Count of subarrays whose products don’t have any repeating prime factor

Given an array of integers. Find the total number of subarrays whose product of all elements doesn’t contain repeating prime factor in prime decomposition of resulting number.

Examples:

Input: 2 3 9
Output: 3
Explanation:
Total sub-array are:-
{2}, {3}, {9}, {2, 3}, {3, 9}, {2, 3, 9}

Subarray which violets the property are:-
{9}       -> {3 * 3}, Since 3 is a repeating prime
             factor in prime decomposition of 9
{3, 9}    -> {3 * 3 * 3}, 3 is a repeating prime 
             factor in prime decomposition of 27
{2, 3, 9} -> {2 * 3 * 3 * 3}, 3 is repeating 
             prime factor in prime decomposition 
             of 54
Hence total subarray remains which satisfies our
condition are 3.

Input: 2, 3, 5, 15, 7, 2
Output: 12

A Naive approach is to run a loop one inside another and generate all subarrays and then take product of all elements such that it’s prime decomposition doesn’t contain repeating elements. This approach would definitely be slow and would lead to overflow for large value of array element.

An efficient approach is to use prime factorization using Sieve of Eratosthenes.

Idea is to store the Smallest Prime Factor(SPF) for all values (till a maximum) using Sieve. We calculate prime factorization of the given number by dividing the given number recursively with its smallest prime factor till it becomes 1.

1. Let ind[] be an array such that ind[i] stores the last index of prime divisor i in arr[], and ‘last_ind’ keeps track the last index of any divisor.
2. Now traverse from left to right(0 to n-1). For particular element of array[i], find prime divisors using above approach, and initialize all the divisors with the latest index ‘i+1’.
3. But before performing step 2, we update the variable of ‘last_ind’ with ind[] of every divisor of array[i].
4. Since the variable ‘last_ind’ contain a last index(less than i) of any divisor of array[i], we can assure that all elements(last_ind+1, last_ind+2 … i) will not have any repeating prime factor of arr[i]. Hence our ans will be (i – last_ind +1)
5. Perform above steps for remaining element of array[] and simultaneously update the answer for every index.

Implementation:

C++

// C++ program to count all sub-arrays whose
// product doesn't contain a repeating prime
// factor.
#include<bits/stdc++.h>
using namespace std;

const int MAXN = 1000001;
int spf[MAXN];

// Calculating SPF (Smallest Prime Factor) for
// every number till MAXN.
// Time Complexity : O(n log log n)
void sieve()
{
    // marking smallest prime factor for every
    // number to be itself.
    for (int i=1; i<MAXN; i++)
        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;
        }
    }
}

// Function to count all sub-arrays whose
// product doesn't contain a repeating prime
// factor.
int countSubArray(int arr[], int n)
{
    // ind[i] is going to store 1 + last index of
    // of an array element which has i as prime
    // factor.
    int ind[MAXN];
    memset(ind, -1, sizeof ind);

    int count = 0; // Initialize result
    int last_ind = 0; // It stores index
    for (int i=0; i < n; ++i)
    {
        while (arr[i] > 1)
        {
            int div = spf[arr[i]];

            // Fetch the last index of prime
            // divisor of element
            last_ind = max(last_ind, ind[div]);

            // Update the current divisor index
            ind[div] = i + 1;

            arr[i] /= div;
        }

        // Update result, we basically include
        // all required subarrays ending with
        // index arr[i].
        count += i - last_ind + 1;
    }
    return count;
}

// Driver code
int main()
{
    sieve();
    int arr[] = {2, 3, 9};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countSubArray(arr, n) << "\n";

    int arr1[] = {2, 3, 5, 15, 7, 2};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    cout << countSubArray(arr1, n1);

    return 0;
}

Java

// Java program to count all sub-arrays whose
// product doesn't contain a repeating prime
// factor
import java.io.*;
import java.util.*;

class GFG
{
    public static int MAXN = 1000001;
    public static int[] spf = new int[MAXN];

    // Calculating SPF (Smallest Prime Factor) for
    // every number till MAXN.
    // Time Complexity : O(n log log n)
    static void sieve()
    {
        // marking smallest prime factor for every
        // number to be itself.
        for (int i=1; i<MAXN; i++)
            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;
            }
        }
    }

    // Function to count all sub-arrays whose
    // product doesn't contain a repeating prime
    // factor
    static int countSubArray(int arr[], int n)
    {
        // ind[i] is going to store 1 + last index of
        // of an array element which has i as prime
        // factor.
        int[] ind = new int[MAXN];
        Arrays.fill(ind, -1);

        int count = 0; // Initialize result
        int last_ind = 0; // It stores index
        for (int i=0; i < n; ++i)
        {
            while (arr[i] > 1)
            {
                int div = spf[arr[i]];

                // Fetch the last index of prime
                // divisor of element
                last_ind = Math.max(last_ind, ind[div]);

                // Update the current divisor index
                ind[div] = i + 1;

                arr[i] /= div;
            }

            // Update result, we basically include
            // all required subarrays ending with
            // index arr[i].
            count += i - last_ind + 1;
        }
        return count;
    }

    // driver program
    public static void main (String[] args)
    {
        sieve();
        int arr[] = {2, 3, 9};
        int n = arr.length;
        System.out.println(countSubArray(arr, n));

        int arr1[] = {2, 3, 5, 15, 7, 2};
        int n1 = arr1.length;
        System.out.println(countSubArray(arr1, n1));
    }
}

// Contributed by Pramod Kumar

Python3

# Python 3 program to count all sub-arrays
# whose product does not contain a repeating
# prime factor.
from math import sqrt

MAXN = 1000001
spf = [0 for i in range(MAXN)]

# Calculating SPF (Smallest Prime Factor)
# for every number till MAXN.
# Time Complexity : O(n log log n)
def sieve():

    # marking smallest prime factor
    # for every number to be itself.
    for i in range(1, MAXN, 1):
        spf[i] = i

    # separately marking spf for
    # every even number as 2
    for i in range(4, MAXN, 2):
        spf[i] = 2

    k = int(sqrt(MAXN))
    for i in range(3, k, 1):

        # checking if i is prime
        if (spf[i] == i):

            # marking SPF for all numbers
            # divisible by i
            for j in range(i * i, MAXN, i):

                # marking spf[j] if it is
                # not previously marked
                if (spf[j] == j):
                    spf[j] = i

# Function to count all sub-arrays whose
# product doesn't contain a repeating
# prime factor.
def countSubArray(arr, n):

    # ind[i] is going to store 1 + last
    # index of an array element which
    # has i as prime factor.
    ind = [-1 for i in range(MAXN)]

    count = 0

    # Initialize result
    last_ind = 0

    # It stores index
    for i in range(0, n, 1):
        while (arr[i] > 1):
            div = spf[arr[i]]

            # Fetch the last index of prime
            # divisor of element
            last_ind = max(last_ind, ind[div])

            # Update the current divisor index
            ind[div] = i + 1

            arr[i] = int(arr[i] / div)

        # Update result, we basically include
        # all required subarrays ending with
        # index arr[i].
        count += i - last_ind + 1
    return count

# Driver code
if __name__ == '__main__':
    sieve()
    arr = [2, 3, 9]
    n = len(arr)
    print(countSubArray(arr, n))

    arr1 = [2, 3, 5, 15, 7, 2]
    n1 = len(arr1)
    print(countSubArray(arr1, n1))

# This code is contributed by
# Shashank_Sharma

C#

// C# program to count all sub-arrays
// whose product doesn't contain a
// repeating prime factor
using System;

public class GFG  {

    public static int MAXN = 1000001;
    public static int[] spf = new int[MAXN];

    // Calculating SPF (Smallest Prime Factor)
    // for every number till MAXN.
    // Time Complexity : O(n log log n)
    static void sieve()
    {
        // marking smallest prime factor
        // for every number to be itself.
        for (int i = 1; i < MAXN; i++)
            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;
            }
        }
    }

    // Function to count all sub-arrays
    // whose product doesn't contain
    // a repeating prime factor
    static int countSubArray(int []arr, int n)
    {

        // ind[i] is going to store 1 + last
        // index of an array element which
        // has i as prime factor.
        int[] ind = new int[MAXN];

        for(int i = 0; i < MAXN; i++)
        {
            ind[i] = -1;
        }


        int count = 0; // Initialize result
        int last_ind = 0; // It stores index
        for (int i = 0; i < n; ++i)
        {
            while (arr[i] > 1)
            {
                int div = spf[arr[i]];

                // Fetch the last index of prime
                // divisor of element
                last_ind = Math.Max(last_ind, ind[div]);

                // Update the current divisor index
                ind[div] = i + 1;

                arr[i] /= div;
            }

            // Update result, we basically include
            // all required subarrays ending with
            // index arr[i].
            count += i - last_ind + 1;
        }
        return count;
    }

    // Driver Code
    public static void Main ()
    {
        sieve();
        int []arr = {2, 3, 9};
        int n = arr.Length;
        Console.WriteLine(countSubArray(arr, n));

        int []arr1 = {2, 3, 5, 15, 7, 2};
        int n1 = arr1.Length;
        Console.WriteLine(countSubArray(arr1, n1));
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP program to count all sub-arrays whose
// product doesn't contain a repeating prime
// factor.
$MAXN = 1000001;
$spf = array_fill(0, $MAXN, NULL);

// Calculating SPF (Smallest Prime Factor)
// for every number till MAXN.
// Time Complexity : O(n log log n)
function sieve()
{
    global $spf, $MAXN;

    // marking smallest prime factor for
    // every number to be itself.
    for ($i = 1; $i < $MAXN; $i++)
        $spf[$i] = $i;

    // separately marking spf for every
    // even number as 2
    for ($i = 4; $i < $MAXN; $i += 2)
        $spf[$i] = 2;

    for ($i = 3; $i * $i < $MAXN; $i++)
    {
        // checking if i is prime
        if ($spf[$i] == $i)
        {
            // marking SPF for all numbers
            // divisible by i
            for ($j = $i * $i; $j < $MAXN; $j += $i)

                // marking spf[j] if it is not
                // previously marked
                if ($spf[$j] == $j)
                    $spf[$j] = $i;
        }
    }
}

// Function to count all sub-arrays whose
// product doesn't contain a repeating
// prime factor.
function countSubArray(&$arr, $n)
{
    global $MAXN, $spf;

    // ind[i] is going to store 1 + last index
    // of an array element which has i as prime
    // factor.
    $ind = array_fill(-1, $MAXN, NULL);

    $count = 0; // Initialize result
    $last_ind = 0; // It stores index
    for ($i = 0; $i < $n; ++$i)
    {
        while ($arr[$i] > 1)
        {
            $div = $spf[$arr[$i]];

            // Fetch the last index of prime
            // divisor of element
            $last_ind = max($last_ind, $ind[$div]);

            // Update the current divisor index
            $ind[$div] = $i + 1;
            if($div != 0)
            $arr[$i] /= $div;
        }

        // Update result, we basically include
        // all required subarrays ending with
        // index arr[i].
        $count += $i - $last_ind + 1;
    }
    return $count;
}

// Driver code
sieve();
$arr = array(2, 3, 9);
$n = sizeof($arr);
echo countSubArray($arr, $n) . "\n";

$arr1 = array(2, 3, 5, 15, 7, 2);
$n1 = sizeof($arr1);
echo countSubArray($arr1, $n1);

// This code is contributed by ita_c
?>

Javascript

<script>
// Javascript program to count all sub-arrays whose
// product doesn't contain a repeating prime
// factor

    let MAXN = 1000001;
    let spf = new Array(MAXN);

     // Calculating SPF (Smallest Prime Factor) for
    // every number till MAXN.
    // Time Complexity : O(n log log n)
    function  sieve()
    {

        // marking smallest prime factor for every
        // number to be itself.
        for (let i = 1; i < MAXN; i++)
            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;
            }
        }
    }

     // Function to count all sub-arrays whose
    // product doesn't contain a repeating prime
    // factor
    function countSubArray(arr,n)
    {

        // ind[i] is going to store 1 + last index of
        // of an array element which has i as prime
        // factor.
        let ind = new Array(MAXN);
        for(let i = 0; i < ind.length; i++)
        {
            ind[i] = -1;
        }

        let count = 0; // Initialize result
        let last_ind = 0; // It stores index
        for (let i = 0; i < n; ++i)
        {
            while (arr[i] > 1)
            {
                let div = spf[arr[i]];

                // Fetch the last index of prime
                // divisor of element
                last_ind = Math.max(last_ind, ind[div]);

                // Update the current divisor index
                ind[div] = i + 1;

                arr[i] /= div;
            }

            // Update result, we basically include
            // all required subarrays ending with
            // index arr[i].
            count += i - last_ind + 1;
        }
        return count;
    }

    // driver program
    sieve();
    let arr = [2, 3, 9];
    let n = arr.length;
    document.write(countSubArray(arr, n)+"<br>");

    let arr1 = [2, 3, 5, 15, 7, 2];
    let n1 = arr1.length;
    document.write(countSubArray(arr1, n1));

    // This code is contributed by rag2127

</script>

Output

3
12

Time complexity: O(MAX*log(log(MAX) + nlog(n)) 
Auxiliary space: O(MAX)