# Smallest subarray with product divisible by k

Given an array of non-negative numbers, find length of smallest subarray whose product is a multiple of k.

Examples :

```Input : arr[] = {1, 9, 16, 5, 4, 3, 2}
k = 720
Output: 3
The smallest subarray is {9, 16, 5} whose
product is 720.

Input : arr[] = {1, 2, 4, 5, 6}
K = 96
Output : No such subarray exists
```

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

The idea is simple. We traverse through all subarrays. For each subarray, we compute its product. If the product is multiple of k, we check if length of the subarray is smaller than current result.

## C++

 `// C++ program to find smallest subarray ` `// with product divisible by k. ` `#include ` `using` `namespace` `std; ` ` `  `// function to find the subarray of ` `// minimum length and end of sub array ` `int` `findsubArray(``int` `arr[], ``int` `k) ` `{ ` `    ``// find the length of array ` `    ``int` `n = 7; ` ` `  `    ``// try of every sub array whether it  ` `    ``// result in multiple of k or not if it ` `    ``// is store it in the result ` `    ``// and find for the minimum using ` `    ``// dynamic programming ` `    ``int` `res = n + 1; ` `     `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Find minimum length product  ` `        ``// beginning with arr[i].  ` `        ``int` `curr_prod = 1;  ` `         `  `        ``for` `(``int` `j = i; j < n; j++) {              ` `            ``curr_prod = curr_prod * arr[j];              ` `             `  `            ``if` `(curr_prod % k == 0 && res ` `                                ``> (j - i + 1))  ` `            ``{ ` `                ``res = min(res, j - i + 1);  ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``return` `(res == n + 1) ? 0 : res; ` `} ` ` `  `// driver Function ` `int` `main(``void``) ` `{ ` `    ``int` `array[] = { 1, 9, 16, 5, 4, 3, 2 }; ` `    ``int` `k = 720; ` `    ``int` `answer = findsubArray(array, k); ` `     `  `    ``if` `(answer != 0) ` `        ``cout<

## Java

 `// Java program to find smallest subarray ` `// with product divisible by k. ` `import` `java.util.*; ` ` `  `// function to find the subarray of ` `// minimum length and end of sub array ` `public` `class` `findSubArray { ` ` `  `    ``public` `static` `int` `findsubArray(``int` `arr[], ``int` `k) ` `    ``{ ` `        ``// find the length of array ` `        ``int` `n = arr.length; ` ` `  `        ``// try of every sub array whether it result ` `        ``// in multiple of k or not if it ` `        ``// is store it in the result ` `        ``// and find for the minimum using ` `        ``// dynamic programming ` `        ``int` `res = n+``1``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `            ``// Find minimum length product beginning ` `            ``// with arr[i].  ` `            ``int` `curr_prod = ``1``;  ` `            ``for` `(``int` `j = i; j < n; j++) {                 ` `                ``curr_prod = curr_prod * arr[j];                 ` `                ``if` `(curr_prod % k == ``0` `&& res > (j-i+``1``))  ` `                ``{ ` `                    ``res = Math.min(res, j-i+``1``);    ` `                    ``break``; ` `                ``} ` `            ``} ` `        ``} ` `        ``return` `(res == n+``1``)? ``0` `: res; ` `    ``} ` ` `  `    ``// driver Function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `array[] = { ``1``, ``9``, ``16``, ``5``, ``4``, ``3``, ``2` `}; ` `        ``int` `k = ``720``; ` `        ``int` `answer = findsubArray(array, k); ` `        ``if` `(answer != ``0``) ` `            ``System.out.println(answer); ` `        ``else` `            ``System.out.println(``"No Such subarray exists."``); ` `    ``} ` `} `

## Python3

 `# Python 3 program to find smallest  ` `# subarray with product divisible by k. ` ` `  ` `  `# function to find the subarray of ` `# minimum length and end of sub array ` `def` `findsubArray(arr, k) : ` `     `  `    ``# find the length of array ` `    ``n``=` `len``(arr) ` ` `  `    ``# try of every sub array whether it  ` `    ``# result in multiple of k or not if  ` `    ``# it is store it in the result ` `    ``# and find for the minimum using ` `    ``# dynamic programming ` `    ``res ``=` `n``+``1` `    ``for` `i ``in` `range``(``0``,n) : ` `         `  `        ``# Find minimum length product  ` `        ``# beginning with arr[i].  ` `        ``curr_prod ``=` `1` `        ``for` `j ``in` `range``( i, n): ` `            ``curr_prod ``=` `curr_prod ``*` `arr[j]          ` `            ``if` `(curr_prod ``%` `k ``=``=` `0` `and`  `                     ``res > (j ``-` `i ``+` `1``)) : ` `                ``res ``=` `min``(res, j ``-` `i ``+` `1``)  ` `                ``break` `             `  `             `  `    ``if``(res ``=``=` `n ``+` `1``) : ` `        ``return` `0` `    ``else` `:  ` `        ``return` `res ` `     `  ` `  `# driver Function ` `array ``=` `[``1``, ``9``, ``16``, ``5``, ``4``, ``3``, ``2` `] ` `k ``=` `720` `answer ``=` `findsubArray(array, k) ` `if` `(answer !``=` `0``): ` `    ``print``(answer) ` `else` `: ` `    ``print``(``"No Such subarray exists."``) ` `     `  `# This code is contributed by Nikita Tiwari. `

## C#

 `// C# program to find smallest subarray ` `// with product divisible by k. ` `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``// function to find the subarray of ` `    ``// minimum length and end of sub array ` `    ``public` `static` `int` `findsubArray(``int` `[]arr, ``int` `k) ` `    ``{ ` `        ``// find the length of array ` `        ``int` `n = arr.Length; ` ` `  `        ``// try of every sub array whether it result ` `        ``// in multiple of k or not if it ` `        ``// is store it in the result ` `        ``// and find for the minimum using ` `        ``// dynamic programming ` `        ``int` `res = n+1; ` `        ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `            ``// Find minimum length product beginning ` `            ``// with arr[i].  ` `            ``int` `curr_prod = 1;  ` `            ``for` `(``int` `j = i; j < n; j++) { ` `                 `  `                ``curr_prod = curr_prod * arr[j];     ` `                 `  `                ``if` `(curr_prod % k == 0 && res > (j-i+1))  ` `                ``{ ` `                    ``res = Math.Min(res, j-i+1);  ` `                    ``break``; ` `                ``} ` `            ``} ` `        ``} ` `         `  `        ``return` `(res == n+1) ? 0 : res; ` `    ``} ` ` `  `    ``// driver Function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]array = { 1, 9, 16, 5, 4, 3, 2 }; ` `        ``int` `k = 720; ` `        ``int` `answer = findsubArray(array, k); ` `        ``if` `(answer != 0) ` `            ``Console.WriteLine(answer); ` `        ``else` `            ``Console.WriteLine(``"No Such subarray"` `                                  ``+ ``" exists."``); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` (``\$j` `- ``\$i` `+ 1))  ` `            ``{ ` `                ``\$res` `= min(``\$res``, ``\$j` `- ``\$i` `+ 1);  ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``return` `(``\$res` `== ``\$n` `+ 1) ? 0 : ``\$res``; ` `} ` ` `  `    ``// Driver Code ` `    ``\$arr` `= ``array``(1, 9, 16, 5, 4, 3, 2); ` `    ``\$k` `= 720; ` `    ``\$answer` `= findsubArray(``\$arr``, ``\$k``); ` `     `  `    ``if` `(``\$answer` `!= 0) ` `        ``echo` `\$answer``.``"\n"``; ` `    ``else` `        ``echo` `"No Such subarray exists."``.``"\n"``; ` `         `  `// This code is contributed by Sam007 ` `?> `

Output:

```3
```

Time Complexity = O(n^2)

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