Count elements in the given range which have maximum number of divisors
Last Updated :
11 Jul, 2025
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;
}
// C implementation of above approach
#include <stdio.h>
#include <math.h>
#include <limits.h>
int max(int a,int b)
{
int max = a;
if(max < b)
max = b;
return max;
}
// 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
printf("%d\n",MaximumDivisors(X, Y));
return 0;
}
// This code is contributed by kothavvsaakash.
// 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
# 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# 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 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
?>
<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>
Time Complexity: O(sqrt(n))
Auxiliary Space: O(n)
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;
}
// C implementation of above approach
#include <stdio.h>
#include <limits.h>
#include <string.h>
// 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;
printf("%d\n",MaximumDivisors(X, Y));
return 0;
}
// This code is contributed by kothavvsaakash.
// 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
# 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# 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 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
?>
<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>
Time Complexity : O((Y - X + 1) * sqrt(Y))
Space Complexity : O(Y - X + 1)
Similar Reads
Count the numbers < N which have equal number of divisors as K Given two integers N and K, the task is to count all the numbers < N which have equal number of positive divisors as K. Examples: Input: n = 10, k = 5 Output: 3 2, 3 and 7 are the only numbers < 10 which have 2 divisors (equal to the number of divisors of 5) Input: n = 500, k = 6 Output: 148 A
9 min read
Count elements which divide all numbers in range L-R The problem statement describes a scenario where we are given an array of N numbers and Q queries. Each query consists of two integers L and R, representing a range of indices in the array. The task is to find the count of numbers in the array that divide all the numbers in the range L-R. Examples :
15+ min read
Querying maximum number of divisors that a number in a given range has Given Q queries, of type: L R, for each query you must print the maximum number of divisors that a number x (L <= x <= R) has. Examples: L = 1 R = 10: 1 has 1 divisor. 2 has 2 divisors. 3 has 2 divisors. 4 has 3 divisors. 5 has 2 divisors. 6 has 4 divisors. 7 has 2 divisors. 8 has 4 divisors.
15+ min read
Program to find count of numbers having odd number of divisors in given range Given two integers A and B. The task is to count how many numbers in the interval [ A, B ] have an odd number of divisors. Examples: Input : A = 1, B = 10Output : 3 Input : A = 5, B = 15Output : 1 Naive Approach :The simple approach would be to iterate through all the numbers between range [A, B] an
12 min read
Count the numbers divisible by 'M' in a given range A and B are two numbers which define a range, where A <= B. Find the total numbers in the given range [A ... B] divisible by 'M'Examples: Input : A = 25, B = 100, M = 30 Output : 3 Explanation : In the given range [25 - 100], 30, 60 and 90 are divisible by 30 Input : A = 6, B = 15, M = 3 Output :
9 min read
Find numbers with K odd divisors in a given range Given two numbers a and b, and a number k which is odd. The task is to find all the numbers between a and b (both inclusive) having exactly k divisors.Examples: Input : a = 2, b = 49, k = 3 Output: 4 // Between 2 and 49 there are four numbers // with three divisors // 4 (Divisors 1, 2, 4), 9 (Diviso
8 min read