Given two integers L and R, the task is to find the greatest divisor that divides all the natural numbers in the range [L, R].
Examples:
Input: L = 3, R = 12
Output: 1
Input: L = 24, R = 24
Output: 24
Approach: For a range of consecutive integer elements, there are two cases:
- If L = R then the answer will L.
- If L < R then all consecutive natural numbers in this range are co-primes. So, 1 is the only number that will be able to divide all the elements of the range.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the greatest divisor that // divides all the natural numbers in the range [l, r] int find_greatest_divisor( int l, int r)
{ if (l == r)
return l;
return 1;
} // Driver Code int main()
{ int l = 2, r = 12;
cout << find_greatest_divisor(l, r);
} |
C
// C implementation of the approach #include <stdio.h> // Function to return the greatest divisor that // divides all the natural numbers in the range [l, r] int find_greatest_divisor( int l, int r)
{ if (l == r)
return l;
return 1;
} // Driver Code int main()
{ int l = 2, r = 12;
printf ( "%d" ,find_greatest_divisor(l, r));
} // This code is contributed by kothvvsaakash. |
Java
// Java implementation of the approach class GFG {
// Function to return the greatest divisor that // divides all the natural numbers in the range [l, r] static int find_greatest_divisor( int l, int r) {
if (l == r) {
return l;
}
return 1 ;
}
// Driver Code public static void main(String[] args) {
int l = 2 , r = 12 ;
System.out.println(find_greatest_divisor(l, r));
}
} // This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach # Function to return the greatest divisor that # divides all the natural numbers in the range [l, r] def find_greatest_divisor(l, r):
if (l = = r):
return l;
return 1 ;
# Driver Code l = 2 ;
r = 12 ;
print (find_greatest_divisor(l, r));
#This code is contributed by Shivi_Aggarwal |
C#
// C# implementation of the approach using System;
class GFG {
// Function to return the greatest divisor that
// divides all the natural numbers in the range [l, r]
static int find_greatest_divisor( int l, int r) {
if (l == r) {
return l;
}
return 1;
}
// Driver Code
public static void Main() {
int l = 2, r = 12 ;
Console.WriteLine(find_greatest_divisor(l, r));
}
// This code is contributed by Ryuga
} |
PHP
<?php // PHP implementation of the approach // Function to return the greatest // divisor that divides all the natural // numbers in the range [l, r] function find_greatest_divisor( $l , $r )
{ if ( $l == $r )
return $l ;
return 1;
} // Driver Code $l = 2;
$r = 12;
echo find_greatest_divisor( $l , $r );
// This code is contributed by jit_t ?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the greatest // divisor that divides all the natural // numbers in the range [l, r] function find_greatest_divisor(l, r)
{ if (l == r)
return l;
return 1;
} // Driver Code let l = 2; let r = 12; document.write( find_greatest_divisor(l, r)); // This code is contributed // by bobby </script> |
Output:
1
Time Complexity: O(1)
Auxiliary Space: O(1)
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