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)
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