Given a range L to R, the task is to find the maximum possible value of GCD(X, Y) such that X and Y belongs to the given range, i.e. L ? X < Y ? R.
Examples:
Input: L = 101, R = 139
Output:
34
Explanation:
For X = 102 and Y = 136, the GCD of x and y is 34, which is the maximum possible.Input: L = 8, R = 14
Output:
7Â
Naive Approach: Every pair that can be formed from L to R, can be iterated over using two nested loops and the maximum GCD can be found.
Time Complexity: O((R-L)2Log(R))
Auxiliary Space: O(1)
Efficient Approach: Follow the below steps to solve the problem:
- Let the maximum GCD be Z, therefore, X and Y are both multiples of Z. Conversely if there are two or more multiples of Z in the segment [L, R], then (X, Y) can be chosen such that GCD(x, y) is maximum by choosing consecutive multiples of Z in [L, R].
- Iterate from R to 1 and find whether any of them has at least two multiples in the range [L, R]
- The multiples of Z between L and R can be calculated using the following formula:
- Number of Multiples of Z in [L, R] = Number of multiples of Z in [1, R] – Number of Multiples of Z in [1, L-1]
- This can be written as :
- No. of Multiples of Z in [L, R] = floor(R/Z) – floor((L-1)/Z)
- We can further optimize this by limiting the iteration from R/2 to 1 as the greatest possible GCD is R/2 (with multiples R/2 and R)
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;Â
// Function to calculate GCDint GCD(int a, int b){Â Â Â Â if (b == 0)Â Â Â Â Â Â Â Â return a;Â Â Â Â return GCD(b, a % b);}Â
// Function to calculate// maximum GCD in a rangeint maxGCDInRange(int L, int R){    // Variable to store the answer    int ans = 1;Â
    for (int Z = R/2; Z >= 1; Z--) {Â
        // If Z has two multiples in [L, R]        if ((R / Z) - ((L - 1) / Z) > 1) {Â
            // Update ans            ans = Z;            break;        }    }Â
    // Return the value    return ans;}// Driver codeint main(){    // Input    int L = 102;    int R = 139;Â
    // Function Call    cout << maxGCDInRange(L, R);Â
    return 0;} |
Java
// Java program for the above approachimport java.io.*;Â
class GFG {Â Â Â Â // Function to calculate GCDÂ Â Â Â public static int GCD(int a, int b)Â Â Â Â {Â Â Â Â Â Â Â Â if (b == 0)Â Â Â Â Â Â Â Â Â Â Â Â return a;Â Â Â Â Â Â Â Â return GCD(b, a % b);Â Â Â Â }Â
    // Function to calculate    // maximum GCD in a range    public static int maxGCDInRange(int L, int R)    {        // Variable to store the answer        int ans = 1;Â
        for (int Z = R/2; Z >= 1; Z--) {Â
            // If Z has two multiples in [L, R]            if ((R / Z) - ((L - 1) / Z) > 1) {Â
                // Update ans                ans = Z;                break;            }        }Â
        // Return the value        return ans;    }     // Driver code    public static void main(String[] args)    {        // Input        int L = 102;        int R = 139;Â
        // Function Call        System.out.println(maxGCDInRange(L, R));    }Â
      // This code is contributed by Potta Lokesh |
Python3
# Python3 program for the above approachÂ
# Function to calculate GCDdef GCD(a, b):         if (b == 0):        return a             return GCD(b, a % b)Â
# Function to calculate# maximum GCD in a rangedef maxGCDInRange(L, R):         # Variable to store the answer    ans = 1Â
    for Z in range(R//2, 1, -1):                 # If Z has two multiples in [L, R]        if (((R // Z) - ((L - 1) // Z )) > 1):                         # Update ans            ans = Z            break             # Return the value    return ansÂ
# Driver codeÂ
# InputL = 102R = 139Â
# Function Callprint(maxGCDInRange(L, R))Â
# This code is contributed by SoumikMondal |
C#
// C# program for the above approachusing System;Â
class GFG{Â Â Â Â Â // Function to calculate GCDpublic static int GCD(int a, int b){Â Â Â Â if (b == 0)Â Â Â Â Â Â Â Â return a;Â Â Â Â Â Â Â Â Â Â Â Â Â return GCD(b, a % b);}Â
// Function to calculate// maximum GCD in a rangepublic static int maxGCDInRange(int L, int R){         // Variable to store the answer    int ans = 1;Â
    for(int Z = R/2; Z >= 1; Z--)    {                 // If Z has two multiples in [L, R]        if ((R / Z) - ((L - 1) / Z) > 1)        {                         // Update ans            ans = Z;            break;        }    }Â
    // Return the value    return ans;}Â
// Driver codepublic static void Main(){         // Input    int L = 102;    int R = 139;Â
    // Function Call    Console.Write(maxGCDInRange(L, R));}}Â
// This code is contributed by rishavmahato348 |
Javascript
<script>Â
// JavaScript program for the above approachÂ
// Function to calculate GCDfunction GCD( a, b){Â Â Â Â if (b == 0)Â Â Â Â Â Â Â Â return a;Â Â Â Â return GCD(b, a % b);}Â
// Function to calculate// maximum GCD in a rangefunction maxGCDInRange(L, R){Â
    // Variable to store the answer    let ans = 1;Â
    for (let Z = parseInt((R / 2)); Z >= 1; Z--)     {             // If Z has two multiples in [L, R]        if (parseInt((R / Z)) - parseInt((L - 1) / Z ) > 1)         {                     // Update ans            ans = Z;            break;        }    }Â
    // Return the value    return ans;}Â
// Driver codeÂ
// Inputlet L = 102;let R = 139;Â
// Function Calldocument.write(maxGCDInRange(L, R));Â
// This code is contributed by Potta LokeshÂ
</script> |
Output
34
Time Complexity: O(R)
Auxiliary Space: O(1)
Â
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