Given an array, find elements that appear a prime number of times in the array with a minimum k frequency (frequency >= k).
Examples :
Input : int[] arr = { 11, 11, 11, 23, 11, 37, 51,
37, 37, 51, 51, 51, 51 };
k = 2
Output : 37, 51
Explanation :
11's count is 4, 23 count 1, 37 count 3, 51 count 5.
37 and 51 are two number that appear prime number of
time and frequencies greater than or equal to k.
Input : int[] arr = { 11, 22, 33 } min Occurrence = 1
Output : -1
None of the count is prime number of times
Approach :
- Create a Map that holds the number as Key and value as its occurrences in the input array.
- Iterate the Map keys and look for the values corresponding to their keys, return the key which has minimum value fulfilling condition key’s value is a prime number and >= min occurrence provided
as input.
Implementation:
C++
// C++ code to find number // occurring prime number // of times with frequency >= k#include <bits/stdc++.h>using namespace std;// Check if the number of // occurrences are primes// or notbool isPrime(int n){ // Corner case if (n <= 1) return false; // Check from 2 to n-1 for (int i = 2; i < n; i++) if (n % i == 0) return false; return true;}// Function to find number// with prime occurrencesvoid primeOccurrences(int arr[], int k){ unordered_map<int, int> map; // Insert values and // their frequencies for (int i = 0; i < 12; i++) map[arr[i]]++; // Traverse map and find // elements with prime // frequencies and frequency // at least k for (auto x : map) { if (isPrime(x.second) && x.second >= k) cout << x.first << endl; }}// Driver codeint main(){ int arr[] = {11, 11, 11, 23, 11, 37, 37, 51, 51, 51, 51, 51}; int k = 2; primeOccurrences(arr, k); return 0;}// This code is contributed by// Manish Shaw(manishshaw1) |
Java
// Java code to find number occurring prime// number of times with frequency >= kimport java.util.*;public class PrimeNumber { // Function to find number with prime occurrences static void primeOccurrences(int[] arr, int k) { Map<Integer, Integer> map = new HashMap<>(); // Insert values and their frequencies for (int i = 0; i < arr.length; i++) { int val = arr[i]; int freq; if (map.containsKey(val)) { freq = map.get(val); freq++; } else freq = 1; map.put(val, freq); } // Traverse map and find elements with // prime frequencies and frequency at // least k for (Map.Entry<Integer, Integer> entry : map.entrySet()) { int value = entry.getValue(); if (isPrime(value) && value >= k) System.out.println(entry.getKey()); } } // Check if the number of occurrences // are primes or not private static boolean isPrime(int n) { if ((n > 2 && n % 2 == 0) || n == 1) return false; for (int i = 3; i <= (int)Math.sqrt(n); i += 2) { if (n % i == 0) return false; } return true; } // Driver code public static void main(String[] args) { int[] arr = { 11, 11, 11, 23, 11, 37, 37, 51, 51, 51, 51, 51 }; int k = 2; primeOccurrences(arr, k); }} |
Python3
# Python3 code to find number # occurring prime number of# times with frequency >= k# Function to find number # with prime occurrencesdef primeOccurrences(arr, k): map = {} # Insert values and their frequencies for val in arr: freq = 0 if val in map : freq = map[val] freq += 1 else : freq = 1 map[val] = freq # Traverse map and find elements # with prime frequencies and # frequency at least k for entry in map : value = map[entry] if isPrime(value) and value >= k: print(entry)# Check if the number of occurrences# are primes or notdef isPrime(n): if (n > 2 and not n % 2) or n == 1: return False for i in range(3, int(n**0.5 + 1), 2): if not n % i: return False return True# Driver codearr = [ 11, 11, 11, 23, 11, 37, 37, 51, 51, 51, 51, 51 ]k = 2primeOccurrences(arr, k)# This code is contributed by Ansu Kumari. |
C#
// C# code to find number // occurring prime number// of times with frequency >= kusing System;using System.Collections.Generic;class GFG{ // Function to find number // with prime occurrences static void primeOccurrences(int[] arr, int k) { Dictionary<int, int> map = new Dictionary<int, int>(); // Insert values and // their frequencies for (int i = 0; i < arr.Length; i++) { int val = arr[i]; int freq; if (map.ContainsKey(val)) { freq = map[val]; freq++; map.Remove(val); } else freq = 1; map.Add(val, freq); } // Traverse map and find elements // with prime frequencies and // frequency atleast k foreach (KeyValuePair<int, int> pair in map) { int value = pair.Value; if (isPrime(value) && value >= k) Console.WriteLine(pair.Key); } } // Check if the number // of occurrences // are primes or not static bool isPrime(int n) { if ((n > 2 && n % 2 == 0) || n == 1) return false; for (int i = 3; i <= (int)Math.Sqrt(n); i += 2) { if (n % i == 0) return false; } return true; } // Driver code static void Main() { int[] arr = new int[]{11, 11, 11, 23, 11, 37, 37, 51, 51, 51, 51, 51}; int k = 2; primeOccurrences(arr, k); }}// This code is contributed by // Manish Shaw(manishshaw1) |
Javascript
<script>// Javascript code to find number// occurring prime number// of times with frequency >= k// Check if the number of// occurrences are primes// or notfunction isPrime(n) { // Corner case if (n <= 1) return false; // Check from 2 to n-1 for (let i = 2; i < n; i++) if (n % i == 0) return false; return true;}// Function to find number// with prime occurrencesfunction primeOccurrences(arr, k) { let map = new Map(); // Insert values and // their frequencies for (let i = 0; i < arr.length; i++) { let val = arr[i]; let freq; if (map.has(val)) { freq = map.get(val); freq++; } else freq = 1; map.set(val, freq); } // Traverse map and find // elements with prime // frequencies and frequency // at least k for (let x of map) { if (isPrime(x[1]) && x[1] >= k) document.write(x[0] + "<br>"); }}// Driver codelet arr = [11, 11, 11, 23, 11, 37, 37, 51, 51, 51, 51, 51];let k = 2;primeOccurrences(arr, k);// This code is contributed by// gfgking</script> |
Output :
37 51
Time Complexity: O(n), where n is number of elements in the array.
Auxiliary Space: O(n), where n is number of elements in the array.
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