Given an integer X, the task is to find the three distinct integers greater than 1 i.e. A, B and C such that (A * B * C) = X. If no such triplet exists then print -1.
Examples:
Input: X = 64
Output: 2 4 8
(2 * 4 * 8) = 64
Input: X = 32
Output: -1
No such triplet exists.
Approach: Suppose we have a triplet (A, B, C). Notice that, for their product to be equal to X, each of the integer has to be a factor of X. So, store all the factors of X in O(sqrt(X)) time using the approach discussed in this article.
There will be at most sqrt(X) factors now. Next, iterate on each factor by running two loops, one picking A and another picking B. Now if this triplet is valid i.e. C = X / (A * B) where C is also a factor of X. To check that, store all the factors in an unordered_set. If a valid triplet is found then print the triplet else print -1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;
// Function to find the required tripletsvoid findTriplets(int x)
{ // To store the factors
vector<int> fact;
unordered_set<int> factors;
// Find factors in sqrt(x) time
for (int i = 2; i <= sqrt(x); i++) {
if (x % i == 0) {
fact.push_back(i);
if (x / i != i)
fact.push_back(x / i);
factors.insert(i);
factors.insert(x / i);
}
}
bool found = false;
int k = fact.size();
for (int i = 0; i < k; i++) {
// Choose a factor
int a = fact[i];
for (int j = 0; j < k; j++) {
// Choose another factor
int b = fact[j];
// These conditions need to be
// met for a valid triplet
if ((a != b) && (x % (a * b) == 0)
&& (x / (a * b) != a)
&& (x / (a * b) != b)
&& (x / (a * b) != 1)) {
// Print the valid triplet
cout << a << " " << b << " "
<< (x / (a * b));
found = true;
break;
}
}
// Triplet found
if (found)
break;
}
// Triplet not found
if (!found)
cout << "-1";
}// Driver codeint main()
{ int x = 105;
findTriplets(x);
return 0;
} |
Java
// Java implementation of the approachimport java.util.*;
class GFG
{// Function to find the required tripletsstatic void findTriplets(int x)
{ // To store the factors
Vector<Integer> fact = new Vector<Integer>();
HashSet<Integer> factors = new HashSet<Integer>();
// Find factors in Math.sqrt(x) time
for (int i = 2; i <= Math.sqrt(x); i++)
{
if (x % i == 0)
{
fact.add(i);
if (x / i != i)
fact.add(x / i);
factors.add(i);
factors.add(x / i);
}
}
boolean found = false;
int k = fact.size();
for (int i = 0; i < k; i++)
{
// Choose a factor
int a = fact.get(i);
for (int j = 0; j < k; j++)
{
// Choose another factor
int b = fact.get(j);
// These conditions need to be
// met for a valid triplet
if ((a != b) && (x % (a * b) == 0)
&& (x / (a * b) != a)
&& (x / (a * b) != b)
&& (x / (a * b) != 1))
{
// Print the valid triplet
System.out.print(a+ " " + b + " "
+ (x / (a * b)));
found = true;
break;
}
}
// Triplet found
if (found)
break;
}
// Triplet not found
if (!found)
System.out.print("-1");
}// Driver codepublic static void main(String[] args)
{ int x = 105;
findTriplets(x);
}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach from math import sqrt
# Function to find the required triplets def findTriplets(x) :
# To store the factors
fact = [];
factors = set();
# Find factors in sqrt(x) time
for i in range(2, int(sqrt(x))) :
if (x % i == 0) :
fact.append(i);
if (x / i != i) :
fact.append(x // i);
factors.add(i);
factors.add(x // i);
found = False;
k = len(fact);
for i in range(k) :
# Choose a factor
a = fact[i];
for j in range(k) :
# Choose another factor
b = fact[j];
# These conditions need to be
# met for a valid triplet
if ((a != b) and (x % (a * b) == 0)
and (x / (a * b) != a)
and (x / (a * b) != b)
and (x / (a * b) != 1)) :
# Print the valid triplet
print(a,b,x // (a * b));
found = True;
break;
# Triplet found
if (found) :
break;
# Triplet not found
if (not found) :
print("-1");
# Driver code if __name__ == "__main__" :
x = 105;
findTriplets(x);
# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;
using System.Collections.Generic;
class GFG
{// Function to find the required tripletsstatic void findTriplets(int x)
{ // To store the factors
List<int> fact = new List<int>();
HashSet<int> factors = new HashSet<int>();
// Find factors in Math.Sqrt(x) time
for (int i = 2; i <= Math.Sqrt(x); i++)
{
if (x % i == 0)
{
fact.Add(i);
if (x / i != i)
fact.Add(x / i);
factors.Add(i);
factors.Add(x / i);
}
}
bool found = false;
int k = fact.Count;
for (int i = 0; i < k; i++)
{
// Choose a factor
int a = fact[i];
for (int j = 0; j < k; j++)
{
// Choose another factor
int b = fact[j];
// These conditions need to be
// met for a valid triplet
if ((a != b) && (x % (a * b) == 0)
&& (x / (a * b) != a)
&& (x / (a * b) != b)
&& (x / (a * b) != 1))
{
// Print the valid triplet
Console.Write(a+ " " + b + " "
+ (x / (a * b)));
found = true;
break;
}
}
// Triplet found
if (found)
break;
}
// Triplet not found
if (!found)
Console.Write("-1");
}// Driver codepublic static void Main(String[] args)
{ int x = 105;
findTriplets(x);
}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to find the required tripletsfunction findTriplets(x)
{ // To store the factors
let fact = [];
let factors = new Set();
// Find factors in Math.sqrt(x) time
for (let i = 2; i <= Math.sqrt(x); i++)
{
if (x % i == 0)
{
fact.push(i);
if (x / i != i)
fact.push(x / i);
factors.add(i);
factors.add(x / i);
}
}
let found = false;
let k = fact.length;
for (let i = 0; i < k; i++)
{
// Choose a factor
let a = fact[i];
for (let j = 0; j < k; j++)
{
// Choose another factor
let b = fact[j];
// These conditions need to be
// met for a valid triplet
if ((a != b) && (x % (a * b) == 0)
&& (x / (a * b) != a)
&& (x / (a * b) != b)
&& (x / (a * b) != 1))
{
// Print the valid triplet
document.write(a+ " " + b + " "
+ (x / (a * b)));
found = true;
break;
}
}
// Triplet found
if (found)
break;
}
// Triplet not found
if (!found)
document.write("-1");
} // Driver code let x = 105;
findTriplets(x);
</script> |
Output:
3 5 7
Time Complexity: O(N), N=X
Auxiliary Space: O(N)