Given a positive integer N, the task is to check if N can be represented as a sum of a Prime Number and a Perfect Square or not. If it is possible to represent N in required form, then print “Yes”. Otherwise, print “No”.
Examples:
Input: N = 27
Output: Yes
Explanation: 27 can be expressed as sum of 2 (prime) and 25 (perfect square).Input: N = 64
Output: No
Naive Approach: The simplest approach to solve the given problem is to store all perfect squares which are less than or equal to N in an array. For every perfect square in the array, say X, check if (N – X) is a prime number or not. If found to be true, then print “Yes”. Otherwise, print “No”.
Algorithm:
- Create a function isPrime(n) that accepts an integer n as input and returns true if the supplied number is prime, otherwise false.
- Inside the isPrime(n) :
- If the given number n is less than or equal to 1, return false as it is not a prime number.
- Return true if the given number n is a prime number and is less than or equal to 3.
- If the number n is not a prime number and is divisible by 2 or 3, return false.
- Repeat this process for each 6th number in the range [5, sqrt(N)]. Return false if n is discovered to be non-prime. Otherwise, return true.
- Create a function sumOfPrimeSquare(n) that takes an integer n as input and returns nothing.
- Inside sumOfPrimeSquare(n) function.
- Initialize the integer variable i to 0
- Create squares ArrayList to hold all perfect squares with a size smaller than N.
- Place the ideal square in the squares ArrayList while i * i < n.
- Set a boolean variable flag’s initial value to false.
- Go through each perfect square in the squares ArrayList iteratively:
- Put the perfect square difference from n in the difference variable.
- Update the flag to true and exit the loop if the difference is prime.
- Print “Yes” if N is the product of a prime number and a perfect square. Print “No” if not.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if a// number is prime or notbool isPrime(int n){ // Base Cases if (n <= 1) return false; if (n <= 3) return true; // Check if n is divisible by 2 or 3 if (n % 2 == 0 || n % 3 == 0) return false; // Iterate over every 6 number // from the range [5, sqrt(N)] for (int i = 5; i * i <= n; i = i + 6) { // If n is found to be non-prime if (n % i == 0 || n % (i + 2) == 0) { return false; } } // Otherwise, return true return true;}// Function to check if a number can// be represented as the sum of a prime// number and a perfect square or notvoid sumOfPrimeSquare(int n){ int i = 0; // Stores all perfect // squares less than N vector<int> squares; while (i * i < n) { // Store the perfect square // in the array squares.push_back(i * i); i++; } bool flag = false; // Iterate over all perfect squares for (i = 0; i < squares.size(); i++) { // Store the difference of // perfect square from n int difference = n - squares[i]; // If difference is prime if (isPrime(difference)) { // Update flag flag = true; // Break out of the loop break; } } // If N is the sum of a prime // number and a perfect square if (flag) { cout << "Yes"; } else cout << "No";}// Driver Codeint main(){ int N = 27; sumOfPrimeSquare(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to check if a// number is prime or notstatic boolean isPrime(int n){ // Base Cases if (n <= 1) return false; if (n <= 3) return true; // Check if n is divisible by 2 or 3 if (n % 2 == 0 || n % 3 == 0) return false; // Iterate over every 6 number // from the range [5, sqrt(N)] for(int i = 5; i * i <= n; i = i + 6) { // If n is found to be non-prime if (n % i == 0 || n % (i + 2) == 0) { return false; } } // Otherwise, return true return true;}// Function to check if a number can// be represented as the sum of a prime// number and a perfect square or notstatic void sumOfPrimeSquare(int n){ int i = 0; // Stores all perfect // squares less than N ArrayList<Integer> squares = new ArrayList<Integer>(); while (i * i < n) { // Store the perfect square // in the array squares.add(i * i); i++; } boolean flag = false; // Iterate over all perfect squares for(i = 0; i < squares.size(); i++) { // Store the difference of // perfect square from n int difference = n - squares.get(i); // If difference is prime if (isPrime(difference)) { // Update flag flag = true; // Break out of the loop break; } } // If N is the sum of a prime // number and a perfect square if (flag) { System.out.print("Yes"); } else System.out.print("No");}// Driver Codepublic static void main(String[] args){ int N = 27; sumOfPrimeSquare(N);}}// This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approachfrom math import sqrt# Function to check if a# number is prime or notdef isPrime(n): # Base Cases if (n <= 1): return False if (n <= 3): return True # Check if n is divisible by 2 or 3 if (n % 2 == 0 or n % 3 == 0): return False # Iterate over every 6 number # from the range [5, sqrt(N)] for i in range(5, int(sqrt(n)) + 1, 6): # If n is found to be non-prime if (n % i == 0 or n % (i + 2) == 0): return False # Otherwise, return true return True# Function to check if a number can# be represented as the sum of a prime# number and a perfect square or notdef sumOfPrimeSquare(n): i = 0 # Stores all perfect # squares less than N squares = [] while (i * i < n): # Store the perfect square # in the array squares.append(i * i) i += 1 flag = False # Iterate over all perfect squares for i in range(len(squares)): # Store the difference of # perfect square from n difference = n - squares[i] # If difference is prime if (isPrime(difference)): # Update flag flag = True # Break out of the loop break # If N is the sum of a prime # number and a perfect square if (flag): print("Yes") else: print("No")# Driver Codeif __name__ == '__main__': N = 27 sumOfPrimeSquare(N)# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to check if a// number is prime or notstatic bool isPrime(int n){ // Base Cases if (n <= 1) return false; if (n <= 3) return true; // Check if n is divisible by 2 or 3 if (n % 2 == 0 || n % 3 == 0) return false; // Iterate over every 6 number // from the range [5, sqrt(N)] for (int i = 5; i * i <= n; i = i + 6) { // If n is found to be non-prime if (n % i == 0 || n % (i + 2) == 0) { return false; } } // Otherwise, return true return true;}// Function to check if a number can// be represented as the sum of a prime// number and a perfect square or notstatic void sumOfPrimeSquare(int n){ int i = 0; // Stores all perfect // squares less than N List<int> squares = new List<int>(); while (i * i < n) { // Store the perfect square // in the array squares.Add(i * i); i++; } bool flag = false; // Iterate over all perfect squares for (i = 0; i < squares.Count; i++) { // Store the difference of // perfect square from n int difference = n - squares[i]; // If difference is prime if (isPrime(difference)) { // Update flag flag = true; // Break out of the loop break; } } // If N is the sum of a prime // number and a perfect square if (flag) { Console.Write("Yes"); } else Console.Write("No");}// Driver Codepublic static void Main(){ int N = 27; sumOfPrimeSquare(N);}}// This code is contributed by ipg2016107. |
Javascript
<script>// Javascript program for the above approach// Function to check if a// number is prime or notfunction isPrime(n){ // Base Cases if (n <= 1) return false; if (n <= 3) return true; // Check if n is divisible by 2 or 3 if (n % 2 == 0 || n % 3 == 0) return false; // Iterate over every 6 number // from the range [5, sqrt(N)] for(let i = 5; i * i <= n; i = i + 6) { // If n is found to be non-prime if (n % i == 0 || n % (i + 2) == 0) { return false; } } // Otherwise, return true return true;}// Function to check if a number can// be represented as the sum of a prime// number and a perfect square or notfunction sumOfPrimeSquare(n){ let i = 0; // Stores all perfect // squares less than N let squares = []; while (i * i < n) { // Store the perfect square // in the array squares.push(i * i); i++; } let flag = false; // Iterate over all perfect squares for(i = 0; i < squares.length; i++) { // Store the difference of // perfect square from n let difference = n - squares[i]; // If difference is prime if (isPrime(difference)) { // Update flag flag = true; // Break out of the loop break; } } // If N is the sum of a prime // number and a perfect square if (flag) { document.write("Yes"); } else document.write("No");}// Driver Codelet N = 27;sumOfPrimeSquare(N);// This code is contributed by rishavmahato348</script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(?N)
Efficient Approach: The above approach can be optimized by storing all the prime numbers smaller than N in an array, using Sieve of Eratosthenes. If there exists any prime number, say X, check if (N – X) is a perfect square or not. If found to be true, then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to store all prime// numbers less than or equal to Nvoid SieveOfEratosthenes(bool prime[], int n){ // Update prime[0] and // prime[1] as false prime[0] = false; prime[1] = false; // Iterate over the range [2, sqrt(N)] for (int p = 2; p * p <= n; p++) { // If p is a prime if (prime[p] == true) { // Update all multiples of p // which are <= n as non-prime for (int i = p * p; i <= n; i += p) { prime[i] = false; } } }}// Function to check whether a number// can be represented as the sum of a// prime number and a perfect squarevoid sumOfPrimeSquare(int n){ bool flag = false; // Stores all the prime numbers // less than or equal to n bool prime[n + 1]; memset(prime, true, sizeof(prime)); // Update the array prime[] SieveOfEratosthenes(prime, n); // Iterate over the range [0, n] for (int i = 0; i <= n; i++) { // If current number // is non-prime if (!prime[i]) continue; // Update difference int dif = n - i; // If difference is a // perfect square if (ceil((double)sqrt(dif)) == floor((double)sqrt(dif))) { // If true, update flag // and break out of loop flag = true; break; } } // If N can be expressed as sum // of prime number and perfect square if (flag) { cout << "Yes"; } else cout << "No";}// Driver Codeint main(){ int N = 27; sumOfPrimeSquare(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to store all prime// numbers less than or equal to Nstatic void SieveOfEratosthenes(boolean prime[], int n){ // Update prime[0] and // prime[1] as false prime[0] = false; prime[1] = false; // Iterate over the range [2, Math.sqrt(N)] for(int p = 2; p * p <= n; p++) { // If p is a prime if (prime[p] == true) { // Update all multiples of p // which are <= n as non-prime for(int i = p * p; i <= n; i += p) { prime[i] = false; } } }}// Function to check whether a number// can be represented as the sum of a// prime number and a perfect squarestatic void sumOfPrimeSquare(int n){ boolean flag = false; // Stores all the prime numbers // less than or equal to n boolean []prime = new boolean[n + 1]; for(int i = 0; i < prime.length; i++) prime[i] = true; // Update the array prime[] SieveOfEratosthenes(prime, n); // Iterate over the range [0, n] for(int i = 0; i <= n; i++) { // If current number // is non-prime if (!prime[i]) continue; // Update difference int dif = n - i; // If difference is a // perfect square if (Math.ceil((double)Math.sqrt(dif)) == Math.floor((double)Math.sqrt(dif))) { // If true, update flag // and break out of loop flag = true; break; } } // If N can be expressed as sum // of prime number and perfect square if (flag) { System.out.print("Yes"); } else System.out.print("No");}// Driver Codepublic static void main(String[] args){ int N = 27; sumOfPrimeSquare(N);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approachimport math# Function to store all prime# numbers less than or equal to Ndef SieveOfEratosthenes(prime, n): # Update prime[0] and # prime[1] as false prime[0] = False prime[1] = False # Iterate over the range [2, sqrt(N)] for p in range(2, int(n ** (1 / 2))): # If p is a prime if (prime[p] == True): # Update all multiples of p # which are <= n as non-prime for i in range(p ** 2, n + 1, p): prime[i] = False# Function to check whether a number# can be represented as the sum of a# prime number and a perfect squaredef sumOfPrimeSquare(n): flag = False # Stores all the prime numbers # less than or equal to n prime = [True] * (n + 1) # Update the array prime[] SieveOfEratosthenes(prime, n) # Iterate over the range [0, n] for i in range(n + 1): # If current number # is non-prime if (not prime[i]): continue # Update difference dif = n - i # If difference is a # perfect square if (math.ceil(dif ** (1 / 2)) == math.floor(dif ** (1 / 2))): # If true, update flag # and break out of loop flag = True break # If N can be expressed as sum # of prime number and perfect square if (flag): print("Yes") else: print("No")# Driver Codeif __name__ == "__main__": N = 27 sumOfPrimeSquare(N) # This code is contributed by AnkThon |
C#
// C# program for the above approachusing System;class GFG{ // Function to store all prime// numbers less than or equal to Nstatic void SieveOfEratosthenes(bool[] prime, int n){ // Update prime[0] and // prime[1] as false prime[0] = false; prime[1] = false; // Iterate over the range [2, sqrt(N)] for(int p = 2; p * p <= n; p++) { // If p is a prime if (prime[p] == true) { // Update all multiples of p // which are <= n as non-prime for(int i = p * p; i <= n; i += p) { prime[i] = false; } } }}// Function to check whether a number// can be represented as the sum of a// prime number and a perfect squarestatic void sumOfPrimeSquare(int n){ bool flag = false; // Stores all the prime numbers // less than or equal to n bool[] prime = new bool[n + 1]; Array.Fill(prime, true); // Update the array prime[] SieveOfEratosthenes(prime, n); // Iterate over the range [0, n] for(int i = 0; i <= n; i++) { // If current number // is non-prime if (!prime[i]) continue; // Update difference int dif = n - i; // If difference is a // perfect square if (Math.Ceiling((double)Math.Sqrt(dif)) == Math.Floor((double)Math.Sqrt(dif))) { // If true, update flag // and break out of loop flag = true; break; } } // If N can be expressed as sum // of prime number and perfect square if (flag) { Console.WriteLine("Yes"); } else Console.WriteLine("No");}// Driver Codepublic static void Main(){ int N = 27; sumOfPrimeSquare(N);}}// This code is contributed by ukasp |
Javascript
<script>// Javascript program for the above approach// Function to store all prime// numbers less than or equal to Nfunction SieveOfEratosthenes(prime, n){ // Update prime[0] and // prime[1] as false prime[0] = false; prime[1] = false; // Iterate over the range [2, sqrt(N)] for(let p = 2; p * p <= n; p++) { // If p is a prime if (prime[p] == true) { // Update all multiples of p // which are <= n as non-prime for(let i = p * p; i <= n; i += p) { prime[i] = false; } } }}// Function to check whether a number// can be represented as the sum of a// prime number and a perfect squarefunction sumOfPrimeSquare(n){ let flag = false; // Stores all the prime numbers // less than or equal to n let prime = new Array(n + 1).fill(true); // Update the array prime[] SieveOfEratosthenes(prime, n); // Iterate over the range [0, n] for(let i = 0; i <= n; i++) { // If current number // is non-prime if (!prime[i]) continue; // Update difference let dif = n - i; // If difference is a // perfect square if (Math.ceil(Math.sqrt(dif)) == Math.floor(Math.sqrt(dif))) { // If true, update flag // and break out of loop flag = true; break; } } // If N can be expressed as sum // of prime number and perfect square if (flag) { document.write("Yes"); } else document.write("No");}// Driver Codelet N = 27;sumOfPrimeSquare(N);// This code is contributed by subhammahato348</script> |
Output:
Yes
Time Complexity: O(N * log(log(N)))
Auxiliary Space: O(N)
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