Dada una array arr[] de N enteros. Los números enteros representan todos los divisores de un número X excepto el 1 y el propio X. La tarea es encontrar el número X. Si no es posible tal elemento, imprima -1 .
Ejemplos:
Entrada: arr[] = {2, 10, 5, 4}
Salida: 20Entrada: arr[] = {2, 10, 5}
Salida: 20Entrada: arr[] = {2, 15}
Salida: -1
Enfoque: ordene los N divisores dados y el número X será el primer número * último número en la array ordenada. Verifique si la X contradice la declaración dada o no almacenando todos los divisores de X excepto 1 y X en otra array y si la array formada y la array dada no son las mismas, imprima -1 , de lo contrario imprima X.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns X
int findX(int a[], int n)
{
// Sort the given array
sort(a, a + n);
// Get the possible X
int x = a[0] * a[n - 1];
// Container to store divisors
vector<int> vec;
// Find the divisors of x
for (int i = 2; i * i <= x; i++)
{
// Check if divisor
if (x % i == 0)
{
vec.push_back(i);
if ((x / i) != i)
vec.push_back(x / i);
}
}
// sort the vec because a is sorted
// and we have to compare all the elements
sort(vec.begin(), vec.end());
// if size of both vectors is not same
// then we are sure that both vectors
// can't be equal
if (vec.size() != n)
return -1;
else
{
// Check if a and vec have
// same elements in them
int i = 0;
for (auto it : vec)
{
if (a[i++] != it)
return -1;
}
}
return x;
}
// Driver code
int main()
{
int a[] = { 2, 5, 4, 10 };
int n = sizeof(a) / sizeof(a[0]);
// Function call
cout << findX(a, n);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG {
// Function that returns X
static int findX(int a[], int n)
{
// Sort the given array
Arrays.sort(a);
// Get the possible X
int x = a[0] * a[n - 1];
// Container to store divisors
Vector<Integer> vec = new Vector<Integer>();
// Find the divisors of x
for (int i = 2; i * i <= x; i++)
{
// Check if divisor
if (x % i == 0)
{
vec.add(i);
if ((x / i) != i)
vec.add(x / i);
}
}
// sort the vec because a is sorted
// and we have to compare all the elements
Collections.sort(vec);
// if size of both vectors is not same
// then we are sure that both vectors
// can't be equal
if (vec.size() != n)
return -1;
else {
// Check if a and vec have
// same elements in them
int i = 0;
for (int it : vec) {
if (a[i++] != it)
return -1;
}
}
return x;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 2, 5, 4, 10 };
int n = a.length;
// Function call
System.out.print(findX(a, n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach # Function that returns X import math def findX(list, int): # Sort the given array list.sort() # Get the possible X x = list[0]*list[int-1] # Container to store divisors vec = [] # Find the divisors of x i = 2 while(i * i <= x): # Check if divisor if(x % i == 0): vec.append(i) if ((x//i) != i): vec.append(x//i) i += 1 # sort the vec because a is sorted # and we have to compare all the elements vec.sort() # if size of both vectors is not same # then we are sure that both vectors # can't be equal if(len(vec) != int): return -1 else: # Check if a and vec have # same elements in them j = 0 for it in range(int): if(a[j] != vec[it]): return -1 else: j += 1 return x # Driver code a = [2, 5, 4, 10] n = len(a) # Function call print(findX(a, n))
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
// Function that returns X
static int findX(int[] a, int n)
{
// Sort the given array
Array.Sort(a);
// Get the possible X
int x = a[0] * a[n - 1];
// Container to store divisors
List<int> vec = new List<int>();
// Find the divisors of a number
for (int i = 2; i * i <= x; i++)
{
// Check if divisor
if (x % i == 0) {
vec.Add(i);
if ((x / i) != i)
vec.Add(x / i);
}
}
// sort the vec because a is sorted
// and we have to compare all the elements
vec.Sort();
// if size of both vectors is not same
// then we are sure that both vectors
// can't be equal
if (vec.Count != n)
{
return -1;
}
else
{
// Check if a and vec have
// same elements in them
int i = 0;
foreach(int it in vec)
{
if (a[i++] != it)
return -1;
}
}
return x;
}
// Driver code
public static void Main(String[] args)
{
int[] a = { 2, 5, 4, 10 };
int n = a.Length;
// Function call
Console.Write(findX(a, n));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript implementation of the approach
// Function that returns X
function findX(a, n)
{
// Sort the given array
a.sort((x,y) => x - y);
// Get the possible X
let x = a[0] * a[n - 1];
// Container to store divisors
let vec = [];
// Find the divisors of x
for (let i = 2; i * i <= x; i++)
{
// Check if divisor
if (x % i == 0)
{
vec.push(i);
if (parseInt(x / i) != i)
vec.push(parseInt(x / i));
}
}
// sort the vec because a is sorted
// and we have to compare all the elements
vec.sort((x,y) => x - y);
// if size of both vectors is not same
// then we are sure that both vectors
// can't be equal
if (vec.length != n)
return -1;
else
{
// Check if a and vec have
// same elements in them
let i = 0;
for (let j = 0; j < vec.length; j++)
{
if (a[i++] != vec[j])
return -1;
}
}
return x;
}
// Driver code
let a = [ 2, 5, 4, 10 ];
let n = a.length;
// Function call
document.write(findX(a, n));
</script>
Producción
20
Complejidad de tiempo: O (nlogn)
Espacio auxiliar: O(sqrt(n))