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Greatest divisor which divides all natural number in range [L, R]

  • Difficulty Level : Easy
  • Last Updated : 15 Apr, 2021

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 
 

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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);
}

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)
 




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