乘积等于2的幂的最长子数组的长度

给定一个由N个正整数组成的数组arr [] ，任务是找到最长子数组的长度，该子数组的元素乘积等于2的幂次幂。

例子：

Input: arr[] = {2, 5, 4, 4, 6}
Output: 2
Explanation:
The subarray having maximum length whose product is perfect power of 2 is {4, 4}.
Product = 4 * 4 = 16, which is a perfect power of 2.

Input: arr[] = {2, 5, 4, 6, 8, 8, 2}
Output: 3
Explanation:
The subarray having maximum length whose product is perfect power of 2 is {8, 8, 2}.
Product = 8 * 8 * 2 = 128, which is a perfect power of 2.

天真的方法：最简单的方法是生成给定数组的所有可能的子数组，并检查任何子数组的元素乘积是否为2的完美幂。如果发现为真，则将最大长度更新为此子数组的长度。最后，打印检查所有可能的子数组后获得的子数组的最大长度。

时间复杂度： O(N ³ )
辅助空间： O(1)

高效方法：为了优化上述方法，该思想基于以下事实：对于任何数为2的幂，它也可以表示为具有2的完美幂的数的乘积。因此，该思想是使用用Kadane算法找到所有元素的完美幂为2的子数组的最大长度。步骤如下：

• 将maxLength和ans初始化为0 ，这将分别存储子数组的最大长度和结果得到的子数组的最大长度。
• 遍历给定数组并执行以下操作：
  • 如果当前元素是2的理想幂，则将maxLength增加1 。否则将maxLength更新为0 。
  • 更新ANS上述步骤后的最大ANS和最大长度的。
• 完成上述步骤后，将ans的值打印为最大长度。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check whether a number
// is power of 2 or not
bool isPower(int x)
{
    return (x && (!(x & (x - 1))));
}
 
// Function to find maximum length
// subarray having product of element
// as a perfect power of 2
int maximumlength(int arr[], int N)
{
    // Stores current subarray length
    int max_length = 0;
 
    // Stores maximum subarray length
    int max_len_subarray = 0;
 
    // Traverse the given array
    for (int i = 0; i < N; i++) {
 
        // If arr[i] is power of 2
        if (isPower(arr[i]) == 1) {
 
            // Increment max_length
            max_length++;
 
            // Update max_len_subarray
            max_len_subarray
                = max(max_length,
                      max_len_subarray);
        }
 
        // Otherwise
        else {
            max_length = 0;
        }
    }
 
    // Print the maximum length
    cout << max_len_subarray;
}
 
// Driver Code
int main()
{
    // Given arr[]
    int arr[] = { 2, 5, 4, 6, 8, 8, 2 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maximumlength(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to check whether a number
// is power of 2 or not
public static boolean isPower(int x)
{
    if (x != 0 && (x & (x - 1)) == 0)
        return true;
         
    return false;
}
 
// Function to find maximum length
// subarray having product of element
// as a perfect power of 2
public static void maximumlength(int arr[],
                                 int N)
{
     
    // Stores current subarray length
    int max_length = 0;
 
    // Stores maximum subarray length
    int max_len_subarray = 0;
 
    // Traverse the given array
    for(int i = 0; i < N; i++)
    {
         
        // If arr[i] is power of 2
        if (isPower(arr[i]))
        {
             
            // Increment max_length
            max_length++;
 
            // Update max_len_subarray
            max_len_subarray = Math.max(max_length,
                                        max_len_subarray);
        }
 
        // Otherwise
        else
        {
            max_length = 0;
        }
    }
 
    // Print the maximum length
    System.out.println(max_len_subarray);
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given arr[]
    int arr[] = { 2, 5, 4, 6, 8, 8, 2 };
 
    int N = arr.length;
 
    // Function Call
    maximumlength(arr, N);
}
}
 
// This code is contributed by hemanthswarna1506

Python3

# Python3 program for the
# above approach
 
# Function to check whether
# a number is power of 2
# or not
def isPower(x):
   
    if (x != 0 and
       (x & (x - 1)) == 0):
        return True;
 
    return False;
 
# Function to find maximum
# length subarray having
# product of element
# as a perfect power of 2
def maximumlength(arr, N):
   
    # Stores current
    # subarray length
    max_length = 0;
 
    # Stores maximum
    # subarray length
    max_len_subarray = 0;
 
    # Traverse the given array
    for i in range(N):
 
        # If arr[i] is power of 2
        if (isPower(arr[i])):
 
            # Increment max_length
            max_length += 1;
 
            # Update max_len_subarray
            max_len_subarray = max(max_length,
                                   max_len_subarray);
 
        # Otherwise
        else:
            max_length = 0;
 
    # Prthe maximum length
    print(max_len_subarray);
 
# Driver Code
if __name__ == '__main__':
   
    # Given arr
    arr = [2, 5, 4,
           6, 8, 8, 2];
 
    N = len(arr);
 
    # Function Call
    maximumlength(arr, N);
 
# This code is contributed by gauravrajput1

C#

// C# program for the above approach
using System;
  
class GFG{
      
// Function to check whether a number
// is power of 2 or not
public static bool isPower(int x)
{
    if (x != 0 && (x & (x - 1)) == 0)
        return true;
          
    return false;
}
  
// Function to find maximum length
// subarray having product of element
// as a perfect power of 2
public static void maximumlength(int[] arr,
                                 int N)
{
     
    // Stores current subarray length
    int max_length = 0;
  
    // Stores maximum subarray length
    int max_len_subarray = 0;
  
    // Traverse the given array
    for(int i = 0; i < N; i++)
    {
         
        // If arr[i] is power of 2
        if (isPower(arr[i]))
        {
             
            // Increment max_length
            max_length++;
  
            // Update max_len_subarray
            max_len_subarray = Math.Max(max_length,
                                        max_len_subarray);
        }
  
        // Otherwise
        else
        {
            max_length = 0;
        }
    }
  
    // Print the maximum length
    Console.WriteLine(max_len_subarray);
}
  
// Driver Code
public static void Main()
{
     
    // Given arr[]
    int[] arr = { 2, 5, 4, 6, 8, 8, 2 };
  
    int N = arr.Length;
  
    // Function Call
    maximumlength(arr, N);
}
}
 
// This code is contributed by sanjoy_62

输出

3

时间复杂度： O(N)
辅助空间： O(1)