Given an positive integer N, the task is to find the number of triplets (X, Y, Z) such that the sum of product of any two numbers with the third number is N.
Examples:
Input: N = 2
Output: 1
Explanation:
The only triplets satisfying the given criteria is (1, 1, 1). Therefore, the count is 1.Input: N = 3
Output: 3
Approach: This given problem can be solved by rearranging the equation as:
Consider a triplet as (X, Y, Z) then
=>
=>
From the above equation the idea is to iterate over all possible value of Z over the range [1, N] and add the count of all possible of X and Y by calculating the prime factor of (N – Z) satisfying the above equation. Follow the below steps to solve the problem:
- Initialize a variable, countTriplets to store the resultant count of triplets satisfying the given criteria.
- Find the smallest prime factor for all the elements over the range [1, 105] using the Sieve Of Eratosthenes.
- Iterate over the range [1, N] using the variable, say K and perform the following steps:
- Find the number of pairs of whose product is (N – K) using the approach discussed in this article and add the count obtained to the variable countTriplets.
- After completing the above steps, print the value of countTriplets as the resultant count of triplets.
Below is the implementation of the above approach:
// C++ program for the above approach #include<bits/stdc++.h> using namespace std;
vector< int > s(11,0);
// Function to find the SPF[i] using the // Sieve Of Eratosthenes void sieveOfEratosthenes( int N){
// Stores whether i is prime or not
bool prime[N+1];
memset (prime, false , sizeof ( false ));
// Initializing smallest factor as
// 2 for all even numbers
for ( int i=2;i< N + 1; i+=2)
s[i] = 2;
// Iterate for all odd numbers < N
for ( int i =3;i<N + 1;i+=2){
if (prime[i] == false ){
// SPF of i for a prime is
// the number itself
s[i] = i;
// Iterate for all the multiples
// of the current prime number
for ( int j= i;j< N / i + 1; j+=2){
if (prime[i * j] == false ){
prime[i * j] = true ;
// The value i is smallest
// prime factor for i * j
s[i * j] = i;
}
}
}
}
} // Function to generate prime factors // and its power int generatePrimeFactors( int N){
// Current prime factor of N
int curr = s[N];
// Stores the powers of the current
// prime factor
map< int , int > cnt;
cnt[s[N]] = 1;
// Find all the prime factors and
// their powers
while (N > 1){
N /= s[N];
if (N and s[N])
if (cnt.find(s[N]) == cnt.end())
cnt[s[N]] = 1;
else
cnt[s[N]] += 1;
}
if (cnt.find(0) != cnt.end())
cnt.erase(0);
int totfactor = 1;
for ( auto i: cnt)
totfactor *= i.second + 1;
// Return the total count of factors
return totfactor;
} // Function to count the number of triplets // satisfying the given criteria int countTriplets( int N){
// Stores the count of resultant triplets
int CountTriplet = 0;
for ( int z=1;z<N + 1;z++){
// Add the count all factors of N-z
// to the variable CountTriplet
int p = generatePrimeFactors(N-z);
if (p > 1)
CountTriplet += p;
}
// Return total count of triplets
return CountTriplet + 1;
}
// Driver Code int main(){
int N = 10;
// S[i] stores the smallest prime factor // for each element i // Find the SPF[i] sieveOfEratosthenes(N); // Function Call cout<<countTriplets(N); } // This code is contributed by SURENDRA_GANGWAR. |
// Java program for the above approach import java.util.*;
class GFG{
static int [] s = new int [ 11 ];
// Function to find the SPF[i] using the // Sieve Of Eratosthenes static void sieveOfEratosthenes( int N){
// Stores whether i is prime or not
boolean []prime = new boolean [N+ 1 ];
// Initializing smallest factor as
// 2 for all even numbers
for ( int i = 2 ; i < N + 1 ; i += 2 )
s[i] = 2 ;
// Iterate for all odd numbers < N
for ( int i = 3 ; i < N + 1 ; i += 2 ){
if (prime[i] == false ){
// SPF of i for a prime is
// the number itself
s[i] = i;
// Iterate for all the multiples
// of the current prime number
for ( int j= i; j < N / i + 1 ; j += 2 ){
if (prime[i * j] == false ){
prime[i * j] = true ;
// The value i is smallest
// prime factor for i * j
s[i * j] = i;
}
}
}
}
} // Function to generate prime factors // and its power static int generatePrimeFactors( int N){
// Current prime factor of N
int curr = s[N];
// Stores the powers of the current
// prime factor
HashMap<Integer,Integer> cnt = new HashMap<>();
cnt.put(s[N], 1 );
// Find all the prime factors and
// their powers
while (N > 1 ){
N /= s[N];
if (N != 0 && s[N] != 0 )
if (!cnt.containsKey(s[N]))
cnt.put(s[N], 1 );
else
cnt.put(s[N], cnt.get(s[N]) + 1 );
}
if (cnt.containsKey( 0 ))
cnt.remove( 0 );
int totfactor = 1 ;
for (Map.Entry<Integer,Integer> i : cnt.entrySet())
totfactor *= i.getValue() + 1 ;
// Return the total count of factors
return totfactor;
} // Function to count the number of triplets // satisfying the given criteria static int countTriplets( int N){
// Stores the count of resultant triplets
int CountTriplet = 0 ;
for ( int z= 1 ;z<N + 1 ;z++){
// Add the count all factors of N-z
// to the variable CountTriplet
int p = generatePrimeFactors(N-z);
if (p > 1 )
CountTriplet += p;
}
// Return total count of triplets
return CountTriplet + 1 ;
}
// Driver Code public static void main(String[] args){
int N = 10 ;
// S[i] stores the smallest prime factor // for each element i // Find the SPF[i] sieveOfEratosthenes(N); // Function Call System.out.print(countTriplets(N)); } } // This code is contributed by gauravrajput1 |
# Python program for the above approach # Function to find the SPF[i] using the # Sieve Of Eratosthenes def sieveOfEratosthenes(N, s):
# Stores whether i is prime or not
prime = [ False ] * (N + 1 )
# Initializing smallest factor as
# 2 for all even numbers
for i in range ( 2 , N + 1 , 2 ):
s[i] = 2
# Iterate for all odd numbers < N
for i in range ( 3 , N + 1 , 2 ):
if (prime[i] = = False ):
# SPF of i for a prime is
# the number itself
s[i] = i
# Iterate for all the multiples
# of the current prime number
for j in range (i, int (N / i) + 1 , 2 ):
if (prime[i * j] = = False ):
prime[i * j] = True
# The value i is smallest
# prime factor for i * j
s[i * j] = i
# Function to generate prime factors # and its power def generatePrimeFactors(N):
# Current prime factor of N
curr = s[N]
# Stores the powers of the current
# prime factor
cnt = {s[N]: 1 }
# Find all the prime factors and
# their powers
while (N > 1 ):
N / / = s[N]
if N and s[N]:
if cnt.get(s[N], 0 ) = = 0 :
cnt[s[N]] = 1
else :
cnt[s[N]] + = 1
if 0 in cnt:
cnt.pop( 0 )
totfactor = 1
for i in cnt.values():
totfactor * = i + 1
# Return the total count of factors
return totfactor
# Function to count the number of triplets # satisfying the given criteria def countTriplets(N):
# Stores the count of resultant triplets
CountTriplet = 0
for z in range ( 1 , N + 1 ):
# Add the count all factors of N-z
# to the variable CountTriplet
p = generatePrimeFactors(N - z)
if p > 1 :
CountTriplet + = p
# Return total count of triplets
return CountTriplet + 1 # Driver Code N = 10
# S[i] stores the smallest prime factor # for each element i s = [ 0 ] * (N + 1 )
# Find the SPF[i] sieveOfEratosthenes(N, s) # Function Call print (countTriplets(N))
|
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG
{ static int [] s = new int [11];
// Function to find the SPF[i] using the
// Sieve Of Eratosthenes
static void sieveOfEratosthenes( int N)
{
// Stores whether i is prime or not
bool [] prime = new bool [N + 1];
// Initializing smallest factor as
// 2 for all even numbers
for ( int i = 2; i < N + 1; i += 2)
s[i] = 2;
// Iterate for all odd numbers < N
for ( int i = 3; i < N + 1; i += 2)
{
if (prime[i] == false )
{
// SPF of i for a prime is
// the number itself
s[i] = i;
// Iterate for all the multiples
// of the current prime number
for ( int j = i; j < N / i + 1; j += 2)
{
if (prime[i * j] == false )
{
prime[i * j] = true ;
// The value i is smallest
// prime factor for i * j
s[i * j] = i;
}
}
}
}
}
// Function to generate prime factors
// and its power
static int generatePrimeFactors( int N)
{
// Current prime factor of N
// int curr = s[N];
// Stores the powers of the current
// prime factor
Dictionary< int , int > cnt = new Dictionary< int , int >();
cnt.Add(s[N], 1);
// Find all the prime factors and
// their powers
while (N > 1)
{
N /= s[N];
if (N != 0 && s[N] != 0)
if (!cnt.ContainsKey(s[N]))
cnt.Add(s[N], 1);
else
cnt[s[N]]++;
}
if (cnt.ContainsKey(0))
cnt.Remove(0);
int totfactor = 1;
// get the factor count and multiply them
foreach ( var entry in cnt)
totfactor *= entry.Value + 1;
// Return the total count of factors
return totfactor;
}
// Function to count the number of triplets
// satisfying the given criteria
static int countTriplets( int N)
{
// Stores the count of resultant triplets
int countTriplet = 0;
for ( int z = 1; z < N + 1; z++)
{
// Add the count all factors of N-z
// to the variable CountTriplet
int p = generatePrimeFactors(N - z);
if (p > 1)
countTriplet += p;
}
// Return total count of triplets
return countTriplet + 1;
}
static void Main( string [] args)
{
int N = 10;
// Find the SPF[i]
sieveOfEratosthenes(N);
// Function Call
Console.WriteLine(countTriplets(N));
}
} |
<script> // Javascript program for the above approach let s = new Array(11).fill(0)
// Function to find the SPF[i] using the // Sieve Of Eratosthenes function sieveOfEratosthenes(N){
// Stores whether i is prime or not
let prime = new Array(N+1).fill( false );
// Initializing smallest factor as
// 2 for all even numbers
for (let i = 2; i < N + 1; i += 2)
s[i] = 2;
// Iterate for all odd numbers < N
for (let i = 3; i < N + 1; i += 2){
if (prime[i] == false ){
// SPF of i for a prime is
// the number itself
s[i] = i;
// Iterate for all the multiples
// of the current prime number
for (let j= i; j < Math.floor(N / i + 1); j += 2){
if (prime[i * j] == false ){
prime[i * j] = true ;
// The value i is smallest
// prime factor for i * j
s[i * j] = i;
}
}
}
}
} // Function to generate prime factors // and its power function generatePrimeFactors(N)
{ // Current prime factor of N
let curr = s[N];
// Stores the powers of the current
// prime factor
let cnt = new Map();
cnt.set(s[N],1);
// Find all the prime factors and
// their powers
while (N > 1){
N = Math.floor(N / s[N]);
if (N != 0 && s[N] != 0)
if (!cnt.has(s[N]))
cnt.set(s[N], 1);
else
cnt.set(s[N], cnt.get(s[N]) + 1);
}
if (cnt.has(0))
cnt. delete (0);
let totfactor = 1;
for (i of cnt.values())
totfactor *= i + 1;
// Return the total count of factors
return totfactor;
} // Function to count the number of triplets // satisfying the given criteria function countTriplets(N){
// Stores the count of resultant triplets
let CountTriplet = 0;
for (let z=1;z<N + 1;z++){
// Add the count all factors of N-z
// to the variable CountTriplet
let p = generatePrimeFactors(N-z);
if (p > 1)
CountTriplet += p;
}
// Return total count of triplets
return CountTriplet + 1;
}
// Driver Code let N = 10; // S[i] stores the smallest prime factor // for each element i // Find the SPF[i] sieveOfEratosthenes(N); // Function Call document.write((countTriplets(N))); // This code is contributed by _saurabh_jaiswal </script> |
Output:
23
Time Complexity: O(N*log N)
Auxiliary Space: O(N)