# 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  ` `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 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 ``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

 ` 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

 ``

Output

```3
12```

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

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