Largest integer upto N having greatest prime factor greater than its square root

Given a positive integer N, the task is to find the largest number in the range [1, N] such that the square root of the number is less than its greatest prime factor.

Input: N = 15
Output: 15
Explanation: The prime factors of 15 are {3, 5}. The square root of 15 is 3.87 (i.e, 3.87 < 5). Therefore 15 is the largest valid integer in the given range.

Input: N = 25
Output: 23

Approach: The given problem can be solved by using the Sieve of Eratosthenes with a few modifications. Create an array gpf[], which stores the Greatest Prime Factor of all integers in the given range. Initially, gpf[] = {0}. Using Sieve, initialize all the indices of the array gpf[] with the greatest prime factor of the respective index similar to the algorithm discussed in this article.

Now, iterate over the range [N, 1] in a reverse manner and print the first integer whose square root of the number is less than its greatest prime factor.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100001;

// Stores the Greatest Prime Factor
int gpf[maxn];

// Modified Sieve to find the Greatest
// Prime Factor of all integers in the
// range [1, maxn]
void modifiedSieve()
{
    // Initialize the array with 0
    memset(gpf, 0, sizeof(gpf));
    gpf[0] = 0;
    gpf[1] = 1;

    // Iterate through all values of i
    for (int i = 2; i < maxn; i++) {

        // If i is not a prime number
        if (gpf[i] > 0)
            continue;

        // Update the multiples of i
        for (int j = i; j < maxn; j += i) {
            gpf[j] = max(i, gpf[j]);
        }
    }
}

// Function to find integer in the range
// [1, N] such that its Greatest Prime
// factor is greater than its square root
int greatestValidInt(int N)
{

    modifiedSieve();

    // Iterate through all values of
    // i in the range [N, 1]
    for (int i = N; i > 0; i--) {

        // If greatest prime factor of i
        // is greater than its square root
        if (gpf[i] > sqrt(i)) {

            // Return answer
            return i;
        }
    }

    // If no valid integer exist
    return -1;
}

// Driver Code
int main()
{
    int N = 25;
    cout << greatestValidInt(N);

    return 0;
}

Java

// Java program for the above approach
public class GFG {

    final static int maxn = 100001;

    // Stores the Greatest Prime Factor
    static int gpf[] = new int[maxn];

    // Modified Sieve to find the Greatest
    // Prime Factor of all integers in the
    // range [1, maxn]
    static void modifiedSieve()
    {

        // Initialize the array with 0
        for (int i = 0; i < maxn; i++ )
            gpf[i] = 0;

        gpf[0] = 0;
        gpf[1] = 1;

        // Iterate through all values of i
        for (int i = 2; i < maxn; i++) {

            // If i is not a prime number
            if (gpf[i] > 0)
                continue;

            // Update the multiples of i
            for (int j = i; j < maxn; j += i) {
                gpf[j] = Math.max(i, gpf[j]);
            }
        }
    }

    // Function to find integer in the range
    // [1, N] such that its Greatest Prime
    // factor is greater than its square root
    static int greatestValidInt(int N)
    {

        modifiedSieve();

        // Iterate through all values of
        // i in the range [N, 1]
        for (int i = N; i > 0; i--) {

            // If greatest prime factor of i
            // is greater than its square root
            if (gpf[i] > Math.sqrt(i)) {

                // Return answer
                return i;
            }
        }

        // If no valid integer exist
        return -1;
    }

    // Driver Code
    public static void main (String[] args)
    {
        int N = 25;
        System.out.println(greatestValidInt(N));
    }
}

// This code is contributed by AnkThon

Python3

# python program for the above approach

import math

maxn = 100001

# Stores the Greatest Prime Factor
gpf = [0 for _ in range(maxn)]

# Modified Sieve to find the Greatest
# Prime Factor of all integers in the
# range [1, maxn]


def modifiedSieve():

    # Initialize the array with 0
    gpf[0] = 0
    gpf[1] = 1

    # Iterate through all values of i
    for i in range(2, maxn):

        # If i is not a prime number
        if (gpf[i] > 0):
            continue

        # Update the multiples of i
        for j in range(i, maxn, i):
            gpf[j] = max(i, gpf[j])


# Function to find integer in the range
# [1, N] such that its Greatest Prime
# factor is greater than its square root
def greatestValidInt(N):

    modifiedSieve()

    # Iterate through all values of
    # i in the range [N, 1]
    for i in range(N, 0, -1):

        # If greatest prime factor of i
        # is greater than its square root
        if (gpf[i] > math.sqrt(i)):

            # Return answer
            return i

    # If no valid integer exist
    return -1


# Driver Code
if __name__ == "__main__":

    N = 25
    print(greatestValidInt(N))

# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
public class GFG {

    static int maxn = 100001;

    // Stores the Greatest Prime Factor
    static int[] gpf = new int[maxn];

    // Modified Sieve to find the Greatest
    // Prime Factor of all integers in the
    // range [1, maxn]
    static void modifiedSieve()
    {

        // Initialize the array with 0
        for (int i = 0; i < maxn; i++)
            gpf[i] = 0;

        gpf[0] = 0;
        gpf[1] = 1;

        // Iterate through all values of i
        for (int i = 2; i < maxn; i++) {

            // If i is not a prime number
            if (gpf[i] > 0)
                continue;

            // Update the multiples of i
            for (int j = i; j < maxn; j += i) {
                gpf[j] = Math.Max(i, gpf[j]);
            }
        }
    }

    // Function to find integer in the range
    // [1, N] such that its Greatest Prime
    // factor is greater than its square root
    static int greatestValidInt(int N)
    {

        modifiedSieve();

        // Iterate through all values of
        // i in the range [N, 1]
        for (int i = N; i > 0; i--) {

            // If greatest prime factor of i
            // is greater than its square root
            if (gpf[i] > Math.Sqrt(i)) {

                // Return answer
                return i;
            }
        }

        // If no valid integer exist
        return -1;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int N = 25;
        Console.WriteLine(greatestValidInt(N));
    }
}

// This code is contributed by ukasp.

Javascript

<script>
// Javascript program for the above approach

const maxn = 100001;

// Stores the Greatest Prime Factor
let gpf = new Array(maxn);

// Modified Sieve to find the Greatest
// Prime Factor of all integers in the
// range [1, maxn]
function modifiedSieve()
{

  // Initialize the array with 0
  gpf.fill(0);
  gpf[0] = 0;
  gpf[1] = 1;

  // Iterate through all values of i
  for (let i = 2; i < maxn; i++)
  {

    // If i is not a prime number
    if (gpf[i] > 0) continue;

    // Update the multiples of i
    for (let j = i; j < maxn; j += i) {
      gpf[j] = Math.max(i, gpf[j]);
    }
  }
}

// Function to find integer in the range
// [1, N] such that its Greatest Prime
// factor is greater than its square root
function greatestValidInt(N) {
  modifiedSieve();

  // Iterate through all values of
  // i in the range [N, 1]
  for (let i = N; i > 0; i--)
  {

    // If greatest prime factor of i
    // is greater than its square root
    if (gpf[i] > Math.sqrt(i))
    {

      // Return answer
      return i;
    }
  }

  // If no valid integer exist
  return -1;
}

// Driver Code
let N = 25;
document.write(greatestValidInt(N));

// This code is contributed by gfgking.
</script>

Output:

23

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