Find number from its divisors in C++

C++Server Side ProgrammingProgramming

In this problem, we are given an array divisors[] consisting of N integers which are the divisors of a number Num. Our task is to find the number from its divisors.

The divisor array does not include 1 and the number.

Let’s take an example to understand the problem,


divisors[] = {3, 25, 5, 15}




The number 75 has divisors {3, 25, 5, 15}

Solution Approach

To solve the problem, we need to find the number Num using the smallest and largest divisors of the number.

Num = smallest * largest

For this, we need to sort the array divisors[] and then find the product of elements at the 1st and last index of the array.

For the number Num, find all the factors of the number. And check the divisors of the number are the same as in the divisor array. If Yes, return Num. Else, return -1, denoting the number cannot be found.

Program to illustrate the working of our solution,


 Live Demo

#include <bits/stdc++.h>
using namespace std;
int findNumberFromDiv(int divisors[], int n){
   sort(divisors, divisors + n);
   int num = divisors[0] * divisors[n - 1];
   int numDiv[2*n];
   int count = 0;
   for (int i = 2; i * i <= num; i++){
      if (num % i == 0){
         numDiv[count] = i;
         count ++ ;
         numDiv[count] = num/i;
   sort(numDiv, numDiv + count);
   if (count != n)
      return -1;
      for (int i = 0; i < count; i++) {
         if (divisors[i] != numDiv[i])
            return -1;
   return num;
int main(){
   int divisors[] = { 3, 25, 5, 15 };
   int n = sizeof(divisors) / sizeof(divisors[0]);
   cout<<"The number is "<<findNumberFromDiv(divisors,n);
   return 0;


The number is 75
Updated on 15-Mar-2021 10:46:52