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

Last Updated : 12 Jul, 2021

Given an array arr[] of N integers and an integer 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) ` `{ `   `    ``// Initialize the minimum ` `    ``// subarray size to N + 1 ` `    ``int` `res = N + 1; `   `    ``// Generate all possible subarray ` `    ``for` `(``int` `i = 0; i < N; i++) { `   `        ``// Initialize 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)` `{` `    `  `    ``// Initialize the minimum` `    ``// subarray size to N + 1` `    ``int` `res = N + ``1``;`   `    ``// Generate all possible subarray` `    ``for``(``int` `i = ``0``; i < N; i++)` `    ``{` `        `  `        ``// Initialize 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): ` `    `  `    ``# Initialize the minimum ` `    ``# subarray size to N + 1 ` `    ``res ``=` `N ``+` `1` `    `  `    ``# Generate all possible subarray ` `    ``for` `i ``in` `range``(``0``, N): ` `        `  `        ``# Initialize 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)` `{` `    `  `    ``// Initialize the minimum` `    ``// subarray size to N + 1` `    ``int` `res = N + 1;`   `    ``// Generate all possible subarray` `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        `  `        ``// Initialize 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`

## Javascript

 ``

Output:

`2`

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

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