Find number from its divisors

Given an array arr[] of N integers. The integers represent all the divisors of a number X except 1 and X itself. The task is to find the number X. If no such element is possible then print -1.

Examples:

Input: arr[] = {2, 10, 5, 4}
Output: 20

Input: arr[] = {2, 10, 5}
Output: 20

Input: arr[] = {2, 15}
Output: -1



Approach: Sort the given N divisors and the number X will be the first number * last number in the sorted array. Cross check if the X contradicts the given statement or not by storing all the divisors of X except 1 and X in another array and if the formed array and given array are not same then print -1, else print X.

Below is the implementation of the above approach:

C++

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// 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 a number 
    for (int i = 2; i * i <= x; i++) { 
    
        // Check if divisor 
        if (n % i == 0) { 
            vec.push_back(i); 
            if ((n / i) != i) 
                vec.push_back(n / i); 
        
    
    
    // Check if a and vec have 
    // same elements in them 
    if (vec.size() == n) { 
        int i = 0; 
        for (auto it : vec) { 
            if (a[i++] != it) 
                return -1; 
        
    
    
    return x; 
    
// Drivers code 
int main() 
    int a[] = { 2, 5, 4, 10 }; 
    int n = sizeof(a) / sizeof(a[0]); 
    cout << findX(a, n); 
    
    return 0; 

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Python3

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# 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 a number
    i = 2
    while(i * i <= x):
        #Check if divisor
        if(n % i == 0):
            vec.append(i)
            if ((int//i) != i):
                vec.append(int//i)
        i +=1
  
    # Check if a and vec have
    # same elements in them
    if(len(vec) == int):
        j = 0
        for it in range(int):
            if(a[j] != it):
                return -1
            else:
                j +=1
    return x
  
# Drivers code
a = [2, 5, 4, 10]
n = len(a)
print(findX(a,n))
  
# This code is contributed by souviksarkhelkgp

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Output:

20


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

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Improved By : souviksarkhelkgp