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Find largest factor of N such that N/F is less than K

  • Last Updated : 11 May, 2021

Given two numbers N and K, the task is to find the minimum value X such that N < X*K.

Examples: 

Input: N = 8, K = 7 
Output:
Explanation: 
Numbers less than K divisible by N are 1, 2 and 4. 
So the minimum value of X is 2 such that 8 < 2*7 = 14.
Input: N = 999999733, K = 999999732 
Output: 999999733 
Explanation: 
Since 999999733 is a prime number, so 999999733 is divisible by 1 and the number itself. Since K is less than 999999733. 
So the minimum value of X is 999999733 such that 999999733 < 999999733*999999732.

Naive Approach: The given problem statement can be visualized as equation K * X = N. In this equation, the objective is to minimize X. So we have to find the maximum K which divides N. Below are the steps: 

  • Iterate over [1, K].
  • Check for each number i such that (N % i) = 0. Keep updating the max variable that stores the maximum divisor of N traversed up to i.
  • The required answer is N/max.

Below is the implementation of the above approach:



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the value of X
void findMaxValue(int N, int K)
{
    int packages;
    int maxi = 1;
 
    // Loop to check all the numbers
    // divisible by N that yield
    // minimum N/i value
    for (int i = 1; i <= K; i++) {
        if (N % i == 0)
            maxi = max(maxi, i);
    }
 
    packages = N / maxi;
 
    // Print the value of packages
    cout << packages << endl;
}
 
// Driver Code
int main()
{
    // Given N and K
    int N = 8, K = 7;
 
    // Function Call
    findMaxValue(N, K);
    return 0;
}

Java




// Java program for the above approach
import java.util.Arrays;
 
class GFG{
     
// Function to find the value of X
static void findMaxValue(int N, int K)
{
    int packages;
    int maxi = 1;
 
    // Loop to check all the numbers
    // divisible by N that yield
    // minimum N/i value
    for(int i = 1; i <= K; i++)
    {
        if (N % i == 0)
            maxi = Math.max(maxi, i);
    }
    packages = N / maxi;
 
    // Print the value of packages
    System.out.println(packages);
}
 
// Driver code
public static void main (String[] args)
{
     
    // Given N and K
    int N = 8, K = 7;
 
    // Function call
    findMaxValue(N, K);
}
}
 
// This code is contributed by Shubham Prakash

Python3




# Python3 program for the above approach
 
# Function to find the value of X
def findMaxValue(N, K):
    packages = 0;
    maxi = 1;
 
    # Loop to check all the numbers
    # divisible by N that yield
    # minimum N/i value
    for i in range(1, K + 1):
        if (N % i == 0):
            maxi = max(maxi, i);
 
    packages = N // maxi;
 
    # Prthe value of packages
    print(packages);
 
# Driver code
if __name__ == '__main__':
   
    # Given N and K
    N = 8;
    K = 7;
     
    # Function call
    findMaxValue(N, K);
 
# This code is contributed by sapnasingh4991

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the value of X
static void findMaxValue(int N, int K)
{
    int packages;
    int maxi = 1;
 
    // Loop to check all the numbers
    // divisible by N that yield
    // minimum N/i value
    for(int i = 1; i <= K; i++)
    {
        if (N % i == 0)
            maxi = Math.Max(maxi, i);
    }
    packages = N / maxi;
 
    // Print the value of packages
    Console.WriteLine(packages);
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given N and K
    int N = 8, K = 7;
 
    // Function call
    findMaxValue(N, K);
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
// Javascript program for the above approach
 
// Function to find the value of X
function findMaxValue(N, K)
{
    let packages;
    let maxi = 1;
 
    // Loop to check all the numbers
    // divisible by N that yield
    // minimum N/i value
    for (let i = 1; i <= K; i++) {
        if (N % i == 0)
            maxi = Math.max(maxi, i);
    }
 
    packages = parseInt(N / maxi);
 
    // Print the value of packages
    document.write(packages);
}
 
// Driver Code
// Given N and K
let N = 8, K = 7;
 
// Function Call
findMaxValue(N, K);
 
// This code is contributed by rishavmahato348.
</script>
Output: 
2

 

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

Efficient Approach: To optimize the above approach we will find the factor using the efficient approach discussed in this article. Below are the steps:

  1. Initialize the ans variable to store the largest factor of N.
  2. Iterate over [1, sqrt(N)] and do the following: 
    • Check if N is divisible by i or not.
    • If not then check for the next number.
    • Else if i ≤ K and N/i ≤ K then update ans to the maximum (i, N/i).
  3. The value of X will be N/ans.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to find the largest
// factor of N which is less than
// or equal to K
int solve(int n, int k)
{
    // Initialise the variable to
    // store the largest factor of
    // N <= K
    int ans = 0;
 
    // Loop to find all factors of N
    for (int j = 1;
        j * j <= n; j++) {
 
        // Check if j is a
        // factor of N or not
        if (n % j == 0) {
 
            // Check if j <= K
            // If yes, then store
            // the larger value between
            // ans and j in ans
            if (j <= k) {
                ans = max(ans, j);
            }
 
            // Check if N/j <= K
            // If yes, then store
            // the larger value between
            // ans and j in ans
            if (n / j <= k) {
                ans = max(ans, n / j);
            }
        }
    }
 
    // Since max value is always
    // stored in ans, the maximum
    // value divisible by N less than
    // or equal to K will be returned.
    return ans;
}
 
// Driver Code
int main()
{
    // Given N and K
    int N = 8, K = 7;
 
    // Function Call
    cout << (N / solve(N, K));
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the largest
// factor of N which is less than
// or equal to K
static int solve(int n, int k)
{
     
    // Initialise the variable to
    // store the largest factor of
    // N <= K
    int ans = 0;
 
    // Loop to find all factors of N
    for(int j = 1; j * j <= n; j++)
    {
         
        // Check if j is a
        // factor of N or not
        if (n % j == 0)
        {
             
            // Check if j <= K
            // If yes, then store
            // the larger value between
            // ans and j in ans
            if (j <= k)
            {
                ans = Math.max(ans, j);
            }
 
            // Check if N/j <= K
            // If yes, then store
            // the larger value between
            // ans and j in ans
            if (n / j <= k)
            {
                ans = Math.max(ans, n / j);
            }
        }
    }
 
    // Since max value is always
    // stored in ans, the maximum
    // value divisible by N less than
    // or equal to K will be returned.
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given N and K
    int N = 8, K = 7;
 
    // Function call
    System.out.print((N / solve(N, K)));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program for the above approach
 
# Function to find the largest
# factor of N which is less than
# or equal to K
def solve(n, k):
 
    # Initialise the variable to
    # store the largest factor of
    # N <= K
    ans = 0;
 
    # Loop to find all factors of N
    for j in range(1, n + 1):
        if (j * j > n):
            break;
 
        # Check if j is a
        # factor of N or not
        if (n % j == 0):
 
            # Check if j <= K
            # If yes, then store
            # the larger value between
            # ans and j in ans
            if (j <= k):
                ans = max(ans, j);
             
            # Check if N/j <= K
            # If yes, then store
            # the larger value between
            # ans and j in ans
            if (n // j <= k):
                ans = max(ans, n // j);
             
    # Since max value is always
    # stored in ans, the maximum
    # value divisible by N less than
    # or equal to K will be returned.
    return ans;
 
# Driver Code
if __name__ == '__main__':
 
    # Given N and K
    N = 8; K = 7;
 
    # Function call
    print((N // solve(N, K)));
 
# This code is contributed by gauravrajput1

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the largest
// factor of N which is less than
// or equal to K
static int solve(int n, int k)
{
     
    // Initialise the variable to
    // store the largest factor of
    // N <= K
    int ans = 0;
 
    // Loop to find all factors of N
    for(int j = 1; j * j <= n; j++)
    {
         
        // Check if j is a
        // factor of N or not
        if (n % j == 0)
        {
             
            // Check if j <= K
            // If yes, then store
            // the larger value between
            // ans and j in ans
            if (j <= k)
            {
                ans = Math.Max(ans, j);
            }
 
            // Check if N/j <= K
            // If yes, then store
            // the larger value between
            // ans and j in ans
            if (n / j <= k)
            {
                ans = Math.Max(ans, n / j);
            }
        }
    }
 
    // Since max value is always
    // stored in ans, the maximum
    // value divisible by N less than
    // or equal to K will be returned.
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given N and K
    int N = 8, K = 7;
 
    // Function call
    Console.Write((N / solve(N, K)));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript program implementation
// of the approach
 
// Function to find the largest
// factor of N which is less than
// or equal to K
function solve(n, k)
{
      
    // Initialise the variable to
    // store the largest factor of
    // N <= K
    let ans = 0;
  
    // Loop to find all factors of N
    for(let j = 1; j * j <= n; j++)
    {
          
        // Check if j is a
        // factor of N or not
        if (n % j == 0)
        {
              
            // Check if j <= K
            // If yes, then store
            // the larger value between
            // ans and j in ans
            if (j <= k)
            {
                ans = Math.max(ans, j);
            }
  
            // Check if N/j <= K
            // If yes, then store
            // the larger value between
            // ans and j in ans
            if (n / j <= k)
            {
                ans = Math.max(ans, n / j);
            }
        }
    }
  
    // Since max value is always
    // stored in ans, the maximum
    // value divisible by N less than
    // or equal to K will be returned.
    return ans;
}
 
// Driver Code
     
       // Given N and K
    let N = 8, K = 7;
  
    // Function call
    document.write((N / solve(N, K)));
          
</script>
Output: 
2

 

Time Complexity: O(sqrt(N)) 
Auxillary Space: O(1)
 

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