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Count of triplets having sum of product of any two numbers with the third number equal to N

  • Last Updated : 13 Sep, 2021

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

=> X*Y + Z = N
=> X*Y = N - Z

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 Erastosthenes.
  • 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++




// 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 Erastothenes
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




// 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 Erastothenes
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

Python3




# Python program for the above approach
 
# Function to find the SPF[i] using the
# Sieve Of Erastothenes
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 +=
     
    # 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))
Output: 
23

 

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

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