# Maximum possible GCD for a pair of integers with product N

Given an integer N, the task is to find the maximum possible GCD among all pair of integers with product N.
Examples:

Input: N=12
Output:
Explanation:
All possible pairs with product 12 are {1, 12}, {2, 6}, {3, 4}
GCD(1, 12) = 1
GCD(2, 6) = 2
GCD(3, 4) = 1
Therefore, the maximum possible GCD = maximum(1, 2, 1) = 2

Input:
Output:
Explanation:
All possible pairs with product 4 are {1, 4}, {2, 2}
GCD(1, 4) = 1
GCD(2, 2) = 2
Hence, maximum possible GCD = maximum(1, 2) = 2

Naive Approach:
The simplest approach to solve this problem is to generate all possible pairs with product N and calculate GCD of all such pairs. Finally, print the maximum GCD obtained.

Time Complexity: O(NlogN)
Auxiliary Space: O(1)

Efficient Approach:
The above approach can be optimized by finding all divisors of the given number N. For each pair of divisors obtained, calculate their GCD. Finally, print the maximum GCD obtained.
Follow the steps below to solve the problem:

• Declare a variable maxGcd to keep track of maximum GCD.
• Iterate up to √N and for every integer, check if it is a factor of N.
• If N is divisible by i, calculate GCD of the pair of factors (i, N / i).
• Comapare with GCD(i, N / i) and update maxGcd.
• Finally, print maxGcd.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return ` `/// the maximum GCD ` `int` `getMaxGcd(``int` `N) ` `{ ` `    ``int` `maxGcd = INT_MIN, A, B; ` ` `  `    ``// To find all divisors of N ` `    ``for` `(``int` `i = 1; i <= ``sqrt``(N); i++) { ` ` `  `        ``// If i is a factor ` `        ``if` `(N % i == 0) { ` `            ``// Store the pair of factors ` `            ``A = i, B = N / i; ` ` `  `            ``// Store the maximum GCD ` `            ``maxGcd = max(maxGcd, __gcd(A, B)); ` `        ``} ` `    ``} ` ` `  `    ``// Return the maximum GCD ` `    ``return` `maxGcd; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `N = 18; ` `    ``cout << getMaxGcd(N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to implement ` `// the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == ``0``)  ` `        ``return` `a; ` `    ``return` `gcd(b, a % b); ` `} ` `     `  `// Function to return ` `// the maximum GCD ` `static` `int` `getMaxGcd(``int` `N) ` `{ ` `    ``int` `maxGcd = Integer.MIN_VALUE, A, B; ` `     `  `    ``// To find all divisors of N ` `    ``for``(``int` `i = ``1``; i <= Math.sqrt(N); i++)  ` `    ``{ ` `         `  `        ``// If i is a factor ` `        ``if` `(N % i == ``0``)  ` `        ``{ ` `             `  `            ``// Store the pair of factors ` `            ``A = i; ` `            ``B = N / i; ` `         `  `            ``// Store the maximum GCD ` `            ``maxGcd = Math.max(maxGcd, gcd(A, B)); ` `        ``} ` `    ``} ` `     `  `    ``// Return the maximum GCD ` `    ``return` `maxGcd; ` `} ` `     `  `// Driver Code ` `public` `static` `void` `main(String s[]) ` `{ ` `    ``int` `N = ``18``; ` `     `  `    ``System.out.println(getMaxGcd(N)); ` `}  ` `} ` ` `  `// This code is contributed by rutvik_56 `

## Python3

 `# Python3 program to implement ` `# the above approach ` `import` `sys ` `import` `math ` ` `  `# Function to return ` `# the maximum GCD ` `def` `getMaxGcd(N): ` `     `  `    ``maxGcd ``=` `-``sys.maxsize ``-` `1` ` `  `    ``# To find all divisors of N ` `    ``for` `i ``in` `range``(``1``, ``int``(math.sqrt(N)) ``+` `1``): ` ` `  `        ``# If i is a factor ` `        ``if` `(N ``%` `i ``=``=` `0``): ` `             `  `            ``# Store the pair of factors ` `            ``A ``=` `i ` `            ``B ``=` `N ``/``/` `i ` ` `  `            ``# Store the maximum GCD ` `            ``maxGcd ``=` `max``(maxGcd, math.gcd(A, B)) ` `         `  `    ``# Return the maximum GCD ` `    ``return` `maxGcd ` ` `  `# Driver Code ` `N ``=` `18` ` `  `print``(getMaxGcd(N)) ` ` `  `# This code is contributed by code_hunt `

## C#

 `// C# program to implement ` `// the above approach ` `using` `System; ` `class` `GFG{ ` `     `  `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == 0)  ` `        ``return` `a; ` `    ``return` `gcd(b, a % b); ` `} ` `     `  `// Function to return ` `// the maximum GCD ` `static` `int` `getMaxGcd(``int` `N) ` `{ ` `    ``int` `maxGcd = ``int``.MinValue, A, B; ` `     `  `    ``// To find all divisors of N ` `    ``for``(``int` `i = 1; i <= Math.Sqrt(N); i++)  ` `    ``{ ` `         `  `        ``// If i is a factor ` `        ``if` `(N % i == 0)  ` `        ``{ ` `             `  `            ``// Store the pair of factors ` `            ``A = i; ` `            ``B = N / i; ` `         `  `            ``// Store the maximum GCD ` `            ``maxGcd = Math.Max(maxGcd, gcd(A, B)); ` `        ``} ` `    ``} ` `     `  `    ``// Return the maximum GCD ` `    ``return` `maxGcd; ` `} ` `     `  `// Driver Code ` `public` `static` `void` `Main(String []s) ` `{ ` `    ``int` `N = 18; ` `     `  `    ``Console.WriteLine(getMaxGcd(N)); ` `}  ` `} ` ` `  `// This code is contributed by sapnasingh4991`

Output:

```3
```

Time Complexity: O(√N*log(N))
Auxiliary Space: O(1)

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