# Smallest subarray whose product leaves remainder K when divided by size of the array

Given an array arr[] of N integers and an inetger K, the task is to find the length of the smallest subarray whose product when divided by N gives remainder K. If no such subarray exists the print “-1”.
Examples:

Input: N = 3, arr = {2, 2, 6}, K = 1
Output:
Explanation:
All possible subarrays are:
{2} -> 2(mod 3) = 2
{2} -> 2(mod 3) = 2
{6} -> 6(mod 3) = 0
{2, 2} -> (2 * 2)(mod 3) = 1
{2, 6} -> (2 * 6)(mod 3) = 0
{2, 2, 6} -> (2 * 2 * 6)(mod 3) = 0
Only subarray {2, 2} leaves the remainder K( = 1).
Therefore, the length of the minimum subarray is 2.
Input: N = 4, arr = {2, 2, 3, 3}, K = 1
Output:
Explanation:
Only subarray {3, 3} satisfies the property, thus the length of the minimum subarray is 2.

Approach:
The idea is to generate all possible subarrays of the given array and print the length of the smallest subarray whose product of all element gives remainder K when divided by N.
Below is the implementation of the above approach:

## C++

 `// C++ Program to implement  ` `// the above approach  ` `#include   ` `using` `namespace` `std;  ` ` `  `// Function to find the subarray of  ` `// minimum length  ` `int` `findsubArray(``int` `arr[], ``int` `N, ``int` `K)  ` `{  ` ` `  `    ``// Initialise the minimum  ` `    ``// subarray size to N + 1  ` `    ``int` `res = N + 1;  ` ` `  `    ``// Generate all possible subarray  ` `    ``for` `(``int` `i = 0; i < N; i++) {  ` ` `  `        ``// Intialise the product  ` `        ``int` `curr_prod = 1;  ` ` `  `        ``for` `(``int` `j = i; j < N; j++) {  ` ` `  `            ``// Find the product  ` `            ``curr_prod = curr_prod * arr[j];  ` ` `  `            ``if` `(curr_prod % N == K  ` `                ``&& res > (j - i + 1)) {  ` ` `  `                ``res = min(res, j - i + 1);  ` `                ``break``;  ` `            ``}  ` `        ``}  ` `    ``}  ` ` `  `    ``// Return the minimum size of subarray  ` `    ``return` `(res == N + 1) ? 0 : res;  ` `}  ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` `    ``// Given array  ` `    ``int` `arr[] = { 2, 2, 3 };  ` ` `  `    ``int` `N = ``sizeof``(arr)  ` `            ``/ ``sizeof``(arr[0]);  ` ` `  `    ``int` `K = 1;  ` ` `  `    ``int` `answer = findsubArray(arr, N, K);  ` ` `  `    ``if` `(answer != 0)  ` `        ``cout << answer;  ` `    ``else` `        ``cout << ``"-1"``;  ` ` `  `    ``return` `0;  ` `}  `

## Java

 `// Java program to implement the ` `// above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to find the subarray of ` `// minimum length ` `static` `int` `findsubArray(``int` `arr[],  ` `                        ``int` `N, ``int` `K) ` `{ ` `     `  `    ``// Initialise the minimum ` `    ``// subarray size to N + 1 ` `    ``int` `res = N + ``1``; ` ` `  `    ``// Generate all possible subarray ` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `         `  `        ``// Initialise the product ` `        ``int` `curr_prod = ``1``; ` ` `  `        ``for``(``int` `j = i; j < N; j++)  ` `        ``{ ` `             `  `            ``// Find the product  ` `            ``curr_prod = curr_prod * arr[j];  ` `   `  `            ``if` `(curr_prod % N == K && ` `                 ``res > (j - i + ``1``)) ` `            ``{  ` `                ``res = Math.min(res, j - i + ``1``); ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``// Return the minimum size of subarray ` `    ``return` `(res == N + ``1``) ? ``0` `: res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `     `  `    ``// Given array ` `    ``int` `arr[] = { ``2``, ``2``, ``3` `}; ` `     `  `    ``int` `N = arr.length; ` `    ``int` `K = ``1``; ` `     `  `    ``int` `answer = findsubArray(arr, N, K); ` `     `  `    ``if` `(answer != ``0``) ` `        ``System.out.println(answer); ` `    ``else` `        ``System.out.println(``"-1"``); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 program to implement  ` `# the above approach  ` ` `  `# Function to find the subarray of  ` `# minimum length  ` `def` `findsubArray(arr, N, K):  ` `     `  `    ``# Initialise the minimum  ` `    ``# subarray size to N + 1  ` `    ``res ``=` `N ``+` `1` `     `  `    ``# Generate all possible subarray  ` `    ``for` `i ``in` `range``(``0``, N):  ` `         `  `        ``# Intialise the product  ` `        ``curr_prad ``=` `1` `         `  `        ``for` `j ``in` `range``(i, N):  ` `             `  `            ``# Find the product  ` `            ``curr_prad ``=` `curr_prad ``*` `arr[j]  ` ` `  `            ``if` `(curr_prad ``%` `N ``=``=` `K ``and` `                ``res > (j ``-` `i ``+` `1``)):  ` `                ``res ``=` `min``(res, j ``-` `i ``+` `1``)  ` `                ``break` `                 `  `    ``# Return the minimum size of subarray  ` `    ``if` `res ``=``=` `N ``+` `1``:  ` `        ``return` `0` `    ``else``:  ` `        ``return` `res  ` `     `  `# Driver code  ` `if` `__name__ ``=``=` `'__main__'``:  ` `     `  `    ``# Given array  ` `    ``arr ``=` `[ ``2``, ``2``, ``3` `]  ` `    ``N ``=` `len``(arr)  ` `    ``K ``=` `1` `     `  `    ``answer ``=` `findsubArray(arr, N, K)  ` `     `  `    ``if` `(answer !``=` `0``):  ` `        ``print``(answer)  ` `    ``else``:  ` `        ``print``(``-``1``)  ` `         `  `# This code is contributed by virusbuddah_  `

## C#

 `// C# program to implement the ` `// above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find the subarray of ` `// minimum length ` `static` `int` `findsubArray(``int` `[]arr,  ` `                        ``int` `N, ``int` `K) ` `{ ` `     `  `    ``// Initialise the minimum ` `    ``// subarray size to N + 1 ` `    ``int` `res = N + 1; ` ` `  `    ``// Generate all possible subarray ` `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{ ` `         `  `        ``// Initialise the product ` `        ``int` `curr_prod = 1; ` ` `  `        ``for``(``int` `j = i; j < N; j++)  ` `        ``{ ` `             `  `            ``// Find the product  ` `            ``curr_prod = curr_prod * arr[j];  ` ` `  `            ``if` `(curr_prod % N == K && ` `                ``res > (j - i + 1)) ` `            ``{  ` `                ``res = Math.Min(res, j - i + 1); ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``// Return the minimum size of subarray ` `    ``return` `(res == N + 1) ? 0 : res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `     `  `    ``// Given array ` `    ``int` `[]arr = { 2, 2, 3 }; ` `     `  `    ``int` `N = arr.Length; ` `    ``int` `K = 1; ` `     `  `    ``int` `answer = findsubArray(arr, N, K); ` `     `  `    ``if` `(answer != 0) ` `        ``Console.WriteLine(answer); ` `    ``else` `        ``Console.WriteLine(``"-1"``); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey `

Output:

```2
```

Time Complexity: O(N2
Auxiliary Space: O(1)

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