Find two numbers with the given LCM and minimum possible difference
Given an integer X, the task is to find two integers A and B such that LCM(A, B) = X and the difference between the A and B is minimum possible.
Examples:
Input: X = 6
Output: 2 3
LCM(2, 3) = 6 and (3 – 2) = 1
which is the minimum possible.
Input X = 7
Output: 1 7
Approach: An approach to solve this problem is to find all the factors of the given number using the approach discussed in this article and then find the pair (A, B) that satisfies the given conditions and has the minimum possible difference.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int lcm( int a, int b)
{
return (a / __gcd(a, b) * b);
}
void findNums( int x)
{
int ans;
for ( int i = 1; i <= sqrt (x); i++) {
if (x % i == 0 && lcm(i, x / i) == x) {
ans = i;
}
}
cout << ans << " " << (x / ans);
}
int main()
{
int x = 12;
findNums(x);
return 0;
}
|
Java
class GFG
{
static int lcm( int a, int b)
{
return (a / __gcd(a, b) * b);
}
static int __gcd( int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
static void findNums( int x)
{
int ans = - 1 ;
for ( int i = 1 ; i <= Math.sqrt(x); i++)
{
if (x % i == 0 && lcm(i, x / i) == x)
{
ans = i;
}
}
System.out.print(ans + " " + (x / ans));
}
public static void main(String[] args)
{
int x = 12 ;
findNums(x);
}
}
|
Python3
from math import gcd as __gcd, sqrt, ceil
def lcm(a, b):
return (a / / __gcd(a, b) * b)
def findNums(x):
ans = 0
for i in range ( 1 , ceil(sqrt(x))):
if (x % i = = 0 and lcm(i, x / / i) = = x):
ans = i
print (ans, (x / / ans))
x = 12
findNums(x)
|
C#
using System;
class GFG
{
static int lcm( int a, int b)
{
return (a / __gcd(a, b) * b);
}
static int __gcd( int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
static void findNums( int x)
{
int ans = -1;
for ( int i = 1; i <= Math.Sqrt(x); i++)
{
if (x % i == 0 && lcm(i, x / i) == x)
{
ans = i;
}
}
Console.Write(ans + " " + (x / ans));
}
public static void Main(String[] args)
{
int x = 12;
findNums(x);
}
}
|
Javascript
<script>
function lcm(a,b)
{
return (a / __gcd(a, b) * b);
}
function __gcd(a,b)
{
return b == 0 ? a : __gcd(b, a % b);
}
function findNums(x)
{
let ans = -1;
for (let i = 1; i <= Math.sqrt(x); i++)
{
if (x % i == 0 && lcm(i, Math.floor(x / i)) == x)
{
ans = i;
}
}
document.write(ans + " " + Math.floor(x / ans));
}
let x = 12;
findNums(x);
</script>
|
Time Complexity: O(n1/2 * log(max(a, b)))
Auxiliary Space: O(log(max(a, b)))
Last Updated :
20 Feb, 2022
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