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 Codeint 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 Codeint 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 Codel = 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 Codelet 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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