Find numbers in between [L, R] which are divisible by all Array elements

Given an array arr[] containing N positive integers and two variables L and R indicating a range of integers from L to R (inclusive). The task is to print all the numbers between L to R which are divisible by all array elements. If no such value exists print -1.

Input: arr[] = {3, 5, 12}, L = 90, R = 280
Output: 120 180 240 
Explanation: 120, 180, 240 are the numbers which are divisible by all the arr[] elements.

Input: arr[] = {4, 7, 13, 16}, L = 200, R = 600
Output: -1

Naive Approach: In this approach for every element in range [L, R] check if it is divisible by all the elements of the array.

Time Complexity: O((R-L)*N)
Auxiliary Space: O(1)

Efficient Approach: The given problem can be solved using basic math. Any element divisible by all the elements of the array is a multiple of the LCM of all the array elements. Find the multiples of LCM in the range [L, R] and store in an array. At last print the numbers stored in the array.

Time Complexity: O(N)
Auxiliary Space: O(R – L)

Space Optimized Approach: Below steps can be used to solve the problem:

1. Calculate the LCM of all the elements of given arr[]
2. Now, check the LCM for these conditions:
   1. If (LCM < L and LCM*2 > R), then print -1.
   2. If (LCM > R), then print -1.
3. Now, take the nearest value of L (between L to R) which is divisible by the LCM, say i.
4. Now, start printing i and increment it by LCM every time after printing, until it becomes greater than R.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return Kth smallest
// prime number if it exists
void solve(int* arr, int N, int L, int R)
{
    // For storing the LCM
    int LCM = arr[0];

    // Loop to iterate the array
    for (int i = 1; i < N; i++) {
        // Taking LCM of numbers
        LCM = (LCM * arr[i]) /
            (__gcd(LCM, arr[i]));
    }

    // Checking if no elements is divisible
    // by all elements of given array of given
    // range, print -1
    if ((LCM < L && LCM * 2 > R) || LCM > R) {
        cout << "-1";
        return;
    }

    // Taking nearest value of L which is
    // divisible by whole array
    int k = (L / LCM) * LCM;

    // If k is less than L, make it in the
    // range between L to R
    if (k < L)
        k = k + LCM;

    // Loop to iterate the from L to R
    // and printing the numbers which
    // are divisible by all array elements
    for (int i = k; i <= R; i = i + LCM) {
        cout << i << ' ';
    }
}

// Driver Code
int main()
{
    int L = 90;
    int R = 280;
    int arr[] = { 3, 5, 12 };
    int N = sizeof(arr) / sizeof(arr[0]);
    solve(arr, N, L, R);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{

    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;

    // Base case
    if (a == b)
        return a;

    // a is greater
    if (a > b)
        return __gcd(a - b, b);

    return __gcd(a, b - a);
}

// Function to return Kth smallest
// prime number if it exists
static void solve(int[] arr, int N, int L, int R)
{

    // For storing the LCM
    int LCM = arr[0];

    // Loop to iterate the array
    for(int i = 1; i < N; i++)
    {

        // Taking LCM of numbers
        LCM = (LCM * arr[i]) /
        (__gcd(LCM, arr[i]));
    }

    // Checking if no elements is divisible
    // by all elements of given array of given
    // range, print -1
    if ((LCM < L && LCM * 2 > R) || LCM > R)
    {
        System.out.println("-1");
        return;
    }

    // Taking nearest value of L which is
    // divisible by whole array
    int k = (L / LCM) * LCM;

    // If k is less than L, make it in the
    // range between L to R
    if (k < L)
        k = k + LCM;

    // Loop to iterate the from L to R
    // and printing the numbers which
    // are divisible by all array elements
    for(int i = k; i <= R; i = i + LCM)
    {
        System.out.print(i + " ");
    }
}

// Driver Code
public static void main(String args[])
{
    int L = 90;
    int R = 280;
    int arr[] = { 3, 5, 12 };
    int N = arr.length;

    solve(arr, N, L, R);
}
}

// This code is contributed by sanjoy_62

Python3

# Python program for the above approach

# Recursive function to return gcd of a and b
def __gcd(a, b):
    # Everything divides 0
    if (a == 0):
        return b;
    if (b == 0):
        return a;

    # Base case
    if (a == b):
        return a;

    # a is greater
    if (a > b):
        return __gcd(a - b, b);

    return __gcd(a, b - a);


# Function to return Kth smallest
# prime number if it exists
def solve(arr, N, L, R):

    # For storing the LCM
    LCM = arr[0];

    # Loop to iterate the array
    for i in range(1, N):

        # Taking LCM of numbers
        LCM = (LCM * arr[i]) // (__gcd(LCM, arr[i]));

    # Checking if no elements is divisible
    # by all elements of given array of given
    # range, pr-1
    if ((LCM < L and LCM * 2 > R) or LCM > R):
        print("-1");
        return;

    # Taking nearest value of L which is
    # divisible by whole array
    k = (L // LCM) * LCM;

    # If k is less than L, make it in the
    # range between L to R
    if (k < L):
        k = k + LCM;

    # Loop to iterate the from L to R
    # and printing the numbers which
    # are divisible by all array elements
    for i in range(k,R+1,LCM):
        print(i, end=" ");

# Driver Code
if __name__ == '__main__':
    L = 90;
    R = 280;
    arr = [3, 5, 12];
    N = len(arr);

    solve(arr, N, L, R);

# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;

public class GFG{

  // Recursive function to return gcd of a and b
  static int __gcd(int a, int b)
  {

    // Everything divides 0
    if (a == 0)
      return b;
    if (b == 0)
      return a;

    // Base case
    if (a == b)
      return a;

    // a is greater
    if (a > b)
      return __gcd(a - b, b);

    return __gcd(a, b - a);
  }

  // Function to return Kth smallest
  // prime number if it exists
  static void solve(int[] arr, int N, int L, int R)
  {

    // For storing the LCM
    int LCM = arr[0];

    // Loop to iterate the array
    for(int i = 1; i < N; i++)
    {

      // Taking LCM of numbers
      LCM = (LCM * arr[i]) /
        (__gcd(LCM, arr[i]));
    }

    // Checking if no elements is divisible
    // by all elements of given array of given
    // range, print -1
    if ((LCM < L && LCM * 2 > R) || LCM > R)
    {
      Console.WriteLine("-1");
      return;
    }

    // Taking nearest value of L which is
    // divisible by whole array
    int k = (L / LCM) * LCM;

    // If k is less than L, make it in the
    // range between L to R
    if (k < L)
      k = k + LCM;

    // Loop to iterate the from L to R
    // and printing the numbers which
    // are divisible by all array elements
    for(int i = k; i <= R; i = i + LCM)
    {
      Console.Write(i + " ");
    }
  }

  // Driver Code
  public static void Main(String []args)
  {
    int L = 90;
    int R = 280;
    int []arr = { 3, 5, 12 };
    int N = arr.Length;

    solve(arr, N, L, R);
  }
}

// This code is contributed by 29AjayKumar

Javascript

  <script>
      // JavaScript code for the above approach
      // Recursive function to return gcd of a and b
      function __gcd(a, b) {

          // Everything divides 0
          if (b == 0) {
              return a;
          }

          return __gcd(b, a % b);
      }

      // Function to return Kth smallest
      // prime number if it exists
      function solve(arr, N, L, R)
      {

          // For storing the LCM
          let LCM = arr[0];

          // Loop to iterate the array
          for (let i = 1; i < N; i++)
          {

              // Taking LCM of numbers
              LCM = Math.floor((LCM * arr[i]) /
                  (__gcd(LCM, arr[i])));
          }

          // Checking if no elements is divisible
          // by all elements of given array of given
          // range, print -1
          if ((LCM < L && LCM * 2 > R) || LCM > R) {
              document.write("-1");
              return;
          }

          // Taking nearest value of L which is
          // divisible by whole array
          let k = Math.floor((L / LCM)) * LCM;

          // If k is less than L, make it in the
          // range between L to R
          if (k < L)
              k = k + LCM;

          // Loop to iterate the from L to R
          // and printing the numbers which
          // are divisible by all array elements
          for (let i = k; i <= R; i = i + LCM) {
              document.write(i + " ");
          }
      }

      // Driver Code
      let L = 90;
      let R = 280;
      let arr = [3, 5, 12];
      let N = arr.length;
      solve(arr, N, L, R);

// This code is contributed by Potta Lokesh
  </script>

Output

120 180 240 

Time Complexity: O(N)
Auxiliary Space: O(1)