Find three prime numbers with given sum

Last Updated : 05 Jan, 2021

Given an integer N, the task is to find three prime numbers X, Y, and Z such that the sum of these three numbers is equal to N i.e. X + Y + Z = N.

Examples:

Input: N = 20
Output: 2 5 13
Input: N = 34
Output: 2 3 29

Approach:

Generate prime numbers using Sieve of Eratosthenes
Start from the first prime number.
Take another number from the generated list.
Subtract first number and second number from the original number to obtain the third number.
Check if the third number is a prime number.
If the third number is a prime number then output the three numbers.
Otherwise, repeat the process for the second number and consequently the first number
If the answer does not exist then print -1.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 100001;
 
// The vector primes holds
// the prime numbers
vector<int> primes;
 
// Function to generate prime numbers
void initialize()
{
 
    // Initialize the array elements to 0s
    bool numbers[MAX] = {};
    int n = MAX;
    for (int i = 2; i * i <= n; i++)
        if (!numbers[i])
            for (int j = i * i; j <= n; j += i)
 
                // Set the non-primes to true
                numbers[j] = true;
 
    // Fill the vector primes with prime
    // numbers which are marked as false
    // in the numbers array
    for (int i = 2; i <= n; i++)
        if (numbers[i] == false)
            primes.push_back(i);
}
 
// Function to print three prime numbers
// which sum up to the number N
void findNums(int num)
{
 
    bool ans = false;
    int first = -1, second = -1, third = -1;
    for (int i = 0; i < num; i++) {
 
        // Take the first prime number
        first = primes[i];
        for (int j = 0; j < num; j++) {
 
            // Take the second prime number
            second = primes[j];
 
            // Subtract the two prime numbers
            // from the N to obtain the third number
            third = num - first - second;
 
            // If the third number is prime
            if (binary_search(primes.begin(),
                              primes.end(), third)) {
                ans = true;
                break;
            }
        }
        if (ans)
            break;
    }
    // Print the three prime numbers
    // if the solution exists
    if (ans)
        cout << first << " "
             << second << " " << third << endl;
    else
        cout << -1 << endl;
}
 
// Driver code
int main()
{
    int n = 101;
 
    // Function for generating prime numbers
    // using Sieve of Eratosthenes
    initialize();
 
    // Function to print the three prime
    // numbers whose sum is equal to N
    findNums(n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
 
static int MAX = 100001;
   
// The vector primes holds
// the prime numbers
static ArrayList<Integer> primes;
   
// Function to generate prime numbers
static void initialize()
{
     
    // Initialize the array elements to 0s
    boolean[] numbers = new boolean[MAX + 1];
    int n = MAX;
     
    for(int i = 2; i * i <= n; i++)
        if (!numbers[i])
            for(int j = i * i; j <= n; j += i)
             
                // Set the non-primes to true
                numbers[j] = true;
   
    // Fill the vector primes with prime
    // numbers which are marked as false
    // in the numbers array
    for(int i = 2; i <= n; i++)
        if (numbers[i] == false)
            primes.add(i);
}
   
// Function to print three prime numbers
// which sum up to the number N
static void findNums(int num)
{
    boolean ans = false;
    int first = -1, second = -1, third = -1;
     
    for(int i = 0; i < num; i++)
    {
         
        // Take the first prime number
        first = primes.get(i);
         
        for(int j = 0; j < num; j++)
        {
             
            // Take the second prime number
            second = primes.get(j);
   
            // Subtract the two prime numbers
            // from the N to obtain the third number
            third = num - first - second;
   
            // If the third number is prime
            if (Collections.binarySearch(
                primes, third) >= 0)
            {
                ans = true;
                break;
            }
        }
        if (ans)
            break;
    }
     
    // Print the three prime numbers
    // if the solution exists
    if (ans)
       System.out.println(first + " " +
                          second + " " +
                          third);
    else
        System.out.println(-1 );
}
 
// Driver code
public static void main (String[] args)
{
    int n = 101;
    primes = new ArrayList<>();
     
    // Function for generating prime numbers
    // using Sieve of Eratosthenes
    initialize();
     
    // Function to print the three prime
    // numbers whose sum is equal to N
    findNums(n);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 implementation of the approach
from math import sqrt
 
MAX = 100001;
 
# The vector primes holds
# the prime numbers
primes = [];
 
# Function to generate prime numbers
def initialize() :
 
    # Initialize the array elements to 0s
    numbers = [0]*(MAX + 1);
    n = MAX;
    for i in range(2, int(sqrt(n)) + 1) :
        if (not numbers[i]) :
            for j in range( i * i , n + 1, i) :
 
                # Set the non-primes to true
                numbers[j] = True;
 
    # Fill the vector primes with prime
    # numbers which are marked as false
    # in the numbers array
    for i in range(2, n + 1) :
        if (numbers[i] == False) :
            primes.append(i);
 
# Function to print three prime numbers
# which sum up to the number N
def findNums(num) :
 
    ans = False;
    first = -1;
    second = -1;
    third = -1;
    for i in range(num) :
 
        # Take the first prime number
        first = primes[i];
        for j in range(num) :
 
            # Take the second prime number
            second = primes[j];
 
            # Subtract the two prime numbers
            # from the N to obtain the third number
            third = num - first - second;
 
            # If the third number is prime
            if (third in primes) :
                ans = True;
                break;
     
        if (ans) :
            break;
     
    # Print the three prime numbers
    # if the solution exists
    if (ans) :
        print(first , second , third);
    else :
        print(-1);
 
# Driver code
if __name__ == "__main__" :
 
    n = 101;
 
    # Function for generating prime numbers
    # using Sieve of Eratosthenes
    initialize();
 
    # Function to print the three prime
    # numbers whose sum is equal to N
    findNums(n);
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{  
  static int MAX = 100001;
 
  // The vector primes holds
  // the prime numbers
  static List<int> primes = new List<int>();
 
  // Function to generate prime numbers
  static void initialize()
  {
 
    // Initialize the array elements to 0s
    bool[] numbers = new bool[MAX + 1];
    int n = MAX;
    for (int i = 2; i * i <= n; i++)
      if (!numbers[i])
        for (int j = i * i; j <= n; j += i)
 
          // Set the non-primes to true
          numbers[j] = true;
 
    // Fill the vector primes with prime
    // numbers which are marked as false
    // in the numbers array
    for (int i = 2; i <= n; i++)
      if (numbers[i] == false)
        primes.Add(i);
  }
 
  // Function to print three prime numbers
  // which sum up to the number N
  static void findNums(int num)
  {
 
    bool ans = false;
    int first = -1, second = -1, third = -1;
    for (int i = 0; i < num; i++)
    {
 
      // Take the first prime number
      first = primes[i];
      for (int j = 0; j < num; j++)
      {
 
        // Take the second prime number
        second = primes[j];
 
        // Subtract the two prime numbers
        // from the N to obtain the third number
        third = num - first - second;
 
        // If the third number is prime
        if (Array.BinarySearch(primes.ToArray(), third) >= 0)
        {
          ans = true;
          break;
        }
      }
      if (ans)
        break;
    }
 
    // Print the three prime numbers
    // if the solution exists
    if (ans)
      Console.WriteLine(first + " " + second + " " + third);
    else
      Console.WriteLine(-1);
  }
 
  // Driver code
  static void Main()
  {
    int n = 101;
 
    // Function for generating prime numbers
    // using Sieve of Eratosthenes
    initialize();
 
    // Function to print the three prime
    // numbers whose sum is equal to N
    findNums(n);
  }
}
 
// This code is contributed by divyesh072019

Output:

2 2 97

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