Check if N can be expressed as product of 3 distinct numbers

0 0 6 minutes read

Given a number N. Print three distinct numbers (>=1) whose product is equal to N. print -1 if it is not possible to find three numbers.
Examples:

Input: 64
Output: 2 4 8
Explanation:
(2*4*8 = 64)
Input: 24
Output: 2 3 4
Explanation:
(2*3*4 = 24)
Input: 12
Output: -1
Explanation:
No such triplet exists

Approach:

Make an array which stores all the divisors of the given number using the approach discussed in this article
Let the three number be a, b, c initialize to 1
Traverse the divisors array and check the following condition:
- value of a = value at 1st index of divisor array.
- value of b = product of value at 2nd and 3rd index of divisor array. If divisor array has only one or two elements then no such triplets exists
- After finding a & b, value of c = product of all the rest elements in divisor array.
Check the final condition such that value of a, b, c must be distinct and not equal to 1.

Below is the implementation code:

CPP

// C++ program to find the
// three numbers
#include "bits/stdc++.h"
using namespace std;
 
// function to find 3 distinct number
// with given product
void getnumbers(int n)
{
    // Declare a vector to store
    // divisors
    vector<int> divisor;
 
    // store all divisors of number
    // in array
    for (int i = 2; i * i <= n; i++) {
 
        // store all the occurrence of
        // divisors
        while (n % i == 0) {
            divisor.push_back(i);
            n /= i;
        }
    }
 
    // check if n is not equals to -1
    // then n is also a prime factor
    if (n != 1) {
        divisor.push_back(n);
    }
 
    // Initialize the variables with 1
    int a, b, c, size;
    a = b = c = 1;
    size = divisor.size();
 
    for (int i = 0; i < size; i++) {
 
        // check for first number a
        if (a == 1) {
            a = a * divisor[i];
        }
 
        // check for second number b
        else if (b == 1 || b == a) {
            b = b * divisor[i];
        }
 
        // check for third number c
        else {
            c = c * divisor[i];
        }
    }
 
    // check for all unwanted condition
    if (a == 1 || b == 1 || c == 1
        || a == b || b == c || a == c) {
        cout << "-1" << endl;
    }
    else {
        cout << a << ' ' << b
             << ' ' << c << endl;
    }
}
 
// Driver function
int main()
{
    int n = 64;
    getnumbers(n);
}

Java

// Java program to find the
// three numbers
import java.util.*;
 
class GFG{
  
// function to find 3 distinct number
// with given product
static void getnumbers(int n)
{
    // Declare a vector to store
    // divisors
    Vector<Integer> divisor = new Vector<Integer>();
  
    // store all divisors of number
    // in array
    for (int i = 2; i * i <= n; i++) {
  
        // store all the occurrence of
        // divisors
        while (n % i == 0) {
            divisor.add(i);
            n /= i;
        }
    }
  
    // check if n is not equals to -1
    // then n is also a prime factor
    if (n != 1) {
        divisor.add(n);
    }
  
    // Initialize the variables with 1
    int a, b, c, size;
    a = b = c = 1;
    size = divisor.size();
  
    for (int i = 0; i < size; i++) {
  
        // check for first number a
        if (a == 1) {
            a = a * divisor.get(i);
        }
  
        // check for second number b
        else if (b == 1 || b == a) {
            b = b * divisor.get(i);
        }
  
        // check for third number c
        else {
            c = c * divisor.get(i);
        }
    }
  
    // check for all unwanted condition
    if (a == 1 || b == 1 || c == 1
        || a == b || b == c || a == c) {
        System.out.print("-1" +"\n");
    }
    else {
        System.out.print(a +" "+ b
                +" "+ c +"\n");
    }
}
  
// Driver function
public static void main(String[] args)
{
    int n = 64;
    getnumbers(n);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program to find the
# three numbers
 
# function to find 3 distinct number
# with given product
def getnumbers(n):
      
     # Declare a vector to store
    # divisors
    divisor = []
 
    # store all divisors of number
    # in array
    for i in range(2, n + 1):
 
        # store all the occurrence of
        # divisors
        while (n % i == 0):
            divisor.append(i)
            n //= i
 
    # check if n is not equals to -1
    # then n is also a prime factor
    if (n != 1):
        divisor.append(n)
 
    # Initialize the variables with 1
    a, b, c, size = 0, 0, 0, 0
    a = b = c = 1
    size = len(divisor)
 
    for i in range(size):
 
        # check for first number a
        if (a == 1):
            a = a * divisor[i]
 
        # check for second number b
        elif (b == 1 or b == a):
            b = b * divisor[i]
 
        # check for third number c
        else:
            c = c * divisor[i]
 
    # check for all unwanted condition
    if (a == 1 or b == 1 or c == 1
        or a == b or b == c or a == c):
        print("-1")
    else:
        print(a, b, c)
 
# Driver function
 
n = 64
getnumbers(n)
 
# This code is contributed by mohit kumar 29

C#

// C# program to find the
// three numbers
using System;
using System.Collections.Generic;
 
class GFG{
   
// function to find 3 distinct number
// with given product
static void getnumbers(int n)
{
    // Declare a vector to store
    // divisors
    List<int> divisor = new List<int>();
   
    // store all divisors of number
    // in array
    for (int i = 2; i * i <= n; i++) {
   
        // store all the occurrence of
        // divisors
        while (n % i == 0) {
            divisor.Add(i);
            n /= i;
        }
    }
   
    // check if n is not equals to -1
    // then n is also a prime factor
    if (n != 1) {
        divisor.Add(n);
    }
   
    // Initialize the variables with 1
    int a, b, c, size;
    a = b = c = 1;
    size = divisor.Count;
   
    for (int i = 0; i < size; i++) {
   
        // check for first number a
        if (a == 1) {
            a = a * divisor[i];
        }
   
        // check for second number b
        else if (b == 1 || b == a) {
            b = b * divisor[i];
        }
   
        // check for third number c
        else {
            c = c * divisor[i];
        }
    }
   
    // check for all unwanted condition
    if (a == 1 || b == 1 || c == 1
        || a == b || b == c || a == c) {
        Console.Write("-1" +"\n");
    }
    else {
        Console.Write(a +" "+ b
                +" "+ c +"\n");
    }
}
   
// Driver function
public static void Main(String[] args)
{
    int n = 64;
    getnumbers(n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// JavaScript program to find the
// three numbers
 
// function to find 3 distinct number
// with given product
function getnumbers(n)
{
    // Declare a vector to store
    // divisors
    let divisor = [];
   
    // store all divisors of number
    // in array
    for (let i = 2; i * i <= n; i++) {
   
        // store all the occurrence of
        // divisors
        while (n % i == 0) {
            divisor.push(i);
            n = Math.floor(n/i);
        }
    }
   
    // check if n is not equals to -1
    // then n is also a prime factor
    if (n != 1) {
        divisor.push(n);
    }
   
    // Initialize the variables with 1
    let a, b, c, size;
    a = b = c = 1;
    size = divisor.length;
   
    for (let i = 0; i < size; i++) {
   
        // check for first number a
        if (a == 1) {
            a = a * divisor[i];
        }
   
        // check for second number b
        else if (b == 1 || b == a) {
            b = b * divisor[i];
        }
   
        // check for third number c
        else {
            c = c * divisor[i];
        }
    }
   
    // check for all unwanted condition
    if (a == 1 || b == 1 || c == 1
        || a == b || b == c || a == c) {
        document.write("-1" +"<br>");
    }
    else {
        document.write(a +" "+ b
                +" "+ c +"<br>");
    }
}
 
// Driver function
let n = 64;
getnumbers(n);
 
 
// This code is contributed by patel2127
 
</script>

Output:

2 4 8

Time Complexity: O((log N)*sqrt(N))

Auxiliary Space: O(sqrt(n))

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!