# Count sub-arrays whose product is divisible by k

Given an integer K and an array arr[], the task is to count all the sub-arrays whose product is divisible by K.

Examples:

Input: arr[] = {6, 2, 8}, K = 4
Output: 4
Required sub-arrays are {6, 2}, {6, 2, 8}, {2, 8}and {8}.

Input: arr[] = {9, 1, 14}, K = 6
Output: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: Run nested loops and check for every sub-array whether product % k == 0. Update count = count + 1 when the condition is true for the sub-array. Time complexity of this approach is O(n3).

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long ` ` `  `// Function to count sub-arrays whose ` `// product is divisible by K ` `int` `countSubarrays(``const` `int``* arr, ``int` `n, ``int` `K) ` `{ ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i; j < n; j++) { ` ` `  `            ``// Calculate the product of the ` `            ``// current sub-array ` `            ``ll product = 1; ` `            ``for` `(``int` `x = i; x <= j; x++) ` `                ``product *= arr[x]; ` ` `  `            ``// If product of the current sub-array ` `            ``// is divisible by K ` `            ``if` `(product % K == 0) ` `                ``count++; ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 2, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `K = 4; ` `    ``cout << countSubarrays(arr, n, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to count sub-arrays whose ` `// product is divisible by K ` `static` `int` `countSubarrays(``int` `[]arr, ``int` `n, ``int` `K) ` `{ ` `    ``int` `count = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``for` `(``int` `j = i; j < n; j++) ` `        ``{ ` ` `  `            ``// Calculate the product of the ` `            ``// current sub-array ` `            ``long` `product = ``1``; ` `            ``for` `(``int` `x = i; x <= j; x++) ` `                ``product *= arr[x]; ` ` `  `            ``// If product of the current sub-array ` `            ``// is divisible by K ` `            ``if` `(product % K == ``0``) ` `                ``count++; ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = { ``6``, ``2``, ``8` `}; ` `    ``int` `n = arr.length; ` `    ``int` `K = ``4``; ` `    ``System.out.println(countSubarrays(arr, n, K)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python 3

 `# Python 3 implementation of the approach ` ` `  `# Function to count sub-arrays whose ` `# product is divisible by K ` `def` `countSubarrays(arr, n, K): ` ` `  `    ``count ``=` `0` `    ``for` `i ``in` `range``( n): ` `        ``for` `j ``in` `range``(i, n): ` ` `  `            ``# Calculate the product of  ` `            ``# the current sub-array ` `            ``product ``=` `1` `            ``for` `x ``in` `range``(i, j ``+` `1``): ` `                ``product ``*``=` `arr[x] ` ` `  `            ``# If product of the current  ` `            ``# sub-array is divisible by K ` `            ``if` `(product ``%` `K ``=``=` `0``): ` `                ``count ``+``=` `1` `    ``return` `count ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``arr ``=` `[ ``6``, ``2``, ``8` `] ` `    ``n ``=` `len``(arr) ` `    ``K ``=` `4` `    ``print``(countSubarrays(arr, n, K)) ` ` `  `# This code is contributed by ita_c  `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``// Function to count sub-arrays whose  ` `    ``// product is divisible by K  ` `    ``static` `int` `countSubarrays(``int` `[]arr, ``int` `n, ``int` `K)  ` `    ``{  ` `        ``int` `count = 0;  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `            ``for` `(``int` `j = i; j < n; j++)  ` `            ``{  ` `     `  `                ``// Calculate the product of the  ` `                ``// current sub-array  ` `                ``long` `product = 1;  ` `                ``for` `(``int` `x = i; x <= j; x++)  ` `                    ``product *= arr[x];  ` `     `  `                ``// If product of the current sub-array  ` `                ``// is divisible by K  ` `                ``if` `(product % K == 0)  ` `                    ``count++;  ` `            ``}  ` `        ``}  ` `        ``return` `count;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[]arr = { 6, 2, 8 };  ` `        ``int` `n = arr.Length;  ` `        ``int` `K = 4;  ` `         `  `        ``Console.WriteLine(countSubarrays(arr, n, K));  ` `    ``}  ` `}  ` ` `  `// This code contributed by Rajput-Ji  `

## PHP

 ` `

Output:

```4
```

A better approach is to either use the two-pointer technique or use segment trees to find the product of a sub-array in quick time.

Segment trees: The complexity of the naive solution is cubic. This is because every sub-array is traversed to find the product. Instead of traversing every sub-array, products can be stored in a segment tree and the tree can be queried to obtain the product % k value in O(log n) time. Thus, time complexity of this approach would then be O(n2 log n).

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long ` `#define MAX 100002 ` ` `  `// Segment tree implemented as an array ` `ll tree[4 * MAX]; ` ` `  `// Function to build the segment tree ` `void` `build(``int` `node, ``int` `start, ``int` `end, ``const` `int``* arr, ``int` `k) ` `{ ` `    ``if` `(start == end) { ` `        ``tree[node] = (1LL * arr[start]) % k; ` `        ``return``; ` `    ``} ` `    ``int` `mid = (start + end) >> 1; ` `    ``build(2 * node, start, mid, arr, k); ` `    ``build(2 * node + 1, mid + 1, end, arr, k); ` `    ``tree[node] = (tree[2 * node] * tree[2 * node + 1]) % k; ` `} ` ` `  `// Function to query product of ` `// sub-array[l..r] in O(log n) time ` `ll query(``int` `node, ``int` `start, ``int` `end, ``int` `l, ``int` `r, ``int` `k) ` `{ ` `    ``if` `(start > end || start > r || end < l) { ` `        ``return` `1; ` `    ``} ` `    ``if` `(start >= l && end <= r) { ` `        ``return` `tree[node] % k; ` `    ``} ` `    ``int` `mid = (start + end) >> 1; ` `    ``ll q1 = query(2 * node, start, mid, l, r, k); ` `    ``ll q2 = query(2 * node + 1, mid + 1, end, l, r, k); ` `    ``return` `(q1 * q2) % k; ` `} ` ` `  `// Function to count sub-arrays whose ` `// product is divisible by K ` `ll countSubarrays(``const` `int``* arr, ``int` `n, ``int` `k) ` `{ ` `    ``ll count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i; j < n; j++) { ` ` `  `            ``// Query segment tree to find product % k ` `            ``// of the sub-array[i..j] ` `            ``ll product_mod_k = query(1, 0, n - 1, i, j, k); ` `            ``if` `(product_mod_k == 0) { ` `                ``count++; ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 2, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 4; ` ` `  `    ``// Build the segment tree ` `    ``build(1, 0, n - 1, arr, k); ` ` `  `    ``cout << countSubarrays(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `    ``// Java implementation for above approach ` `class` `GFG ` `{ ` `     `  `static` `int` `MAX = ``100002``; ` ` `  `// Segment tree implemented as an array ` `static` `long` `tree[] = ``new` `long``[``4` `* MAX]; ` ` `  `// Function to build the segment tree ` `static` `void` `build(``int` `node, ``int` `start, ``int` `end,  ` `                            ``int` `[]arr, ``int` `k) ` `{ ` `    ``if` `(start == end) ` `    ``{ ` `        ``tree[node] = (1L * arr[start]) % k; ` `        ``return``; ` `    ``} ` `    ``int` `mid = (start + end) >> ``1``; ` `    ``build(``2` `* node, start, mid, arr, k); ` `    ``build(``2` `* node + ``1``, mid + ``1``, end, arr, k); ` `    ``tree[node] = (tree[``2` `* node] * tree[``2` `* node + ``1``]) % k; ` `} ` ` `  `// Function to query product of ` `// sub-array[l..r] in O(log n) time ` `static` `long` `query(``int` `node, ``int` `start, ``int` `end,  ` `                            ``int` `l, ``int` `r, ``int` `k) ` `{ ` `    ``if` `(start > end || start > r || end < l)  ` `    ``{ ` `        ``return` `1``; ` `    ``} ` `    ``if` `(start >= l && end <= r)  ` `    ``{ ` `        ``return` `tree[node] % k; ` `    ``} ` `    ``int` `mid = (start + end) >> ``1``; ` `    ``long` `q1 = query(``2` `* node, start, mid, l, r, k); ` `    ``long` `q2 = query(``2` `* node + ``1``, mid + ``1``, end, l, r, k); ` `    ``return` `(q1 * q2) % k; ` `} ` ` `  `// Function to count sub-arrays whose ` `// product is divisible by K ` `static` `long` `countSubarrays(``int` `[]arr, ``int` `n, ``int` `k) ` `{ ` `    ``long` `count = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i; j < n; j++)  ` `        ``{ ` ` `  `            ``// Query segment tree to find product % k ` `            ``// of the sub-array[i..j] ` `            ``long` `product_mod_k = query(``1``, ``0``, n - ``1``, i, j, k); ` `            ``if` `(product_mod_k == ``0``)  ` `            ``{ ` `                ``count++; ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``6``, ``2``, ``8` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``4``; ` ` `  `    ``// Build the segment tree ` `    ``build(``1``, ``0``, n - ``1``, arr, k); ` ` `  `    ``System.out.println(countSubarrays(arr, n, k)); ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## C#

 `// C# implementation for above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `MAX = 100002; ` ` `  `// Segment tree implemented as an array ` `static` `long` `[]tree = ``new` `long``[4 * MAX]; ` ` `  `// Function to build the segment tree ` `static` `void` `build(``int` `node, ``int` `start, ``int` `end,  ` `                            ``int` `[]arr, ``int` `k) ` `{ ` `    ``if` `(start == end) ` `    ``{ ` `        ``tree[node] = (1L * arr[start]) % k; ` `        ``return``; ` `    ``} ` `    ``int` `mid = (start + end) >> 1; ` `    ``build(2 * node, start, mid, arr, k); ` `    ``build(2 * node + 1, mid + 1, end, arr, k); ` `    ``tree[node] = (tree[2 * node] * tree[2 * node + 1]) % k; ` `} ` ` `  `// Function to query product of ` `// sub-array[l..r] in O(log n) time ` `static` `long` `query(``int` `node, ``int` `start, ``int` `end,  ` `                            ``int` `l, ``int` `r, ``int` `k) ` `{ ` `    ``if` `(start > end || start > r || end < l)  ` `    ``{ ` `        ``return` `1; ` `    ``} ` `    ``if` `(start >= l && end <= r)  ` `    ``{ ` `        ``return` `tree[node] % k; ` `    ``} ` `    ``int` `mid = (start + end) >> 1; ` `    ``long` `q1 = query(2 * node, start, mid, l, r, k); ` `    ``long` `q2 = query(2 * node + 1, mid + 1, end, l, r, k); ` `    ``return` `(q1 * q2) % k; ` `} ` ` `  `// Function to count sub-arrays whose ` `// product is divisible by K ` `static` `long` `countSubarrays(``int` `[]arr, ``int` `n, ``int` `k) ` `{ ` `    ``long` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i; j < n; j++)  ` `        ``{ ` ` `  `            ``// Query segment tree to find product % k ` `            ``// of the sub-array[i..j] ` `            ``long` `product_mod_k = query(1, 0, n - 1, i, j, k); ` `            ``if` `(product_mod_k == 0)  ` `            ``{ ` `                ``count++; ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 6, 2, 8 }; ` `    ``int` `n = arr.Length; ` `    ``int` `k = 4; ` ` `  `    ``// Build the segment tree ` `    ``build(1, 0, n - 1, arr, k); ` ` `  `    ``Console.WriteLine(countSubarrays(arr, n, k)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

Output:

```4
```

Further optimizing the solution: It is understandable that if the product of a sub-array [i..j] is divisible by k, then the product of all subarrays [i..t] such that j < t < n will also be divisible by k. Hence binary search can be applied to count sub-arrays starting at a particular index i and whose product is divisible by k.

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define MAX 100005 ` `typedef` `long` `long` `ll; ` ` `  `// Segment tree implemented as an array ` `ll tree[MAX << 2]; ` ` `  `// Function to build segment tree ` `void` `build(``int` `node, ``int` `start, ``int` `end, ``const` `int``* arr, ``int` `k) ` `{ ` `    ``if` `(start == end) { ` `        ``tree[node] = arr[start] % k; ` `        ``return``; ` `    ``} ` `    ``int` `mid = (start + end) >> 1; ` `    ``build(2 * node, start, mid, arr, k); ` `    ``build(2 * node + 1, mid + 1, end, arr, k); ` `    ``tree[node] = (tree[2 * node] * tree[2 * node + 1]) % k; ` `} ` ` `  `// Function to query product % k ` `// of sub-array[l..r] ` `ll query(``int` `node, ``int` `start, ``int` `end, ``int` `l, ``int` `r, ``int` `k) ` `{ ` `    ``if` `(start > end || start > r || end < l) ` `        ``return` `1; ` `    ``if` `(start >= l && end <= r) ` `        ``return` `tree[node] % k; ` `    ``int` `mid = (start + end) >> 1; ` `    ``ll q1 = query(2 * node, start, mid, l, r, k); ` `    ``ll q2 = query(2 * node + 1, mid + 1, end, l, r, k); ` `    ``return` `(q1 * q2) % k; ` `} ` ` `  `// Function to return the count of sub-arrays ` `// whose product is divisible by K ` `ll countSubarrays(``int``* arr, ``int` `n, ``int` `k) ` `{ ` ` `  `    ``ll ans = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``int` `low = i, high = n - 1; ` ` `  `        ``// Binary search ` `        ``// Check if sub-array[i..mid] satisfies the constraint ` `        ``// Adjust low and high accordingly ` `        ``while` `(low <= high) { ` `            ``int` `mid = (low + high) >> 1; ` `            ``if` `(query(1, 0, n - 1, i, mid, k) == 0) ` `                ``high = mid - 1; ` `            ``else` `                ``low = mid + 1; ` `        ``} ` `        ``ans += n - low; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 2, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 4; ` ` `  `    ``// Build the segment tree ` `    ``build(1, 0, n - 1, arr, k); ` ` `  `    ``cout << countSubarrays(arr, n, k); ` ` `  `    ``return` `0; ` `} `

Output:

```4
```

Two pointers technique: Analogous to the binary-search discussion, it is clear that if sub-array[i..j] has product divisible by k, then all sub-arrays [i..t] such that j < t < n will also have products divisible by k.

Hence, two-pointer technique can also be applied here corresponding to the above fact. Two pointers l, r are taken with l pointing to the start of the current sub-array and r pointing to the end of the current sub-array. If the sub-array [l..r] has product divisible by k, then all sub-arrays [l..s] such that r < s < n will have product divisible by k. Therefore perform count = count + n – r. Since elements need to be added and removed from the current sub-array, simply products can’t be taken of the whole sub-array since it will be cumbersome to add and remove elements in this way.

Instead, k is prime factorized in O(sqrt(n)) time and its prime-factors are stored in an STL map. Another map is used to maintain counts of primes in the current sub-array which may be called current map in this context. Then whenever an element needs to be added in the current sub-array, then count of those primes is added to the current map which occur in prime factorization of k. Whenever an element needs to be removed from the current map, then count of primes is similarly subtracted.

Below is the implementation of the above approach.

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long ` `#define MAX 100002 ` `#define pb push_back ` ` `  `// Vector to store primes ` `vector<``int``> primes; ` ` `  `// k_cnt stores count of prime factors of k ` `// current_map stores the count of primes ` `// in the current sub-array ` `// cnts[] is an array of maps which stores ` `// the count of primes for element at index i ` `unordered_map<``int``, ``int``> k_cnt, current_map, cnts[MAX]; ` ` `  `// Function to store primes in ` `// the vector primes ` `void` `sieve() ` `{ ` ` `  `    ``int` `prime[MAX]; ` `    ``prime = prime = 1; ` `    ``for` `(``int` `i = 2; i < MAX; i++) { ` `        ``if` `(prime[i] == 0) { ` `            ``for` `(``int` `j = i * 2; j < MAX; j += i) { ` `                ``if` `(prime[j] == 0) { ` `                    ``prime[j] = i; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``for` `(``int` `i = 2; i < MAX; i++) { ` `        ``if` `(prime[i] == 0) { ` `            ``prime[i] = i; ` `            ``primes.pb(i); ` `        ``} ` `    ``} ` `} ` ` `  `// Function to count sub-arrays whose product ` `// is divisible by k ` `ll countSubarrays(``int``* arr, ``int` `n, ``int` `k) ` `{ ` ` `  `    ``// Special case ` `    ``if` `(k == 1) { ` `        ``cout << (1LL * n * (n + 1)) / 2; ` `        ``return` `0; ` `    ``} ` ` `  `    ``vector<``int``> k_primes; ` ` `  `    ``for` `(``auto` `p : primes) { ` `        ``while` `(k % p == 0) { ` `            ``k_primes.pb(p); ` `            ``k /= p; ` `        ``} ` `    ``} ` ` `  `    ``// If k is prime and is more than 10^6 ` `    ``if` `(k > 1) { ` `        ``k_primes.pb(k); ` `    ``} ` ` `  `    ``for` `(``auto` `num : k_primes) { ` `        ``k_cnt[num]++; ` `    ``} ` ` `  `    ``// Two pointers initialized ` `    ``int` `l = 0, r = 0; ` ` `  `    ``ll ans = 0; ` ` `  `    ``while` `(r < n) { ` ` `  `        ``// Add rth element to the current segment ` `        ``for` `(``auto``& it : k_cnt) { ` ` `  `            ``// p = prime factor of k ` `            ``int` `p = it.first; ` ` `  `            ``while` `(arr[r] % p == 0) { ` `                ``current_map[p]++; ` `                ``cnts[r][p]++; ` `                ``arr[r] /= p; ` `            ``} ` `        ``} ` ` `  `        ``// Check if current sub-array's product ` `        ``// is divisible by k ` `        ``int` `flag = 0; ` `        ``for` `(``auto``& it : k_cnt) { ` `            ``int` `p = it.first; ` `            ``if` `(current_map[p] < k_cnt[p]) { ` `                ``flag = 1; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// If for all prime factors p of k, ` `        ``// current_map[p] >= k_cnt[p] ` `        ``// then current sub-array is divisible by k ` ` `  `        ``if` `(!flag) { ` ` `  `            ``// flag = 0 means that after adding rth element ` `            ``// segment's product is divisible by k ` `            ``ans += n - r; ` ` `  `            ``// Eliminate 'l' from the current segment ` `            ``for` `(``auto``& it : k_cnt) { ` `                ``int` `p = it.first; ` `                ``current_map[p] -= cnts[l][p]; ` `            ``} ` ` `  `            ``l++; ` `        ``} ` `        ``else` `{ ` `            ``r++; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 2, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 4; ` ` `  `    ``sieve(); ` ` `  `    ``cout << countSubarrays(arr, n, k); ` ` `  `    ``return` `0; ` `} `

Output:

```4
```

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