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Find the row whose product has maximum count of prime factors
  • Last Updated : 08 Sep, 2020

Given a matrix of size N x M, the task is to print the elements of the row whose product has a maximum count of prime factors.
Examples:  

Input: arr[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; 
Output: 7 8 9 
Explanation: 
Row 1: (1, 2, 3) has product 6 and it has 2 prime factors. 
Row 2: (4, 5, 6) has product 120 and it has 3 prime factors. 
Row 3: (7, 8, 9) has product 504 and it has 6 prime factors. 
Therefore, the output is 7 8 9, as it has maximum count of prime factors.
Input: arr[][] = {{11, 12, 13}, {14, 15, 16}, {17, 18, 19}} 
Output: 14 15 16 
 

Approach:  

  • Find the number of overall occurrences of each prime factor in whole each row by traversing all elements and finding their prime factors. We use hashing to count occurrences.
  • Let the counts of occurrences of prime factors be a1, a2, …aK. If we have K distinct prime factors, then the answer will be:
     

(a_{1}+1)(a_{2}+1)(...)*(a_{K}+1)

  • Compare this with the value that stores the maximum count of prime factors in a row in max_factor. If greater, update the value of the row.
  • Continue until all rows have been traversed.
    Below is the implementation of the above approach: 
     

C++

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// C++ implementation to find the row
// whose product has maximum number
// of prime factors
 
#include <bits/stdc++.h>
using namespace std;
 
#define N 3
#define M 5
 
int Large = 1e6;
 
vector<int> prime;
 
// function for SieveOfEratosthenes
void SieveOfEratosthenes()
{
 
    // Create a boolean array "isPrime[0..N]"
    // and initialize all entries it as true.
    // A value in isPrime[i] will finally be
    // false if i is not a prime, else true.
    bool isPrime[Large + 1];
    memset(isPrime, true, sizeof(isPrime));
 
    for (int p = 2; p * p <= Large; p++) {
 
        // check if isPrime[p] is not changed
        if (isPrime[p] == true) {
 
            // Update all multiples of p
            for (int i = p * 2; i <= Large; i += p)
                isPrime[i] = false;
        }
    }
 
    // Print all isPrime numbers
    for (int p = 2; p <= Large; p++)
 
        if (isPrime[p])
 
            prime.push_back(p);
}
 
// function to display the answer
void Display(int arr[][M], int row)
{
 
    for (int i = 0; i < M; i++)
        cout << arr[row][i] << " ";
}
 
// function to Count the row number of
// divisors in particular row multiplication
void countDivisorsMult(int arr[][M])
{
 
    // Find count of occurrences
    // of each prime factor
    unordered_map<int, int> mp;
    int row_no = 0;
    long long max_factor = 0;
 
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            int no = arr[i][j];
 
            for (int k = 0; k < prime.size(); k++) {
                while (no > 1
                       && no % prime[k] == 0) {
 
                    no /= prime[k];
                    mp[prime[k]]++;
                }
 
                if (no == 1)
                    break;
            }
        }
 
        // Compute count of all divisors
        long long int res = 1;
        for (auto it : mp) {
            res *= (it.second + 1L);
        }
 
        // Update row number if
        // factors of this row is max
        if (max_factor < res) {
            row_no = i;
            max_factor = res;
        }
 
        // Clearing map to store
        // prime factors for next row
        mp.clear();
    }
 
    Display(arr, row_no);
}
 
// Driver code
int main()
{
 
    int arr[N][M] = { { 1, 2, 3, 10, 23 },
                      { 4, 5, 6, 7, 8 },
                      { 7, 8, 9, 15, 45 } };
 
    SieveOfEratosthenes();
 
    countDivisorsMult(arr);
 
    return 0;
}

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Java

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// Java implementation to find the row
// whose product has maximum number
// of prime factors
import java.util.*;
 
class GFG{
  
static final int N = 3;
static final int M = 5;
  
static int Large = (int) 1e6;
  
static Vector<Integer> prime = new Vector<Integer>();
  
// function for SieveOfEratosthenes
static void SieveOfEratosthenes()
{
  
    // Create a boolean array "isPrime[0..N]"
    // and initialize all entries it as true.
    // A value in isPrime[i] will finally be
    // false if i is not a prime, else true.
    boolean []isPrime = new boolean[Large + 1];
    Arrays.fill(isPrime, true);
  
    for (int p = 2; p * p <= Large; p++) {
  
        // check if isPrime[p] is not changed
        if (isPrime[p] == true) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= Large; i += p)
                isPrime[i] = false;
        }
    }
  
    // Print all isPrime numbers
    for (int p = 2; p <= Large; p++)
  
        if (isPrime[p])
  
            prime.add(p);
}
  
// function to display the answer
static void Display(int arr[][], int row)
{
  
    for (int i = 0; i < M; i++)
        System.out.print(arr[row][i]+ " ");
}
  
// function to Count the row number of
// divisors in particular row multiplication
static void countDivisorsMult(int arr[][])
{
  
    // Find count of occurrences
    // of each prime factor
    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
    int row_no = 0;
    long max_factor = 0;
  
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            int no = arr[i][j];
  
            for (int k = 0; k < prime.size(); k++) {
                while (no > 1
                       && no % prime.get(k) == 0) {
  
                    no /= prime.get(k);
                    if(mp.containsKey(prime.get(k)))
                        mp.put(prime.get(k), prime.get(k)+1);
                    else
                        mp.put(prime.get(k), 1);
                }
  
                if (no == 1)
                    break;
            }
        }
  
        // Compute count of all divisors
        int res = 1;
        for (Map.Entry<Integer,Integer> it : mp.entrySet()) {
            res *= (it.getValue() + 1L);
        }
  
        // Update row number if
        // factors of this row is max
        if (max_factor < res) {
            row_no = i;
            max_factor = res;
        }
  
        // Clearing map to store
        // prime factors for next row
        mp.clear();
    }
  
    Display(arr, row_no);
}
  
// Driver code
public static void main(String[] args)
{
  
    int arr[][] = { { 1, 2, 3, 10, 23 },
                      { 4, 5, 6, 7, 8 },
                      { 7, 8, 9, 15, 45 } };
  
    SieveOfEratosthenes();
  
    countDivisorsMult(arr);
  
}
}
 
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation to find the row
# whose product has maximum number
# of prime factors
N = 3
M = 5
 
Large = int(1e6);
 
prime = [];
 
# function for SieveOfEratosthenes
def SieveOfEratosthenes() :
 
    # Create a boolean array "isPrime[0..N]"
    # and initialize all entries it as true.
    # A value in isPrime[i] will finally be
    # false if i is not a prime, else true.
    isPrime = [True]*(Large + 1);
     
    for p in range(2, int(Large**(1/2))) :
 
        # check if isPrime[p] is not changed
        if (isPrime[p] == True) :
 
            # Update all multiples of p
            for i in range(p*2, Large + 1, p) :
                isPrime[i] = False;
 
    # Print all isPrime numbers
    for p in range(2, Large + 1) :
 
        if (isPrime[p]) :
 
            prime.append(p);
 
# function to display the answer
def Display(arr, row) :
 
    for i in range(M) :
        print(arr[row][i], end=" ");
 
# function to Count the row number of
# divisors in particular row multiplication
def countDivisorsMult(arr) :
 
    # Find count of occurrences
    # of each prime factor
    mp = {};
    row_no = 0;max_factor = 0;
 
    for i in range(N) :
        for j in range(M) :
            no = arr[i][j]
             
            for k in range(len(prime)) :
                while (no > 1 and no % prime[k] == 0) :
                     
                    no //= prime[k];
                     
                    if prime[k] not in mp :
                        mp[prime[k]] = 0
                     
                    mp[prime[k]] += 1;
                     
                if (no == 1) :
                    break;
 
        # Compute count of all divisors
        res = 1;
        for it in mp :
            res *= mp[it];
 
        # Update row number if
        # factors of this row is max
        if (max_factor < res) :
            row_no = i;
            max_factor = res;
         
        # Clearing map to store
        # prime factors for next row
        mp.clear();
 
    Display(arr, row_no);
 
# Driver code
if __name__ == "__main__" :
 
 
    arr = [ [ 1, 2, 3, 10, 23 ],
            [ 4, 5, 6, 7, 8 ],
            [ 7, 8, 9, 15, 45 ] ];
 
    SieveOfEratosthenes();
 
    countDivisorsMult(arr);
 
# This code is contributed by Yash_R

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C#

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// C# implementation to find the row
// whose product has maximum number
// of prime factors
using System;
using System.Collections.Generic;
class GFG{
static readonly int N = 3;
static readonly int M = 5;
static int Large = (int) 1e6;
static List<int> prime = new List<int>();
  
// function for SieveOfEratosthenes
static void SieveOfEratosthenes()
{
    // Create a bool array "isPrime[0..N]"
    // and initialize all entries it as true.
    // A value in isPrime[i] will finally be
    // false if i is not a prime, else true.
    bool []isPrime = new bool[Large + 1];
    for (int p = 0; p <= Large; p++)
        isPrime[p] = true;
  
    for (int p = 2; p * p <= Large; p++)
    {
        // check if isPrime[p] is not changed
        if (isPrime[p] == true)
        {
            // Update all multiples of p
            for (int i = p * 2; i <= Large; i += p)
                isPrime[i] = false;
        }
    }
  
    // Print all isPrime numbers
    for (int p = 2; p <= Large; p++)
        if (isPrime[p])
            prime.Add(p);
}
  
// function to display the answer
static void Display(int [, ]arr, int row)
{
    for (int i = 0; i < M; i++)
        Console.Write(arr[row, i] + " ");
}
  
// function to Count the row number of
// divisors in particular row multiplication
static void countDivisorsMult(int [, ]arr)
{
    // Find count of occurrences
    // of each prime factor
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();
    int row_no = 0;
    long max_factor = 0;
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            int no = arr[i,j];
            for (int k = 0; k < prime.Count; k++)
            {
                while (no > 1 && no %
                       prime[k] == 0)
                {
                    no /= prime[k];
                    if(mp.ContainsKey(prime[k]))
                        mp[prime[k]] = prime[k] + 1;
                    else
                        mp.Add(prime[k], 1);
                }
  
                if (no == 1)
                    break;
            }
        }
  
        // Compute count of all divisors
        int res = 1;
        foreach (KeyValuePair<int,int> it in mp)
        {
            res *= (it.Value + 1);
        }
  
        // Update row number if
        // factors of this row is max
        if (max_factor < res)
        {
            row_no = i;
            max_factor = res;
        }
  
        // Clearing map to store
        // prime factors for next row
        mp.Clear();
    }
    Display(arr, row_no);
}
  
// Driver code
public static void Main(String[] args)
{
    int [, ]arr = {{1, 2, 3, 10, 23},
                  {4, 5, 6, 7, 8},
                  {7, 8, 9, 15, 45}};
    SieveOfEratosthenes();
    countDivisorsMult(arr);
}
}
 
// This code is contributed by Rajput-Ji

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Output: 

7 8 9 15 45



 

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