Find a number that divides maximum array elements

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 element 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++

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// 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;
}

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Java

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// 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

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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

Output:

2

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



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