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Find a number that divides maximum array elements
  • Difficulty Level : Expert
  • Last Updated : 31 May, 2021

Given an array A[] of N non-negative integers. Find an Integer greater than 1, such that maximum array elements are divisible by it. In case of same answer print the smaller one.
Examples
 

Input : A[] = { 2, 4, 5, 10, 8, 15, 16 }; 
Output : 2 
Explanation: 2 divides [ 2, 4, 10, 8, 16] no other element divides greater than 5 numbers.
Input : A[] = { 2, 5, 10 } 
Output : 2 
Explanation: 2 divides [2, 10] and 5 divides [5, 10], but 2 is smaller. 
 

 

Naive Approach: Run a for loop upto maximum element of the array. Let it be K. Iterate the array and divide each element of the array by all numbers 1 \leq i \leq K  . Update the result according the maximum number of elements got divided by the element i.
Efficient Approach: We know that a number can be divisible only by elements which can be formed by their prime factors
Thus we find the prime factors of all elements of the array and store their frequency in the hash. Finally, we return the element with maximum frequency among them.
You can use factorization-using-sieve to find prime factors in Log(n).
Below is the implementation of above approach: 
 

C++




// CPP program to find a number that
// divides maximum array elements
 
#include <bits/stdc++.h>
using namespace std;
 
#define MAXN 100001
 
// stores smallest prime factor for every number
int spf[MAXN];
 
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
void sieve()
{
    spf[1] = 1;
    for (int i = 2; i < MAXN; i++)
 
        // marking smallest prime factor for every
        // number to be itself.
        spf[i] = i;
 
    // separately marking spf for every even
    // number as 2
    for (int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
 
    for (int i = 3; i * i < MAXN; i++) {
        // checking if i is prime
        if (spf[i] == i) {
            // marking SPF for all numbers divisible by i
            for (int j = i * i; j < MAXN; j += i)
 
                // marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
        }
    }
}
 
// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
vector<int> getFactorization(int x)
{
    vector<int> ret;
    while (x != 1) {
        int temp = spf[x];
        ret.push_back(temp);
        while (x % temp == 0)
            x = x / temp;
    }
    return ret;
}
 
// Function to find a number that
// divides maximum array elements
int maxElement(int A[], int n)
{
    // precalculating Smallest Prime Factor
    sieve();
 
    // Hash to store frequency of each divisors
    map<int, int> m;
 
    // Traverse the array and get spf of each element
    for (int i = 0; i < n; ++i) {
 
        // calling getFactorization function
        vector<int> p = getFactorization(A[i]);
 
        for (int i = 0; i < p.size(); i++)
            m[p[i]]++;
    }
 
    int cnt = 0, ans = 1e+7;
 
    for (auto i : m) {
        if (i.second >= cnt) {
            cnt = i.second;
            ans > i.first ? ans = i.first : ans = ans;
        }
    }
 
    return ans;
}
 
// Driver program
int main()
{
    int A[] = { 2, 5, 10 };
    int n = sizeof(A) / sizeof(A[0]);
 
    cout << maxElement(A, n);
 
    return 0;
}

Java




// Java program to find a number that
// divides maximum array elements
import java.util.*;
class Solution
{
static final int MAXN=100001;
   
// stores smallest prime factor for every number
static int spf[]= new int[MAXN];
   
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
static void sieve()
{
    spf[1] = 1;
    for (int i = 2; i < MAXN; i++)
   
        // marking smallest prime factor for every
        // number to be itself.
        spf[i] = i;
   
    // separately marking spf for every even
    // number as 2
    for (int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
   
    for (int i = 3; i * i < MAXN; i++) {
        // checking if i is prime
        if (spf[i] == i) {
            // marking SPF for all numbers divisible by i
            for (int j = i * i; j < MAXN; j += i)
   
                // marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
        }
    }
}
   
// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
static Vector<Integer> getFactorization(int x)
{
    Vector<Integer> ret= new Vector<Integer>();
    while (x != 1) {
        int temp = spf[x];
        ret.add(temp);
        while (x % temp == 0)
            x = x / temp;
    }
    return ret;
}
   
// Function to find a number that
// divides maximum array elements
static int maxElement(int A[], int n)
{
    // precalculating Smallest Prime Factor
    sieve();
   
    // Hash to store frequency of each divisors
    Map<Integer, Integer> m= new HashMap<Integer, Integer>();
   
    // Traverse the array and get spf of each element
    for (int j = 0; j < n; ++j) {
   
        // calling getFactorization function
        Vector<Integer> p = getFactorization(A[j]);
   
        for (int i = 0; i < p.size(); i++)
            m.put(p.get(i),m.get(p.get(i))==null?0:m.get(p.get(i))+1);
    }
   
    int cnt = 0, ans = 10000000;
    // Returns Set view     
       Set< Map.Entry< Integer,Integer> > st = m.entrySet();   
   
       for (Map.Entry< Integer,Integer> me:st)
       {
        if (me.getValue() >= cnt) {
            cnt = me.getValue();
            if(ans > me.getKey())
            ans = me.getKey() ;
            else
            ans = ans;
        }
    }
   
    return ans;
}
   
// Driver program
public static void main(String args[])
{
    int A[] = { 2, 5, 10 };
    int n =A.length;
   
    System.out.print(maxElement(A, n));
   
 
}
}
//contributed by Arnab Kundu

Python3




# Python3 program to find a number that
# divides maximum array elements
import math as mt
 
MAXN = 100001
 
# stores smallest prime factor for
# every number
spf = [0 for i in range(MAXN)]
 
# Calculating SPF (Smallest Prime Factor)
# for every number till MAXN.
# Time Complexity : O(nloglogn)
def sieve():
 
    spf[1] = 1
    for i in range(2, MAXN):
 
        # marking smallest prime factor for
        # every number to be itself.
        spf[i] = i
 
    # separately marking spf for every
    # even number as 2
    for i in range(4, MAXN, 2):
        spf[i] = 2
 
    for i in range(3, mt.ceil(mt.sqrt(MAXN + 1))):
         
        # checking if i is prime
        if (spf[i] == i):
             
            # marking SPF for all numbers divisible by i
            for j in range(2 * i, MAXN, i):
 
                # marking spf[j] if it is not
                # previously marked
                if (spf[j] == j):
                    spf[j] = i
         
# A O(log n) function returning primefactorization
# by dividing by smallest prime factor at every step
def getFactorization (x):
 
    ret = list()
    while (x != 1):
        temp = spf[x]
        ret.append(temp)
        while (x % temp == 0):
            x = x //temp
     
    return ret
 
# Function to find a number that
# divides maximum array elements
def maxElement (A, n):
 
    # precalculating Smallest Prime Factor
    sieve()
 
    # Hash to store frequency of each divisors
    m = dict()
 
    # Traverse the array and get spf of each element
    for i in range(n):
 
        # calling getFactorization function
        p = getFactorization(A[i])
 
        for i in range(len(p)):
            if p[i] in m.keys():
                m[p[i]] += 1
            else:
                m[p[i]] = 1
 
    cnt = 0
    ans = 10**9+7
 
    for i in m:
        if (m[i] >= cnt):
            cnt = m[i]
            if ans > i:
                ans = i
            else:
                ans = ans
 
    return ans
 
# Driver Code
A = [2, 5, 10 ]
n = len(A)
 
print(maxElement(A, n))
 
# This code is contributed by Mohit kumar 29

C#




     
// C# program to find a number that
// divides maximum array elements
using System;
using System.Collections.Generic;
 
class Solution
{
     
static readonly int MAXN = 100001;
     
// stores smallest prime factor for every number
static int []spf = new int[MAXN];
     
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
static void sieve()
{
    spf[1] = 1;
    for (int i = 2; i < MAXN; i++)
     
        // marking smallest prime factor for every
        // number to be itself.
        spf[i] = i;
     
    // separately marking spf for every even
    // number as 2
    for (int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
     
    for (int i = 3; i * i < MAXN; i++)
    {
        // checking if i is prime
        if (spf[i] == i)
        {
            // marking SPF for all numbers divisible by i
            for (int j = i * i; j < MAXN; j += i)
     
                // marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
        }
    }
}
     
// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
static List<int> getFactorization(int x)
{
    List<int> ret= new List<int>();
    while (x != 1)
    {
        int temp = spf[x];
        ret.Add(temp);
        while (x % temp == 0)
            x = x / temp;
    }
    return ret;
}
     
// Function to find a number that
// divides maximum array elements
static int maxElement(int []A, int n)
{
    // precalculating Smallest Prime Factor
    sieve();
     
    // Hash to store frequency of each divisors
    Dictionary<int, int> m= new Dictionary<int, int>();
     
    // Traverse the array and get spf of each element
    for (int j = 0; j < n; ++j)
    {
     
        // calling getFactorization function
        List<int> p = getFactorization(A[j]);
     
        for (int i = 0; i < p.Count; i++)
            if(m.ContainsKey(p[i]))
            m[p[i]] = m[p[i]] + 1;
            else
                m.Add(p[i], 1);
    }
     
    int cnt = 0, ans = 10000000;
     
    // Returns Set view    
    foreach(KeyValuePair<int, int> me in m)
    {
        if (me.Value >= cnt)
        {
            cnt = me.Value;
            if(ans > me.Key)
                ans = me.Key ;
            else
                ans = ans;
        }
    }
     
    return ans;
}
     
// Driver program
public static void Main(String []args)
{
    int []A = { 2, 5, 10 };
    int n =A.Length;
     
    Console.Write(maxElement(A, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program to find a number that
// divides maximum array elements
 
let MAXN=100001;
 
// stores smallest prime factor for every number
let spf= new Array(MAXN);
for(let i=0;i<MAXN;i++)
{
    spf[i]=0;
}
 
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
function sieve()
{
    spf[1] = 1;
    for (let i = 2; i < MAXN; i++)
     
        // marking smallest prime factor for every
        // number to be itself.
        spf[i] = i;
     
    // separately marking spf for every even
    // number as 2
    for (let i = 4; i < MAXN; i += 2)
        spf[i] = 2;
     
    for (let i = 3; i * i < MAXN; i++) {
        // checking if i is prime
        if (spf[i] == i) {
            // marking SPF for all numbers divisible by i
            for (let j = i * i; j < MAXN; j += i)
     
                // marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
        }
    }
}
 
// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
function getFactorization(x)
{
    let ret= [];
    while (x != 1) {
        let temp = spf[x];
        ret.push(temp);
        while (x % temp == 0)
            x = Math.floor(x / temp);
    }
    return ret;
}
 
// Function to find a number that
// divides maximum array elements
function maxElement(A,n)
{
    // precalculating Smallest Prime Factor
    sieve();
     
    // Hash to store frequency of each divisors
    let m= new Map();
     
    // Traverse the array and get spf of each element
    for (let j = 0; j < n; ++j) {
     
        // calling getFactorization function
        let p = getFactorization(A[j]);
     
        for (let i = 0; i < p.length; i++)
            m.set(p[i],m.get(p[i])==null?0:m.get(p[i])+1);
    }
     
    let cnt = 0, ans = 10000000;
    // Returns Set view     
           
     
       for (let [key, value] of m.entries())
       {
        if (value >= cnt) {
            cnt = value;
            if(ans > key)
                ans = key ;
            else
                ans = ans;
        }
    }
     
    return ans;
}
 
// Driver program
let A=[ 2, 5, 10];
let n =A.length;
document.write(maxElement(A, n));
 
     
 
// This code is contributed by patel2127
 
</script>
Output: 
2

 

Time Complexity: O(N*log(N))
 

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