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Count elements in the given range which have maximum number of divisors

  • Difficulty Level : Medium
  • Last Updated : 19 May, 2021

Given two numbers X and Y. The task is to find the number of elements in the range [X,Y] both inclusive, that have the maximum number of divisors.

Examples

Input: X = 2, Y = 9 
Output: 2 
6, 8 are numbers with the maximum number of divisors.

Input: X = 1, Y = 10 
Output: 3 
6, 8, 10 are numbers with the maximum number of divisors. 

Method 1:  

  • Traverse all the elements from X to Y one by one.
  • Find the number of divisors of each element.
  • Store the number of divisors in an array and update the maximum number of Divisors(maxDivisors).
  • Traverse the array that contains divisors and counts the number of elements equal to maxDivisors.
  • Return the count.

Below is the implementation of above method:  

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the divisors
int countDivisors(int n)
{
    int count = 0;
 
    // Note that this loop runs till square root
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
 
            // If divisors are equal, print only one
            if (n / i == i)
                count++;
 
            else // Otherwise print both
                count += 2;
        }
    }
 
    return count;
}
 
// Function to count the number with
// maximum divisors
int MaximumDivisors(int X, int Y)
{
    int maxDivisors = INT_MIN, result = 0;
 
    // to store number of divisors
    int arr[Y - X + 1];
 
    // Traverse from X to Y
    for (int i = X; i <= Y; i++) {
 
            // Count the number of divisors of i
             int Div = countDivisors(i);
 
            // Store the value of div in an array
             arr[i - X] = Div;
 
            // Update the value of maxDivisors
             maxDivisors = max(Div, maxDivisors);
 
    }
 
    // Traverse the array
    for (int i = 0; i < (Y - X + 1); i++)
 
        // Count the value equals to maxDivisors
        if (arr[i] == maxDivisors)
            result++;
 
    return result;
}
 
// Driver Code
int main()
{
    int X = 1, Y = 10;
 
    // function call
    cout << MaximumDivisors(X, Y) << endl;
 
    return 0;
}

Java




// Java implementation of above approach
class GFG
{
 
// Function to count the divisors
static int countDivisors(int n)
{
int count = 0;
 
// Note that this loop
// runs till square root
for (int i = 1; i <= Math.sqrt(n); i++)
{
    if (n % i == 0)
    {
 
        // If divisors are equal,
        // print only one
        if (n / i == i)
            count++;
 
        else // Otherwise print both
            count += 2;
    }
}
 
return count;
}
 
// Function to count the number
// with maximum divisors
static int MaximumDivisors(int X, int Y)
{
int maxDivisors = 0, result = 0;
 
// to store number of divisors
int[] arr = new int[Y - X + 1];
 
// Traverse from X to Y
for (int i = X; i <= Y; i++)
{
 
    // Count the number of divisors of i
    int Div = countDivisors(i);
 
    // Store the value of div in an array
    arr[i - X] = Div;
 
    // Update the value of maxDivisors
    maxDivisors = Math.max(Div, maxDivisors);
 
}
 
// Traverse the array
for (int i = 0; i < (Y - X + 1); i++)
 
    // Count the value equals
    // to maxDivisors
    if (arr[i] == maxDivisors)
        result++;
 
return result;
}
 
// Driver Code
public static void main(String[] args)
{
    int X = 1, Y = 10;
 
    // function call
    System.out.println(MaximumDivisors(X, Y));
}
}
 
// This code is contributed
// by ChitraNayal

Python3




# from math module import everything
from math import *
 
# Python 3 implementation of above approach
 
# Function to count the divisors
def countDivisors(n) :
    count = 0
     
    # Note that this loop runs till square root
    for i in range(1,int(sqrt(n)+1)) :
        if n % i == 0 :
 
            # If divisors are equal, print only one
            if n / i == i :
                count += 1
                 
            # Otherwise print both
            else :
                count += 2
 
    return count
 
# Function to count the number with
# maximum divisors
def MaximumDivisors(X,Y) :
    result = 0
    maxDivisors = 0
 
    # create list to store number of divisors
    arr = []
     
    # initialize with 0 upto length Y-X+1
    for i in range(Y - X + 1) :
        arr.append(0)
 
    # Traverse from X to Y  
    for i in range(X,Y+1) :
 
        # Count the number of divisors of i
        Div = countDivisors(i)
 
        # Store the value of div in an array
        arr[i - X] = Div
         
        # Update the value of maxDivisors
        maxDivisors = max(Div,maxDivisors)
         
    # Traverse the array 
    for i in range (Y - X + 1) :
 
        # Count the value equals to maxDivisors
        if arr[i] == maxDivisors :
            result += 1
 
    return result
 
# Driver code
if __name__ == "__main__" :
 
    X, Y = 1, 10
 
    # function call
    print(MaximumDivisors(X,Y))

C#




// C# implementation of above approach
using System;
 
class GFG
{
 
// Function to count the divisors
static int countDivisors(int n)
{
int count = 0;
 
// Note that this loop
// runs till square root
for (int i = 1; i <= Math.Sqrt(n); i++)
{
    if (n % i == 0)
    {
 
        // If divisors are equal,
        // print only one
        if (n / i == i)
            count++;
 
        else // Otherwise print both
            count += 2;
    }
}
 
return count;
}
 
// Function to count the number
// with maximum divisors
static int MaximumDivisors(int X, int Y)
{
int maxDivisors = 0, result = 0;
 
// to store number of divisors
int[] arr = new int[Y - X + 1];
 
// Traverse from X to Y
for (int i = X; i <= Y; i++)
{
 
    // Count the number of divisors of i
    int Div = countDivisors(i);
 
    // Store the value of div in an array
    arr[i - X] = Div;
 
    // Update the value of maxDivisors
    maxDivisors = Math.Max(Div, maxDivisors);
 
}
 
// Traverse the array
for (int i = 0; i < (Y - X + 1); i++)
 
    // Count the value equals
    // to maxDivisors
    if (arr[i] == maxDivisors)
        result++;
 
return result;
}
 
// Driver Code
public static void Main()
{
    int X = 1, Y = 10;
 
    // function call
    Console.Write(MaximumDivisors(X, Y));
}
}
 
// This code is contributed
// by ChitraNayal

PHP




<?php
// PHP implementation of above approach
 
// Function to count the divisors
function countDivisors($n)
{
    $count = 0;
 
    // Note that this loop
    // runs till square root
    for ($i = 1; $i <= sqrt($n); $i++)
    {
        if ($n % $i == 0)
        {
 
            // If divisors are equal,
            // print only one
            if ($n / $i == $i)
                $count++;
 
            else // Otherwise print both
                $count += 2;
        }
    }
 
    return $count;
}
 
// Function to count the number
// with maximum divisors
function MaximumDivisors($X, $Y)
{
    $maxDivisors = PHP_INT_MIN;
    $result = 0;
 
    // to store number of divisors
    $arr = array_fill(0, ($Y - $X + 1), NULL);
 
    // Traverse from X to Y
    for ($i = $X; $i <= $Y; $i++)
    {
 
        // Count the number of divisors of i
        $Div = countDivisors($i);
 
        // Store the value of div in an array
        $arr[$i - $X] = $Div;
 
        // Update the value of maxDivisors
        $maxDivisors = max($Div, $maxDivisors);
    }
 
    // Traverse the array
    for ($i = 0; $i < ($Y - $X + 1); $i++)
 
        // Count the value equals to maxDivisors
        if ($arr[$i] == $maxDivisors)
            $result++;
 
    return $result;
}
 
// Driver Code
$X = 1;
$Y = 10;
 
// function call
echo MaximumDivisors($X, $Y)." ";
 
// This code is contributed
// by ChitraNayal
?>

Javascript




<script>
 
// Javascript implementation of above approach
 
// Function to count the divisors
function countDivisors(n)
{
    let count = 0;
     
    // Note that this loop
    // runs till square root
    for(let i = 1; i <= Math.sqrt(n); i++)
    {
        if (n % i == 0)
        {
             
            // If divisors are equal,
            // print only one
            if (Math.floor(n / i) == i)
                count++;
       
            else
             
                // Otherwise print both
                count += 2;
        }
    }
    return count;
}
 
// Function to count the number
// with maximum divisors
function MaximumDivisors(X, Y)
{
    let maxDivisors = 0, result = 0;
   
    // To store number of divisors
    let arr = new Array(Y - X + 1);
       
    // Traverse from X to Y
    for(let i = X; i <= Y; i++)
    {
         
        // Count the number of divisors of i
        let Div = countDivisors(i);
       
        // Store the value of div in an array
        arr[i - X] = Div;
       
        // Update the value of maxDivisors
        maxDivisors = Math.max(Div, maxDivisors);
    }
       
    // Traverse the array
    for(let i = 0; i < (Y - X + 1); i++)
     
        // Count the value equals
        // to maxDivisors
        if (arr[i] == maxDivisors)
            result++;
       
    return result;
}
 
// Driver Code
let X = 1, Y = 10;
 
// Function call
document.write(MaximumDivisors(X, Y));
 
// This code is contributed by rag2127
 
</script>
Output: 
3

 

Method 2: 

  • Create an array of size Y-X+1 to store number of divisors of X in arr[0], X+1 in arr[1]… up to Y.
  • To get the number of divisors of all numbers from X to Y run two loops(nested).
  • Run an outer loop from 1 to sqrt(Y).
  • Run an inner loop from first_divisible to Y.
  • First_divisible is the number which is the first number to be divisible by I(outer loop) and greater than or equals to X.

Here, first_divisible is calculated by using Find the number closest to n and divisible by m method. Then find divisors of the number.

Below is the implementation of above approach:  

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the elements
// with maximum number of divisors
int MaximumDivisors(int X, int Y)
{
    // to store number of divisors
    int arr[Y - X + 1];
 
    // initialise with zero
    memset(arr, 0, sizeof(arr));
 
    // to store the maximum number of divisors
    int mx = INT_MIN;
 
    // to store required answer
    int cnt = 0;
 
    for (int i = 1; i * i <= Y; i++) {
        int sq = i * i;
        int first_divisible;
 
        // Find the first divisible number
        if ((X / i) * i >= X)
            first_divisible = (X / i) * i;
        else
            first_divisible = (X / i + 1) * i;
 
        // Count number of divisors
        for (int j = first_divisible; j <= Y; j += i) {
            if (j < sq)
                continue;
            else if (j == sq)
                arr[j - X]++;
            else
                arr[j - X] += 2;
        }
    }
 
    // Find number of elements with
    // maximum number of divisors
    for (int i = X; i <= Y; i++) {
        if (arr[i - X] > mx) {
            cnt = 1;
            mx = arr[i - X];
        }
        else if (arr[i - X] == mx)
            cnt++;
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    int X = 1, Y = 10;
    cout << MaximumDivisors(X, Y) << endl;
 
    return 0;
}

Java




// Java implementation of above approach
class GFG
{
 
// Function to count the elements
// with maximum number of divisors
static int MaximumDivisors(int X, int Y)
{
     
// to store number of divisors
int[] arr = new int[Y - X + 1];
 
// initialise with zero
for(int i = 0; i < arr.length; i++)
    arr[i] = 0;
 
// to store the maximum
// number of divisors
int mx = 0;
 
// to store required answer
int cnt = 0;
 
for (int i = 1; i * i <= Y; i++)
{
    int sq = i * i;
    int first_divisible;
 
    // Find the first divisible number
    if ((X / i) * i >= X)
        first_divisible = (X / i) * i;
    else
        first_divisible = (X / i + 1) * i;
 
    // Count number of divisors
    for (int j = first_divisible;
             j <= Y; j += i)
    {
        if (j < sq)
            continue;
        else if (j == sq)
            arr[j - X]++;
        else
            arr[j - X] += 2;
    }
}
 
// Find number of elements with
// maximum number of divisors
for (int i = X; i <= Y; i++)
{
    if (arr[i - X] > mx)
    {
        cnt = 1;
        mx = arr[i - X];
    }
    else if (arr[i - X] == mx)
        cnt++;
}
 
return cnt;
}
 
// Driver code
public static void main(String[] args)
{
    int X = 1, Y = 10;
    System.out.println(MaximumDivisors(X, Y));
}
}
 
// This code is contributed
// by ChitraNayal

Python3




# Python 3 implementation of above approach
 
# Function to count the elements
# with maximum number of divisors
def MaximumDivisors(X, Y):
 
    # to store number of divisors
    # initialise with zero
    arr = [0] * (Y - X + 1)
 
    # to store the maximum
    # number of divisors
    mx = 0
 
    # to store required answer
    cnt = 0
 
    i = 1
    while i * i <= Y :
        sq = i * i
 
        # Find the first divisible number
        if ((X // i) * i >= X) :
            first_divisible = (X // i) * i
        else:
            first_divisible = (X // i + 1) * i
 
        # Count number of divisors
        for j in range(first_divisible, Y + 1, i):
            if j < sq :
                continue
            elif j == sq :
                arr[j - X] += 1
            else:
                arr[j - X] += 2
        i += 1
 
    # Find number of elements with
    # maximum number of divisors
    for i in range(X, Y + 1):
        if arr[i - X] > mx :
            cnt = 1
            mx = arr[i - X]
 
        elif arr[i - X] == mx :
            cnt += 1
 
    return cnt
 
# Driver code
if __name__ == "__main__":
    X = 1
    Y = 10
    print(MaximumDivisors(X, Y))
 
# This code is contributed
# by ChitraNayal

C#




// C# implementation of above approach
using System;
 
class GFG
{
 
// Function to count the elements
// with maximum number of divisors
static int MaximumDivisors(int X, int Y)
{
     
// to store number of divisors
int[] arr = new int[Y - X + 1];
 
// initialise with zero
for(int i = 0; i < arr.Length; i++)
    arr[i] = 0;
 
// to store the maximum
// number of divisors
int mx = 0;
 
// to store required answer
int cnt = 0;
 
for (int i = 1; i * i <= Y; i++)
{
    int sq = i * i;
    int first_divisible;
 
    // Find the first divisible number
    if ((X / i) * i >= X)
        first_divisible = (X / i) * i;
    else
        first_divisible = (X / i + 1) * i;
 
    // Count number of divisors
    for (int j = first_divisible;
             j <= Y; j += i)
    {
        if (j < sq)
            continue;
        else if (j == sq)
            arr[j - X]++;
        else
            arr[j - X] += 2;
    }
}
 
// Find number of elements with
// maximum number of divisors
for (int i = X; i <= Y; i++)
{
    if (arr[i - X] > mx)
    {
        cnt = 1;
        mx = arr[i - X];
    }
    else if (arr[i - X] == mx)
        cnt++;
}
 
return cnt;
}
 
// Driver code
public static void Main()
{
    int X = 1, Y = 10;
    Console.Write(MaximumDivisors(X, Y));
}
}
 
// This code is contributed
// by ChitraNayal

PHP




<?php
// PHP implementation of above approach
 
// Function to count the elements
// with maximum number of divisors
function MaximumDivisors($X, $Y)
{
    // to store number of divisors
    // initialise with zero
    $arr = array_fill(0,($Y - $X + 1), NULL);
  
    // to store the maximum
    // number of divisors
    $mx = PHP_INT_MIN;
 
    // to store required answer
    $cnt = 0;
 
    for ($i = 1; $i * $i <= $Y; $i++)
    {
        $sq = $i * $i;
 
        // Find the first divisible number
        if (($X / $i) * $i >= $X)
            $first_divisible = ($X / $i) * $i;
        else
            $first_divisible = ($X / $i + 1) * $i;
 
        // Count number of divisors
        for ($j = $first_divisible;
             $j < $Y; $j += $i)
        {
            if ($j < $sq)
                continue;
            else if ($j == $sq)
                $arr[$j - $X]++;
            else
                $arr[$j - $X] += 2;
        }
    }
 
    // Find number of elements with
    // maximum number of divisors
    for ($i = $X; $i <= $Y; $i++)
    {
        if ($arr[$i - $X] > $mx
        {
            $cnt = 1;
            $mx = $arr[$i - $X];
        }
        else if ($arr[$i - $X] == $mx)
            $cnt++;
    }
 
    return $cnt;
}
 
// Driver code
$X = 1;
$Y = 10;
echo MaximumDivisors($X, $Y)."\n";
 
// This code is contributed
// by ChitraNayal
?>

Javascript




<script>
 
// Javascript implementation of above approach
 
// Function to count the elements
// with maximum number of divisors
function MaximumDivisors(X, Y)
{
     
    // To store number of divisors
    let arr = new Array(Y - X + 1);
      
    // Initialise with zero
    for(let i = 0; i < arr.length; i++)
        arr[i] = 0;
      
    // To store the maximum
    // number of divisors
    let mx = 0;
      
    // To store required answer
    let cnt = 0;
      
    for(let i = 1; i * i <= Y; i++)
    {
        let sq = i * i;
        let first_divisible;
      
        // Find the first divisible number
        if (Math.floor(X / i) * i >= X)
            first_divisible = Math.floor(X / i) * i;
        else
            first_divisible = (Math.floor(X / i) +
                                         1) * i;
      
        // Count number of divisors
        for(let j = first_divisible;
                j <= Y; j += i)
        {
            if (j < sq)
                continue;
            else if (j == sq)
                arr[j - X]++;
            else
                arr[j - X] += 2;
        }
    }
      
    // Find number of elements with
    // maximum number of divisors
    for(let i = X; i <= Y; i++)
    {
        if (arr[i - X] > mx)
        {
            cnt = 1;
            mx = arr[i - X];
        }
        else if (arr[i - X] == mx)
            cnt++;
    }
    return cnt;
}
 
// Driver code
let X = 1, Y = 10;
 
document.write(MaximumDivisors(X, Y));
 
// This code is contributed by avanitrachhadiya2155
 
</script>
Output: 
3

 


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