# Divide N into K unique parts such that gcd of those parts is maximum

Given a positive integer N, the task is to divide it into K unique parts such that the sum of these parts is equal to the original number and the gcd of all the parts is maximum. Print the maximum gcd if such a division exists else print -1.

Examples:

Input: N = 6, K = 3
Output:
Only possible division with unique
elements are (1, 2, 3) and gcd(1, 2, 3) = 1.
Input: N = 18, K = 3
Output:

Naive approach: Find all the possible divisions of N and compute the maximum gcd of them. But this approach will take exponential time and space.
Efficient approach: Let the divisions of N be A1, A2……..AK
Now, it is known that gcd(a, b) = gcd(a, b, a + b) and hence, gcd(A1, A2….AK) = gcd(A1, A2….AK, A1 + A2…. + AK)
But A1 + A2…. + AK = N and hence, the gcd of the divisions will be one of the factors of N
Let a be the factor of N which can be the possible answer:
Since a is the gcd, all the division will be the multiples of a and hence, for K unique Bi
a * B1 + a * B2……. + a * BK = N
a * (B1 + B2…….+ BK) = N
Since all the Bi are unique,
B1 + B2…….+ BK ? 1 + 2 + 3 ….. + K
B1 + B2…….+ BK ? K * (K + 1) / 2
Hence, all the factors of N whose complementary factor is greater than or equal to K * (K + 1) / 2 can be one of the possible answers, and we have taken to the maximum of all the possible answers.

Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to calculate maximum GCD` `int` `maxGCD(``int` `N, ``int` `K)` `{`   `    ``// Minimum possible sum for` `    ``// K unique positive integers` `    ``int` `minSum = (K * (K + 1)) / 2;`   `    ``// It is not possible to divide` `    ``// N into K unique parts` `    ``if` `(N < minSum)` `        ``return` `-1;`   `    ``// All the factors greater than sqrt(N)` `    ``// are complementary of the factors less` `    ``// than sqrt(N)` `    ``int` `i = ``sqrt``(N);` `    ``int` `res = 1;` `    ``while` `(i >= 1) {`   `        ``// If i is a factor of N` `        ``if` `(N % i == 0) {` `            ``if` `(i >= minSum)` `                ``res = max(res, N / i);`   `            ``if` `(N / i >= minSum)` `                ``res = max(res, i);` `        ``}` `        ``i--;` `    ``}`   `    ``return` `res;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `N = 18, K = 3;`   `    ``cout << maxGCD(N, K);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.io.*;` `import` `java.lang.Math; `   `class` `GFG` `{` `    ``// Function to calculate maximum GCD` `    ``static` `int` `maxGCD(``int` `N, ``int` `K)` `    ``{`   `        ``// Minimum possible sum for` `        ``// K unique positive integers` `        ``int` `minSum = (K * (K + ``1``)) / ``2``;`   `        ``// It is not possible to divide` `        ``// N into K unique parts` `        ``if` `(N < minSum)` `            ``return` `-``1``;`   `        ``// All the factors greater than sqrt(N)` `        ``// are complementary of the factors less` `        ``// than sqrt(N)` `        ``int` `i = (``int``) Math.sqrt(N);` `        ``int` `res = ``1``;` `        ``while` `(i >= ``1``)` `        ``{`   `            ``// If i is a factor of N` `            ``if` `(N % i == ``0``)` `            ``{` `                ``if` `(i >= minSum)` `                    ``res = Math.max(res, N / i);`   `                ``if` `(N / i >= minSum)` `                    ``res = Math.max(res, i);` `            ``}` `            ``i--;` `        ``}`   `        ``return` `res;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main (String[] args)` `    ``{` `        ``int` `N = ``18``, K = ``3``;`   `        ``System.out.println(maxGCD(N, K));` `    ``}` `}`   `// This code is contributed by ApurvaRaj`

## Python

 `# Python3 implementation of the approach` `from` `math ``import` `sqrt,ceil,floor`   `# Function to calculate maximum GCD` `def` `maxGCD(N, K):`   `    ``# Minimum possible sum for` `    ``# K unique positive integers` `    ``minSum ``=` `(K ``*` `(K ``+` `1``)) ``/` `2`   `    ``# It is not possible to divide` `    ``# N into K unique parts` `    ``if` `(N < minSum):` `        ``return` `-``1`   `    ``# All the factors greater than sqrt(N)` `    ``# are complementary of the factors less` `    ``# than sqrt(N)` `    ``i ``=` `ceil(sqrt(N))` `    ``res ``=` `1` `    ``while` `(i >``=` `1``):`   `        ``# If i is a factor of N` `        ``if` `(N ``%` `i ``=``=` `0``):` `            ``if` `(i >``=` `minSum):` `                ``res ``=` `max``(res, N ``/` `i)`   `            ``if` `(N ``/` `i >``=` `minSum):` `                ``res ``=` `max``(res, i)`   `        ``i``-``=``1`   `    ``return` `res`   `# Driver code`   `N ``=` `18` `K ``=` `3`   `print``(maxGCD(N, K))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG` `{` `    ``// Function to calculate maximum GCD` `    ``static` `int` `maxGCD(``int` `N, ``int` `K)` `    ``{`   `        ``// Minimum possible sum for` `        ``// K unique positive integers` `        ``int` `minSum = (K * (K + 1)) / 2;`   `        ``// It is not possible to divide` `        ``// N into K unique parts` `        ``if` `(N < minSum)` `            ``return` `-1;`   `        ``// All the factors greater than sqrt(N)` `        ``// are complementary of the factors less` `        ``// than sqrt(N)` `        ``int` `i = (``int``) Math.Sqrt(N);` `        ``int` `res = 1;` `        ``while` `(i >= 1)` `        ``{`   `            ``// If i is a factor of N` `            ``if` `(N % i == 0)` `            ``{` `                ``if` `(i >= minSum)` `                    ``res = Math.Max(res, N / i);`   `                ``if` `(N / i >= minSum)` `                    ``res = Math.Max(res, i);` `            ``}` `            ``i--;` `        ``}` `        ``return` `res;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `N = 18, K = 3;`   `        ``Console.WriteLine(maxGCD(N, K));` `    ``}` `}`   `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output:

`3`

Time Complexity: O(N1/2)

Auxiliary Space: O(1)

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