Find LCM of two numbers

Problem Statement: Find lcm of two numbers.

Examples:

```Example 1:
Input: num1 = 4,num2 = 8
Output: 8

Example 2:
Input: num1 = 3,num2 = 6
Output: 6
```

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution 1:Brute force

Intuition: Simply traverse from 1 to min(a,b) and check for every number.

Approach

• Find gcd using brute force.
• Traverse from 1 to min(a,b).
• And check if i is divisible by both a and b.If yes store it in the answer.
• Find the maximum value of i which divides both a and b.
• To find lcm simply divide (a*b) by gcd of a and b.

Code:

C++ Code

``````#include<bits/stdc++.h>
using namespace std;
int main()
{
int a = 4, b = 8;
int gcd;
for (int i = 1; i <= min(a, b); i++) {
if (a % i == 0 && b % i == 0) {
gcd = i;
}
}
int lcm = (a * b) / gcd;
cout <<"The LCM of the two given numbers is "<<lcm;

}
``````

Output:

The LCM of the two given numbers is 8

Time Complexity: O(N).

Space Complexity: O(1).

Java Code

``````public class Main {
public static void main(String args[]) {
int a = 4, b = 8;
int gcd = 1;
for (int i = 1; i <= Math.min(a, b); i++) {
if (a % i == 0 && b % i == 0) {
gcd = i;
}
}
int lcm = (a * b) / gcd;
System.out.print("The LCM of the two given numbers is "+lcm);

}
}
``````

Output:

The LCM of the two given numbers is 8

Time Complexity: O(N).

Space Complexity: O(1).

Solution 2: Using Euclidean’s theorem.

Intuition: Gcd is the greatest number which is divided by both a and b.If a number is divided by both a and b, it is should be divided by (a-b) and b as well.Using this we can find gcd in log(min(a,b)) and hence lcm.

Approach:

• First find gcd using euclidean’s algorithm,
• Recursively call gcd(b,a%b) function till the base condition is hit.
• b==0 is the base condition.When base condition is hit return a,as gcd(a,0) is equal to a.
• Once we get gcd,we can find lcm using formula lcm = (a*b)/gcd.

Code:

C++ Code

``````#include<bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
int main()
{
int a = 4, b = 8;
int g = gcd(a, b);
int lcm = (a * b) / g;
cout <<"The LCM of the two given numbers is "<<lcm;
}
``````

Output:

The LCM of the two given numbers is 8

Time Complexity: O(logɸmin(a,b)),where ɸ is 1.61.

Space Complexity: O(1).

Java Code

``````public class Main {
static int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
public static void main(String args[]) {
int a = 4, b = 8;
int gcd = gcd(a, b);
int lcm = (a * b) / gcd;
System.out.print("The LCM of the two given numbers is "+lcm);
}
}
``````

Output:

The LCM of the two given numbers is 8

Time Complexity: O(logɸmin(a,b)),where ɸ is 1.61.

Space Complexity: O(1).