<int> freqSet;
for(int i = 0; i < n; i++)
freq[arr[i]]++;
for(auto it : freq)
freqSet.insert(it.second);
return (freqSet.size() == freq.size());
}
int main()
{
int arr[] = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
int n = sizeof arr / sizeof arr[0];
bool res = checkUniqueFrequency(arr, n);
if (res)
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
import java.util.*;
public class Main {
// Function to check the unique frequency of the array
public static boolean checkUniqueFrequency(int[] arr)
{
// Create an empty HashMap to store element
// frequencies
Map<Integer, Integer> freq = new HashMap<>();
// Create an empty HashSet to store unique
// frequencies
Set<Integer> freqSet = new HashSet<>();
// Loop through each element in the array
for (int i : arr) {
// Increment element frequency in HashMap
freq.put(i, freq.getOrDefault(i, 0) + 1);
}
// Loop through each frequency in HashMap
for (int f : freq.values()) {
// Add frequency to HashSet
freqSet.add(f);
}
// Return true if number of unique frequencies
// equals number of element frequencies
return freqSet.size() == freq.size();
}
public static void main(String[] args)
{
int[] arr = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
boolean res = checkUniqueFrequency(arr);
if (res) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}
# Python program for the above approach
# Function to check the unique frequency
# of the array
def checkUniqueFrequency(arr):
freq = {}
freqSet = set()
for i in arr:
freq[i] = freq.get(i, 0) + 1
for f in freq.values():
freqSet.add(f)
return len(freqSet) == len(freq)
# Driver Code
arr = [2, 2, 5, 10, 1, 2, 10, 5, 10, 2]
res = checkUniqueFrequency(arr)
if res:
print("Yes")
else:
print("No")
using System;
using System.Collections.Generic;
public class GFG
{
// Function to check the unique frequency of the array
public static bool CheckUniqueFrequency(int[] arr)
{
// Create an empty Dictionary to store element frequencies
Dictionary<int, int> freq = new Dictionary<int, int>();
// Create an empty HashSet to store unique frequencies
HashSet<int> freqSet = new HashSet<int>();
// Loop through each element in the array
foreach (int i in arr)
{
// Increment element frequency in Dictionary
if (freq.ContainsKey(i))
{
freq[i]++;
}
else
{
freq[i] = 1;
}
}
// Loop through each frequency in Dictionary
foreach (int f in freq.Values)
{
// Add frequency to HashSet
freqSet.Add(f);
}
// Return true if number of unique frequencies
// equals number of element frequencies
return freqSet.Count == freq.Count;
}
public static void Main(string[] args)
{
int[] arr = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
bool res = CheckUniqueFrequency(arr);
if (res)
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by shivamgupta310570
// Function to check the unique frequency of the array
function checkUniqueFrequency(arr) {
let freq = {}; // Create an empty object to store frequencies
let freqSet = new Set(); // Create an empty Set to store unique frequencies
for (let i = 0; i < arr.length; i++) {
freq[arr[i]] = (freq[arr[i]] || 0) + 1; // Count the frequencies
}
for (let f in freq) {
freqSet.add(freq[f]); // Add frequencies to the set
}
// Check if set size is equal to number of unique elements
return freqSet.size === Object.keys(freq).length;
}
// Driver Code
const arr = [2, 2, 5, 10, 1, 2, 10, 5, 10, 2];
const res = checkUniqueFrequency(arr);
if (res) {
console.log("Yes");
} else {
console.log("No");
}
Time Complexity: O(nlogn) due to the insertion operation in the set.
Auxiliary Space: O(N)
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