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

0 0 5 minutes read

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

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……..A_K
Now, it is known that gcd(a, b) = gcd(a, b, a + b) and hence, gcd(A₁, A₂….A_K) = gcd(A₁, A₂….A_K, A₁ + A₂…. + A_K)
But A₁ + A₂…. + A_K = 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 B_i,
a * B₁ + a * B₂……. + a * B_K = N
a * (B₁ + B₂…….+ B_K) = N
Since all the B_i are unique,
B₁ + B₂…….+ B_K ? 1 + 2 + 3 ….. + K
B₁ + B₂…….+ B_K ? 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 <bits/stdc++.h>
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

<script>
// javascript implementation of the approach
 
    // Function to calculate maximum GCD
    function maxGCD(N , K) {
 
        // Minimum possible sum for
        // K unique positive integers
        var 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)
        var i = parseInt( Math.sqrt(N));
        var 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
     
        var N = 18, K = 3;
 
        document.write(maxGCD(N, K));
 
// This code contributed by Rajput-Ji 
</script>

Output:

Time Complexity: O(N^1/2)

Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!