Find the row whose product has maximum count of prime factors

Last Updated : 12 Jul, 2025

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++

// 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;
}

Java

// 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

Python3

# 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

// 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

JavaScript

<script>

// JavaScript implementation to find the row
// whose product has maximum number
// of prime factors


let N  = 3
let M = 5

let Large = 1e6;

let prime = new Array();

// function for SieveOfEratosthenes
function 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.
    let isPrime = new Array();

    for(let i = 0;  i < Large + 1; i++){
        isPrime.push([])
    }

    isPrime.fill(true);

    for (let p = 2; p * p <= Large; p++) {

        // check if isPrime[p] is not changed
        if (isPrime[p] == true) {

            // Update all multiples of p
            for (let i = p * 2; i <= Large; i += p)
                isPrime[i] = false;
        }
    }

    // Print all isPrime numbers
    for (let p = 2; p <= Large; p++)

        if (isPrime[p])

            prime.push(p);
}

// function to display the answer
function Display(arr, row)
{

    for (let i = 0; i < M; i++)
        document.write(arr[row][i] + " ");
}

// function to Count the row number of
// divisors in particular row multiplication
function countDivisorsMult(arr)
{

    // Find count of occurrences
    // of each prime factor
    let mp = new Map();
    let row_no = 0;
    let max_factor = 0;

    for (let i = 0; i < N; i++) {
        for (let j = 0; j < M; j++) {
            let no = arr[i][j];

            for (let k = 0; k < prime.length; k++) {
                while (no > 1 && no % prime[k] == 0) {

                    no /= prime[k];
                    if(mp.has(prime[k])){
                        mp.set(prime[k], mp.get(prime[k]) + 1)
                    }else{
                        mp.set(prime[k], 1)
                    }
                }

                if (no == 1)
                    break;
            }
        }

        // Compute count of all divisors
        let res = 1;
        for (let it of mp) {
            res *= (it[1] + 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


let arr = [ [ 1, 2, 3, 10, 23 ],
            [ 4, 5, 6, 7, 8 ],
            [ 7, 8, 9, 15, 45 ] ];

SieveOfEratosthenes();

countDivisorsMult(arr);

// This code is contributed by gfgking

</script>

Output:

7 8 9 15 45

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Find the row whose product has maximum count of prime factors

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