Kth largest factor of number N

Given two positive integers N and K, the task is to print the Kth largest factor of N.  

Input: N = 12, K = 3
Output: 4
Explanation: The factors of 12 are {1, 2, 3, 4, 6, 12}. The largest factor is 12 and the 3^rd largest factor is 4.

Input: N = 30, K = 2
Output: 15
Explanation: The factors of 30 are {1, 2, 3, 5, 6, 10, 15, 30} and the 2^nd largest factor is 15.

Approach: The idea is to check for each number in the range [N, 1], and print the Kth number that divides N completely.

Iterate through the loop from N to 0. Now, for each number in this loop:

• Check if it divides N or not.
• If N is divisible by the current number, decrement the value of K by 1.
• When K becomes 0, this means that the current number is the Kth largest factor of N.
• Print the answer according to the above observation.

Below is the implementation of the above approach:

C

// C program for the above approach

#include <stdio.h>

// Function to print Kth largest
// factor of N
int KthLargestFactor(int N, int K)
{

    // Check for numbers
    // in the range [N, 1]
    for (int i = N; i > 0; i--) {

        // Check if i is a factor of N
        if (N % i == 0)

            // If Yes, reduce K by 1
            K--;

        // If K is 0, it means
        // i is the required
        // Kth factor of N
        if (K == 0) {
            return i;
        }
    }

    // When K is more
    // than the factors of N
    return -1;
}

// Driver Code
int main()
{
    int N = 12, K = 3;
    printf("%d", KthLargestFactor(N, K));
    return 0;
}

C++

// C++ program for the above approach

#include <iostream>
using namespace std;

// Function to print Kth largest
// factor of N
int KthLargestFactor(int N, int K)
{
    // Check for numbers
    // in the range [N, 1]
    for (int i = N; i > 0; i--) {

        // Check if i is a factor of N
        if (N % i == 0)

            // If Yes, reduce K by 1
            K--;

        // If K is 0, it means
        // i is the required
        // Kth factor of N
        if (K == 0) {
            return i;
        }
    }

    // When K is more
    // than the factors of N
    return -1;
}

// Driver Code
int main()
{
    int N = 12, K = 3;
    cout << KthLargestFactor(N, K);
}

Java

// Java program for the above approach

import java.io.*;

class GFG {

    // Function to print Kth largest
    // factor of N
    static int KthLargestFactor(int N, int K)
    {
        // Check for numbers
        // in the range [N, 1]
        for (int i = N; i > 0; i--) {

            // Check if i is a factor of N
            if (N % i == 0)

                // If Yes, reduce K by 1
                K--;

            // If K is 0, it means
            // i is the required
            // Kth factor of N
            if (K == 0) {
                return i;
            }
        }

        // When K is more
        // than the factors of N
        return -1;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 12, K = 3;
        System.out.println(KthLargestFactor(N, K));
    }
}

Python

# Python program for the above approach

# Function to print Kth largest
# factor of N
def KthLargestFactor(N, K):
    for i in range(N, 0, -1):
        if N % i == 0:
            K -= 1
        if K == 0:
            return i
    return -1


# Driver Code
N = 12
K = 3
print(KthLargestFactor(N, K))

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to print Kth largest
// factor of N
static int KthLargestFactor(int N, int K)
{

    // Check for numbers
    // in the range [N, 1]
    for (int i = N; i > 0; i--) {

        // Check if i is a factor of N
        if (N % i == 0)

            // If Yes, reduce K by 1
            K--;

        // If K is 0, it means
        // i is the required
        // Kth factor of N
        if (K == 0) {
            return i;
        }
    }

    // When K is more
    // than the factors of N
    return -1;
}

// Driver Code
public static void Main()
{
    int N = 12, K = 3;
    Console.Write(KthLargestFactor(N, K));
}
}

// This code is contributed by ipg2016107.

Javascript

<script>
// JavaScript program for the above approach
// Function to print Kth largest
// factor of N
function KthLargestFactor(N, K)
    {
        // Check for numbers
        // in the range [N, 1]
        for (let i = N; i > 0; i--) {

            // Check if i is a factor of N
            if (N % i == 0)

                // If Yes, reduce K by 1
                K--;

            // If K is 0, it means
            // i is the required
            // Kth factor of N
            if (K == 0) {
                return i;
            }
        }

        // When K is more
        // than the factors of N
        return -1;
    }

// Driver Code
let N = 12, K = 3;
document.write(KthLargestFactor(N, K));

// This code is contributed by shivanisinghss2110
</script>

Output:

4

Time Complexity: 
Auxiliary Space: