最小的子数组，其乘积除以数组的大小时留下余数 K

给定一个由N 个整数组成的数组arr[]和一个整数K ，任务是找到最小子数组的长度，其乘积除以N给出余数K 。如果不存在这样的子数组，则打印“-1” 。

例子：

Input: N = 3, arr = {2, 2, 6}, K = 1 
Output: 2 
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: 2 
Explanation: 
Only subarray {3, 3} satisfies the property, thus the length of the minimum subarray is 2. 

方法：
这个想法是生成给定数组的所有可能子数组，并打印最小子数组的长度，其所有元素的乘积在除以N时给出余数K 。

下面是上述方法的实现：

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++) {
 
        // Intialize 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):
         
        # Intialize 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

 
// Javascript program to implement the
// above approach
 
// Function to find the subarray of
// minimum length
function findsubArray(arr, N, K)
{
     
    // Initialize the minimum
    // subarray size to N + 1
    var res = N + 1;
 
    // Generate all possible subarray
    for(i = 0; i < N; i++)
    {
         
        // Initialize the product
        var curr_prod = 1;
 
        for(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
 
// Given array
var arr = [ 2, 2, 3 ];
 
var N = arr.length;
var K = 1;
 
var answer = findsubArray(arr, N, K);
 
if (answer != 0)
    document.write(answer);
else
    document.write("-1");
 
// This code is contributed by umadevi9616
 

输出：

2

时间复杂度： O(N ² )
辅助空间： O(1)

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