Given a number n such that 1 <= N <= 10^6, the Task is to Find the LCM of First n Natural Numbers.
Examples:
Input : n = 5 Output : 60 Input : n = 6 Output : 60 Input : n = 7 Output : 420
We strongly recommend that you click here and practice it, before moving on to the solution.
We have discussed a simple solution in below article.
Smallest number divisible by first n numbers
The above solution works fine for single input. But if we have multiple inputs, it is a good idea to use Sieve of Eratosthenes to store all prime factors. As we know if LCM(a, b) = X so any prime factor of a or b will also be the prime factor of ‘X’.
- Initialize lcm variable with 1
- Generate a Sieve of Eratosthenes (bool vector isPrime) of length 10^6 (ideally must be equal to no. of digits in factorial)
- Now, for each number in bool vector isPrime, if the number is prime (isPrime[i] is true), find the maximum number which is less than the given number and equal to power of the prime.
- Then multiply this number with lcm variable.
- Repeat step 3 and 4 until prime is less than the given number.
Illustration:
For example, if n = 10 8 will be the first number which is equal to 2^3 then 9 which is equal to 3^2 then 5 which is equal to 5^1 then 7 which is equal to 7^1 Finally, we multiply those numbers 8*9*5*7 = 2520
Below is the implementation of the above idea.
// C++ program to find LCM of First N Natural Numbers. #include <bits/stdc++.h> #define MAX 100000 using namespace std;
vector< bool > isPrime (MAX, true );
// utility function for sieve of sieve of Eratosthenes void sieve()
{ for ( int i = 2; i * i <= MAX; i++)
{
if (isPrime[i] == true )
for ( int j = i*i; j<= MAX; j+=i)
isPrime[j] = false ;
}
} // Function to find LCM of first n Natural Numbers long long LCM( int n)
{ long long lcm = 1;
int i=2;
while (i<=n) {
if (isPrime[i]){
int pp = i;
while (pp * i <= n)
pp = pp * i;
lcm *= pp;
}
i++;
}
return lcm;
} // Driver code int main()
{ // build sieve
sieve();
int N = 7;
// Function call
cout << LCM(N);
return 0;
} |
// Java program to find LCM of First N Natural Numbers. import java.util.*;
class GFG
{ static int MAX = 100000 ;
// array to store all prime less than and equal to 10^6
static ArrayList<Integer> primes
= new ArrayList<Integer>();
// utility function for sieve of sieve of Eratosthenes
static void sieve()
{
boolean [] isComposite = new boolean [MAX + 1 ];
for ( int i = 2 ; i * i <= MAX; i++)
{
if (isComposite[i] == false )
for ( int j = 2 ; j * i <= MAX; j++)
isComposite[i * j] = true ;
}
// Store all prime numbers in vector primes[]
for ( int i = 2 ; i <= MAX; i++)
if (isComposite[i] == false )
primes.add(i);
}
// Function to find LCM of first n Natural Numbers
static long LCM( int n)
{
long lcm = 1 ;
for ( int i = 0 ;
i < primes.size() && primes.get(i) <= n;
i++)
{
// Find the highest power of prime, primes[i]
// that is less than or equal to n
int pp = primes.get(i);
while (pp * primes.get(i) <= n)
pp = pp * primes.get(i);
// multiply lcm with highest power of prime[i]
lcm *= pp;
lcm %= 1000000007 ;
}
return lcm;
}
// Driver code
public static void main(String[] args)
{
sieve();
int N = 7 ;
// Function call
System.out.println(LCM(N));
}
} // This code is contributed by mits |
# Python3 program to find LCM of # First N Natural Numbers. MAX = 100000
# array to store all prime less # than and equal to 10^6 primes = []
# utility function for # sieve of Eratosthenes def sieve():
isComposite = [ False ] * ( MAX + 1 )
i = 2
while (i * i < = MAX ):
if (isComposite[i] = = False ):
j = 2
while (j * i < = MAX ):
isComposite[i * j] = True
j + = 1
i + = 1
# Store all prime numbers in
# vector primes[]
for i in range ( 2 , MAX + 1 ):
if (isComposite[i] = = False ):
primes.append(i)
# Function to find LCM of # first n Natural Numbers def LCM(n):
lcm = 1
i = 0
while (i < len (primes) and primes[i] < = n):
# Find the highest power of prime,
# primes[i] that is less than or
# equal to n
pp = primes[i]
while (pp * primes[i] < = n):
pp = pp * primes[i]
# multiply lcm with highest
# power of prime[i]
lcm * = pp
lcm % = 1000000007
i + = 1
return lcm
# Driver code sieve() N = 7
# Function call print (LCM(N))
# This code is contributed by mits |
// C# program to find LCM of First N // Natural Numbers. using System.Collections;
using System;
class GFG {
static int MAX = 100000;
// array to store all prime less than
// and equal to 10^6
static ArrayList primes = new ArrayList();
// utility function for sieve of
// sieve of Eratosthenes
static void sieve()
{
bool [] isComposite = new bool [MAX + 1];
for ( int i = 2; i * i <= MAX; i++)
{
if (isComposite[i] == false )
for ( int j = 2; j * i <= MAX; j++)
isComposite[i * j] = true ;
}
// Store all prime numbers in vector primes[]
for ( int i = 2; i <= MAX; i++)
if (isComposite[i] == false )
primes.Add(i);
}
// Function to find LCM of first
// n Natural Numbers
static long LCM( int n)
{
long lcm = 1;
for ( int i = 0;
i < primes.Count && ( int )primes[i] <= n;
i++)
{
// Find the highest power of prime, primes[i]
// that is less than or equal to n
int pp = ( int )primes[i];
while (pp * ( int )primes[i] <= n)
pp = pp * ( int )primes[i];
// multiply lcm with highest power of prime[i]
lcm *= pp;
lcm %= 1000000007;
}
return lcm;
}
// Driver code
public static void Main()
{
sieve();
int N = 7;
// Function call
Console.WriteLine(LCM(N));
}
} // This code is contributed by mits |
<script> // Javascript program to find LCM of First N
// Natural Numbers.
let MAX = 100000;
// array to store all prime less than
// and equal to 10^6
let primes = [];
// utility function for sieve of
// sieve of Eratosthenes
function sieve()
{
let isComposite = new Array(MAX + 1);
isComposite.fill( false );
for (let i = 2; i * i <= MAX; i++)
{
if (isComposite[i] == false )
for (let j = 2; j * i <= MAX; j++)
isComposite[i * j] = true ;
}
// Store all prime numbers in vector primes[]
for (let i = 2; i <= MAX; i++)
if (isComposite[i] == false )
primes.push(i);
}
// Function to find LCM of first
// n Natural Numbers
function LCM(n)
{
let lcm = 1;
for (let i = 0;
i < primes.length && primes[i] <= n;
i++)
{
// Find the highest power of prime, primes[i]
// that is less than or equal to n
let pp = primes[i];
while (pp * primes[i] <= n)
pp = pp * primes[i];
// multiply lcm with highest power of prime[i]
lcm *= pp;
lcm %= 1000000007;
}
return lcm;
}
sieve();
let N = 7;
// Function call
document.write(LCM(N));
// This code is contributed by decode2207. </script> |
<?php // PHP program to find LCM of // First N Natural Numbers. $MAX = 100000;
// array to store all prime less // than and equal to 10^6 $primes = array ();
// utility function for // sieve of Eratosthenes function sieve()
{ global $MAX , $primes ;
$isComposite = array_fill (0, $MAX , false);
for ( $i = 2; $i * $i <= $MAX ; $i ++)
{
if ( $isComposite [ $i ] == false)
for ( $j = 2; $j * $i <= $MAX ; $j ++)
$isComposite [ $i * $j ] = true;
}
// Store all prime numbers in
// vector primes[]
for ( $i = 2; $i <= $MAX ; $i ++)
if ( $isComposite [ $i ] == false)
array_push ( $primes , $i );
} // Function to find LCM of // first n Natural Numbers function LCM( $n )
{ global $MAX , $primes ;
$lcm = 1;
for ( $i = 0; $i < count ( $primes ) &&
$primes [ $i ] <= $n ; $i ++)
{
// Find the highest power of prime,
// primes[i] that is less than or
// equal to n
$pp = $primes [ $i ];
while ( $pp * $primes [ $i ] <= $n )
$pp = $pp * $primes [ $i ];
// multiply lcm with highest
// power of prime[i]
$lcm *= $pp ;
$lcm %= 1000000007;
}
return $lcm ;
} // Driver code sieve(); $N = 7;
// Function call echo LCM( $N );
// This code is contributed by mits ?> |
420
Time Complexity: O(n2)
Auxiliary Space: O(n)
Another Approach:
The idea is that if the number is less than 3 then return number. If the number is greater than 2 then find LCM of n,n-1
- Lets say x=LCM(n,n-1)
- again x=LCM(x,n-2)
- again x=LCM(x,n-3) …
- .
- .
- again x=LCM(x,1) …
now the result is x.
For finding LCM(a,b) we use a function hcf(a,b) whichwill return HCF of (a,b)
We know that LCM(a,b)= (a*b)/HCF(a,b)
Illustration:
For example, if n = 7 function call lcm(7,6) now lets say a=7 , b=6 Now , b!= 1 Hence a=lcm(7,6) = 42 and b=6-1=5 function call lcm(42,5) a=lcm(42,5) = 210 and b=5-1=4 function call lcm(210,4) a=lcm(210,4) = 420 and b=4-1=3 function call lcm(420,3) a=lcm(420,3) = 420 and b=3-1=2 function call lcm(420,2) a=lcm(420,2) = 420 and b=2-1=1 Now b=1 Hence return a=420
Below is the implementation of the above approach
// C++ program to find LCM of First N Natural Numbers. #include <bits/stdc++.h> using namespace std;
// to calculate hcf int hcf( int a, int b)
{ if (b == 0)
return a;
return hcf(b, a % b);
} int findlcm( int a, int b)
{ if (b == 1)
// lcm(a,b)=(a*b)/hcf(a,b)
return a;
// assign a=lcm of n,n-1
a = (a * b) / hcf(a, b);
// b=b-1
b -= 1;
return findlcm(a, b);
} // Driver code int main()
{ int n = 7;
if (n < 3)
cout << n; // base case
else
// Function call
// pass n,n-1 in function to find LCM of first n natural
// number
cout << findlcm(n, n - 1);
return 0;
} // contributed by ajaykr00kj |
// Java program to find LCM of First N Natural Numbers public class Main
{ // to calculate hcf
static int hcf( int a, int b)
{
if (b == 0 )
return a;
return hcf(b, a % b);
}
static int findlcm( int a, int b)
{
if (b == 1 )
// lcm(a,b)=(a*b)/hcf(a,b)
return a;
// assign a=lcm of n,n-1
a = (a * b) / hcf(a, b);
// b=b-1
b -= 1 ;
return findlcm(a, b);
}
// Driver code.
public static void main(String[] args)
{
int n = 7 ;
if (n < 3 )
System.out.print(n); // base case
else
// Function call
// pass n,n-1 in function to find LCM of first n natural
// number
System.out.print(findlcm(n, n - 1 ));
}
} // This code is contributed by divyeshrabadiya07. |
# Python3 program to find LCM # of First N Natural Numbers. # To calculate hcf def hcf(a, b):
if (b = = 0 ):
return a
return hcf(b, a % b)
def findlcm(a, b):
if (b = = 1 ):
# lcm(a,b)=(a*b)//hcf(a,b)
return a
# Assign a=lcm of n,n-1
a = (a * b) / / hcf(a, b)
# b=b-1
b - = 1
return findlcm(a, b)
# Driver code n = 7
if (n < 3 ):
print (n)
else :
# Function call
# pass n,n-1 in function
# to find LCM of first n
# natural number
print (findlcm(n, n - 1 ))
# This code is contributed by Shubham_Singh |
// C# program to find LCM of First N Natural Numbers. using System;
class GFG {
// to calculate hcf
static int hcf( int a, int b)
{
if (b == 0)
return a;
return hcf(b, a % b);
}
static int findlcm( int a, int b)
{
if (b == 1)
// lcm(a,b)=(a*b)/hcf(a,b)
return a;
// assign a=lcm of n,n-1
a = (a * b) / hcf(a, b);
// b=b-1
b -= 1;
return findlcm(a, b);
}
// Driver code
static void Main() {
int n = 7;
if (n < 3)
Console.Write(n); // base case
else
// Function call
// pass n,n-1 in function to find LCM of first n natural
// number
Console.Write(findlcm(n, n - 1));
}
} // This code is contributed by divyesh072019. |
<script> // Javascript program to find LCM of First N Natural Numbers.
// to calculate hcf
function hcf(a, b)
{
if (b == 0)
return a;
return hcf(b, a % b);
}
function findlcm(a,b)
{
if (b == 1)
// lcm(a,b)=(a*b)/hcf(a,b)
return a;
// assign a=lcm of n,n-1
a = (a * b) / hcf(a, b);
// b=b-1
b -= 1;
return findlcm(a, b);
}
let n = 7;
if (n < 3)
document.write(n); // base case
else
// Function call
// pass n,n-1 in function to find LCM of first n natural
// number
document.write(findlcm(n, n - 1));
</script> |
420
Time complexity : O(nlog n)
Auxiliary Space: O(1)