Max occurring divisor in an interval

Given an interval [x, y]. Find the divisor that occurs maximum number of times except ‘1’ in the divisors of numbers in the range [x, y], both inclusive.

Examples :

Input : [2, 5]
Output : 2
Explanation : Divisors of 2, 3, 4, 5 except 1 are:
              2 -> 2
              3 -> 3
              4 -> 2, 4
              5 -> 5
              Among all the divisors 2 occurs 
              maximum time, i.e two times.

Input : [3, 3]
Output : 3

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Brute Force Approach: The most simple approach is to traverse all of the numbers from x to y and calculate all of their divisors and store the divisors in a map with their counts. Finally traverse the map to find the divisor occurring maximum times. You may refer to this article on how to efficiently print all divisors of a natural number.

Below is the implementation of above approach:

C++

// Simple CPP program to find maximum occurring 
// factor in an interval 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to find max occurring 
// divisor in interval [x, y] 
int findDivisor(int x, int y) 
{ 
    // map to store count of divisors 
    unordered_map<int, int> m; 
  
    // iterate for every number in the 
    // interval 
    for (int num = x; num <= y; num++) { 
        // find all divisors of num 
        for (int i = 2; i <= sqrt(num) + 1; i++) { 
            if (num % i == 0) { 
  
                // If divisors are equal, print only one 
                if (num / i == i) { 
                    if (m.find(i) == m.end()) 
                        m.insert(make_pair(i, 1)); 
                    else
                        m[i]++; 
                } 
                else { 
  
                    // insert first one to map 
                    if (m.find(i) == m.end()) 
                        m.insert(make_pair(i, 1)); 
                    else
                        m[i]++; 
  
                    // insert second to map 
                    if (m.find(num / i) == m.end()) 
                        m.insert(make_pair(num / i, 1)); 
                    else
                        m[num / i]++; 
                } 
            } 
        } 
    } 
  
    int divisor = 0; 
    int divisorCount = INT_MIN; 
  
    // iterate on map 
    for (auto itr = m.begin(); itr != m.end(); itr++) { 
        if (itr->second > divisorCount) { 
            divisorCount = itr->second; 
            divisor = itr->first; 
        } 
    } 
  
    return divisor; 
} 
  
// Driver code 
int main() 
{ 
    int x = 3, y = 16; 
    cout << findDivisor(x, y); 
    return 0; 
} 

Java

// Java program to find Max 
// occurring divisor in an interval 
import java.io.*; 
import java.util.*; 
  
class GFG { 
  
// function to find max occurring 
// divisor in interval [x, y] 
static int findDivisor(int x, int y) 
{ 
    // map to store count of divisors 
    HashMap<Integer, Integer> m = new HashMap<Integer,  
                                            Integer>(); 
  
    // iterate for every number  
    // in the interval 
    for (int num = x; num <= y; num++) { 
          
        // find all divisors of num 
        for (int i = 2; i <= Math.sqrt(num) + 1; i++) { 
            if (num % i == 0) { 
  
                // If divisors are equal, 
                // print only one 
                if (num / i == i) { 
                    if (m.containsKey(i) != true) 
                        m.put(i, 1); 
                    else
                        { 
                            int val = m.get(i); 
                            m.put(i, ++val); 
                        } 
                } 
                else { 
  
                    // insert first one to map 
                    if (m.containsKey(i) != true) 
                        m.put(i, 1); 
                    else
                        { 
                            int val=m.get(i); 
                            m.put(i,++val); 
                        } 
  
                    // insert second to map 
                    if (m.containsKey(num / i) != true) 
                        m.put(num / i, 1); 
                    else
                        { 
                            int k = num / i; 
                            int val = m.get(k); 
                            m.put( k, ++val); 
                        } 
                } 
            } 
        } 
    } 
  
    int divisor = 0; 
    int divisorCount = Integer.MIN_VALUE; 
  
    // iterate on map 
    for(Map.Entry<Integer, Integer> entry:m.entrySet()){  
        int key = entry.getKey();  
        int value = entry.getValue();  
          
        if (value > divisorCount) { 
            divisorCount = value; 
            divisor = key; 
        }  
    }  
  
    return divisor; 
} 
  
/* Driver program to test above function */
public static void main(String[] args) 
{ 
    int x = 3, y = 16; 
    System.out.println(findDivisor(x, y)); 
} 
} 
  
// This code is contributed by Gitanjali 

Python

# Simple python program to find maximum 
# occurring factor in an interval 
import math  
  
# Function to find max occurring 
# divisor in interval [x, y] 
def findDivisor( x, y): 
    # Map to store count of divisors 
    m = {} 
      
    # Iterate for every number in the interval 
    for num in range(x, y + 1): 
          
        # Find all divisors of num 
        for i in range(2, int(math.sqrt(num)) + 2): 
            if (num % i == 0): 
                  
                # If divisors are equal, print only one 
                if (num / i == i): 
                    if i not in m: 
                        m[i]= 1
                    else: 
                        m[i]+= 1
                  
                else: 
                    # Insert first one to map 
                    if (i not in m): 
                        m[i]= 1
                    else: 
                        m[i]= m[i]+1
  
                    # Insert second to map 
                    if (num / i not in m ): 
                        m[num / i]= 1
                    else: 
                        m[num / i]= m[num / i]+1
                  
    divisor = 0
    divisorCount = -999999
  
    # Iterate on map 
    for itr in m : 
        if m[itr] > divisorCount: 
            divisorCount = m[itr] 
            divisor = itr 
          
    return divisor 
  
# Driver method 
x = 3
y = 16
print (findDivisor(x, y)) 
  
# This code is contributed by 'Gitanjali'. 

C#

// C# program to find Max 
// occurring divisor in an interval 
using System; 
using System.Collections.Generic;  
  
class GFG  
{ 
  
// function to find max occurring 
// divisor in interval [x, y] 
static int findDivisor(int x, int y) 
{ 
    // map to store count of divisors 
    Dictionary<int,  
               int> m = new Dictionary<int, 
                                       int>(); 
  
    // iterate for every number  
    // in the interval 
    for (int num = x; num <= y; num++) 
    { 
          
        // find all divisors of num 
        for (int i = 2; i <= Math.Sqrt(num) + 1; i++)  
        { 
            if (num % i == 0) 
            { 
  
                // If divisors are equal, 
                // print only one 
                if (num / i == i) 
                { 
                    if (m.ContainsKey(i) != true) 
                        m.Add(i, 1); 
                    else
                    { 
                        int val = m[i]; 
                        m[i] = ++val; 
                    } 
                } 
                else
                { 
  
                    // insert first one to map 
                    if (m.ContainsKey(i) != true) 
                        m.Add(i, 1); 
                    else
                    { 
                        int val = m[i]; 
                        m[i] = ++val; 
                    } 
  
                    // insert second to map 
                    if (m.ContainsKey(num / i) != true) 
                        m.Add(num / i, 1); 
                    else
                    { 
                        int k = num / i; 
                        int val = m[k]; 
                        m[k] = ++val; 
                    } 
                } 
            } 
        } 
    } 
  
    int divisor = 0; 
    int divisorCount = int.MinValue; 
  
    // iterate on map 
    foreach(KeyValuePair<int, int> entry in m) 
    {  
        int key = entry.Key;  
        int value = entry.Value;  
          
        if (value > divisorCount)  
        { 
            divisorCount = value; 
            divisor = key; 
        }  
    }  
  
    return divisor; 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
    int x = 3, y = 16; 
    Console.WriteLine(findDivisor(x, y)); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Time Complexity: O(n*sqrt(n)), where n is total number of numbers between interval [x, y].

Efficient Approach: An efficient approach will be to observe carefully that every even number will have 2 as a divisor. And in the interval [x, y] there is atleast floor((y-x)/2) + 1 numbers of 2. That is, half of the total numbers in the interval will have divisor as 2. Clearly this is the maximum number of occurrences of a divisor in the interval. If (x == y), then answer will be x or y.

Below is implementation of above idea:

C++

// Efficient C++ program to 
// find maximum occurring 
// factor in an interval 
#include <iostream> 
using namespace std; 
  
// function to find max 
// occurring divisor 
// interval [x, y] 
int findDivisor(int x, int y) 
{ 
    // if there is only  
    // one number in the 
    // in the interval,  
    // return that number 
    if (x == y) 
        return y; 
  
    // otherwise, 2 is the 
    // max occurring  
    // divisor 
    return 2; 
} 
  
// Driver code 
int main() 
{ 
    int x = 3, y = 16; 
    cout << findDivisor(x, y); 
    return 0; 
} 

Java

// Efficient Java program to 
// find maximum occurring 
// factor in an interval 
import java.io.*; 
  
class GFG { 
          
    // function to find max 
    // occurring divisor 
    // interval [x, y] 
    static int findDivisor(int x, int y) 
    { 
        // if there is only  
        // one number in the 
        // in the interval,  
        // return that number 
        if (x == y) 
            return y; 
      
        // otherwise, 2 is the max  
        // occurring divisor 
        return 2; 
    } 
      
    /* Driver program  */
    public static void main(String[] args) 
    { 
        int x = 3, y = 16; 
        System.out.println(findDivisor(x, y)); 
    } 
} 
  
// This code is contributed by Gitanjali. 

Python3

# Efficient python 3 program  
# to find maximum occurring 
# factor in an interval 
  
# function to find max 
# occurring divisor 
# interval [x, y] 
def findDivisor(x, y): 
  
    # if there is only  
    # one number in the 
    # in the interval,  
    # return that number 
    if (x == y): 
        return y 
  
    # otherwise, 2 is  
    # max occurring  
    # divisor 
    return 2
  
# Driver code 
x = 3
y = 16
print(findDivisor(x, y)) 
  
# This code is contributed by 
# Smitha Dinesh Semwal 

C#

// Efficient C# program to 
// find maximum occurring 
// factor in an interval 
using System; 
  
class GFG { 
          
    // function to find max 
    // occurring divisor 
    // interval [x, y] 
    static int findDivisor(int x, int y) 
    { 
          
        // if there is only  
        // one number in the 
        // in the interval,  
        // return that number 
        if (x == y) 
            return y; 
      
        // otherwise, 2 is the max  
        // occurring divisor 
        return 2; 
    } 
      
    // Driver Code 
    public static void Main() 
    { 
        int x = 3, y = 16; 
        Console.Write(findDivisor(x, y)); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// Efficient PHP program to 
// find maximum occurring 
// factor in an interval 
  
// function to find max 
// occurring divisor 
// interval [x, y] 
function findDivisor($x, $y) 
{ 
    // if there is only  
    // one number in the 
    // in the interval,  
    // return that number 
    if ($x == $y) 
        return $y; 
  
    // otherwise, 2 is the 
    // max occurring  
    // divisor 
    return 2; 
} 
  
// Driver code 
$x = 3; 
$y = 16; 
echo findDivisor($x, $y); 
  
// This code is contributed by mits. 
?> 

Output :

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

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

C++

Java

Python3

C#

PHP

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