# 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:

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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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