Check if a number can be represented as a sum of a Prime Number and a Perfect Square
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++
#include <bits/stdc++.h>
using namespace std;
bool isPrime( int n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n;
i = i + 6) {
if (n % i == 0
|| n % (i + 2) == 0) {
return false ;
}
}
return true ;
}
void sumOfPrimeSquare( int n)
{
int i = 0;
vector< int > squares;
while (i * i < n) {
squares.push_back(i * i);
i++;
}
bool flag = false ;
for (i = 0; i < squares.size(); i++) {
int difference = n - squares[i];
if (isPrime(difference)) {
flag = true ;
break ;
}
}
if (flag) {
cout << "Yes" ;
}
else
cout << "No" ;
}
int main()
{
int N = 27;
sumOfPrimeSquare(N);
return 0;
}
|
Java
import java.util.*;
class GFG{
static boolean isPrime( int n)
{
if (n <= 1 )
return false ;
if (n <= 3 )
return true ;
if (n % 2 == 0 || n % 3 == 0 )
return false ;
for ( int i = 5 ; i * i <= n;
i = i + 6 )
{
if (n % i == 0 || n % (i + 2 ) == 0 )
{
return false ;
}
}
return true ;
}
static void sumOfPrimeSquare( int n)
{
int i = 0 ;
ArrayList<Integer> squares = new ArrayList<Integer>();
while (i * i < n)
{
squares.add(i * i);
i++;
}
boolean flag = false ;
for (i = 0 ; i < squares.size(); i++)
{
int difference = n - squares.get(i);
if (isPrime(difference))
{
flag = true ;
break ;
}
}
if (flag)
{
System.out.print( "Yes" );
}
else
System.out.print( "No" );
}
public static void main(String[] args)
{
int N = 27 ;
sumOfPrimeSquare(N);
}
}
|
Python3
from math import sqrt
def isPrime(n):
if (n < = 1 ):
return False
if (n < = 3 ):
return True
if (n % 2 = = 0 or n % 3 = = 0 ):
return False
for i in range ( 5 , int (sqrt(n)) + 1 , 6 ):
if (n % i = = 0 or n % (i + 2 ) = = 0 ):
return False
return True
def sumOfPrimeSquare(n):
i = 0
squares = []
while (i * i < n):
squares.append(i * i)
i + = 1
flag = False
for i in range ( len (squares)):
difference = n - squares[i]
if (isPrime(difference)):
flag = True
break
if (flag):
print ( "Yes" )
else :
print ( "No" )
if __name__ = = '__main__' :
N = 27
sumOfPrimeSquare(N)
|
C#
using System;
using System.Collections.Generic;
class GFG{
static bool isPrime( int n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n;
i = i + 6) {
if (n % i == 0
|| n % (i + 2) == 0) {
return false ;
}
}
return true ;
}
static void sumOfPrimeSquare( int n)
{
int i = 0;
List< int > squares = new List< int >();
while (i * i < n) {
squares.Add(i * i);
i++;
}
bool flag = false ;
for (i = 0; i < squares.Count; i++) {
int difference = n - squares[i];
if (isPrime(difference)) {
flag = true ;
break ;
}
}
if (flag) {
Console.Write( "Yes" );
}
else
Console.Write( "No" );
}
public static void Main()
{
int N = 27;
sumOfPrimeSquare(N);
}
}
|
Javascript
<script>
function isPrime(n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
if (n % 2 == 0 || n % 3 == 0)
return false ;
for (let i = 5; i * i <= n;
i = i + 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false ;
}
}
return true ;
}
function sumOfPrimeSquare(n)
{
let i = 0;
let squares = [];
while (i * i < n)
{
squares.push(i * i);
i++;
}
let flag = false ;
for (i = 0; i < squares.length; i++)
{
let difference = n - squares[i];
if (isPrime(difference))
{
flag = true ;
break ;
}
}
if (flag)
{
document.write( "Yes" );
}
else
document.write( "No" );
}
let N = 27;
sumOfPrimeSquare(N);
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes( bool prime[],
int n)
{
prime[0] = false ;
prime[1] = false ;
for ( int p = 2; p * p <= n; p++) {
if (prime[p] == true ) {
for ( int i = p * p; i <= n;
i += p) {
prime[i] = false ;
}
}
}
}
void sumOfPrimeSquare( int n)
{
bool flag = false ;
bool prime[n + 1];
memset (prime, true , sizeof (prime));
SieveOfEratosthenes(prime, n);
for ( int i = 0; i <= n; i++) {
if (!prime[i])
continue ;
int dif = n - i;
if ( ceil (( double ) sqrt (dif))
== floor (( double ) sqrt (dif))) {
flag = true ;
break ;
}
}
if (flag) {
cout << "Yes" ;
}
else
cout << "No" ;
}
int main()
{
int N = 27;
sumOfPrimeSquare(N);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void SieveOfEratosthenes( boolean prime[],
int n)
{
prime[ 0 ] = false ;
prime[ 1 ] = false ;
for ( int p = 2 ; p * p <= n; p++)
{
if (prime[p] == true )
{
for ( int i = p * p; i <= n; i += p)
{
prime[i] = false ;
}
}
}
}
static void sumOfPrimeSquare( int n)
{
boolean flag = false ;
boolean []prime = new boolean [n + 1 ];
for ( int i = 0 ; i < prime.length; i++)
prime[i] = true ;
SieveOfEratosthenes(prime, n);
for ( int i = 0 ; i <= n; i++)
{
if (!prime[i])
continue ;
int dif = n - i;
if (Math.ceil(( double )Math.sqrt(dif)) ==
Math.floor(( double )Math.sqrt(dif)))
{
flag = true ;
break ;
}
}
if (flag)
{
System.out.print( "Yes" );
}
else
System.out.print( "No" );
}
public static void main(String[] args)
{
int N = 27 ;
sumOfPrimeSquare(N);
}
}
|
Python3
import math
def SieveOfEratosthenes(prime, n):
prime[ 0 ] = False
prime[ 1 ] = False
for p in range ( 2 , int (n * * ( 1 / 2 ))):
if (prime[p] = = True ):
for i in range (p * * 2 , n + 1 , p):
prime[i] = False
def sumOfPrimeSquare(n):
flag = False
prime = [ True ] * (n + 1 )
SieveOfEratosthenes(prime, n)
for i in range (n + 1 ):
if ( not prime[i]):
continue
dif = n - i
if (math.ceil(dif * * ( 1 / 2 )) = =
math.floor(dif * * ( 1 / 2 ))):
flag = True
break
if (flag):
print ( "Yes" )
else :
print ( "No" )
if __name__ = = "__main__" :
N = 27
sumOfPrimeSquare(N)
|
C#
using System;
class GFG{
static void SieveOfEratosthenes( bool [] prime, int n)
{
prime[0] = false ;
prime[1] = false ;
for ( int p = 2; p * p <= n; p++)
{
if (prime[p] == true )
{
for ( int i = p * p; i <= n; i += p)
{
prime[i] = false ;
}
}
}
}
static void sumOfPrimeSquare( int n)
{
bool flag = false ;
bool [] prime = new bool [n + 1];
Array.Fill(prime, true );
SieveOfEratosthenes(prime, n);
for ( int i = 0; i <= n; i++)
{
if (!prime[i])
continue ;
int dif = n - i;
if (Math.Ceiling(( double )Math.Sqrt(dif)) ==
Math.Floor(( double )Math.Sqrt(dif)))
{
flag = true ;
break ;
}
}
if (flag)
{
Console.WriteLine( "Yes" );
}
else
Console.WriteLine( "No" );
}
public static void Main()
{
int N = 27;
sumOfPrimeSquare(N);
}
}
|
Javascript
<script>
function SieveOfEratosthenes(prime, n)
{
prime[0] = false ;
prime[1] = false ;
for (let p = 2; p * p <= n; p++)
{
if (prime[p] == true )
{
for (let i = p * p; i <= n; i += p)
{
prime[i] = false ;
}
}
}
}
function sumOfPrimeSquare(n)
{
let flag = false ;
let prime = new Array(n + 1).fill( true );
SieveOfEratosthenes(prime, n);
for (let i = 0; i <= n; i++)
{
if (!prime[i])
continue ;
let dif = n - i;
if (Math.ceil(Math.sqrt(dif)) ==
Math.floor(Math.sqrt(dif)))
{
flag = true ;
break ;
}
}
if (flag)
{
document.write( "Yes" );
}
else
document.write( "No" );
}
let N = 27;
sumOfPrimeSquare(N);
</script>
|
Time Complexity: O(N * log(log(N)))
Auxiliary Space: O(N)
Last Updated :
17 Mar, 2023
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