大小为K的子数组中的山谷元素的最大数量

给定数组arr [] ，任务是选择大小为K的子数组，该子数组包含相对于相邻元素的最大谷点数。

如果一个元素arr [i]的两个相邻元素都大于该谷点，则称为谷点，即$arr[i-1] > arr[i]$和$arr[i] < arr[i+1]$ 。

例子：

Input: arr[] = {5, 4, 6, 4, 5, 2, 3, 1}, K = 7
Output: 3
Explanation:
In subarray arr[0-6] = {5, 4, 6, 4, 5, 2, 3}
There are 3 Valley points in the subarray which is maximum.

Input: arr[] = {2, 1, 4, 2, 3, 4, 1, 2}, K = 4
Output: 1
Explanation:
In subarray arr[0-3] = {2, 1, 4, 2}
There is only one valley point in the subarray which is maximum.

方法：想法是使用滑动窗口技术来解决此问题。

下面是该方法步骤的说明：

• 在大小为K的第一个子数组中找到谷点的总数。
• 对可能的子数组的所有起点(即数组的NK点)进行迭代，并应用包含和排除原理来计算当前窗口中的谷点数。
• 在每个步骤中，更新最终答案以计算每个子数组的全局最大值。

下面是上述方法的实现：

C++

// C++ implementation to find the 
// maximum number of valley elements
// in the subarrays of size K
  
#include
using namespace std;
  
// Function to find the valley elements
// in the array which contains 
// in the subarrays of the size K
void minpoint(int arr[],int n, int k)
{
    int min_point = 0;
    for (int i = 1; i < k-1 ; i++)
    {
        // Increment min_point
        // if element at index i 
        // is smaller than element
        // at index i + 1 and i-1
        if(arr[i] < arr[i - 1] && arr[i] < arr[i + 1])
            min_point += 1;
    }
    // final_point to maintain maximum
    // of min points of subarray
    int final_point = min_point;
      
    // Iterate over array
    // from kth element
    for(int i = k ; i < n; i++)
    {
        // Leftmost element of subarray
        if(arr[i - ( k - 1 )] < arr[i - ( k - 1 ) + 1]&&
        arr[i - ( k - 1 )] < arr[i - ( k - 1 ) - 1])
            min_point -= 1;
          
        // Rightmost element of subarray
        if(arr[i - 1] < arr[i] && arr[i - 1] < arr[i - 2])
            min_point += 1;
          
        // if new subarray have greater
        // number of min points than previous 
        // subarray, then final_point is modified
        if(min_point > final_point)
            final_point = min_point;
    }
      
    // Max minimum points in 
    // subarray of size k
    cout<<(final_point);
}
  
// Driver Code
int main()
{
    int arr[] = {2, 1, 4, 2, 3, 4, 1, 2};
    int n = sizeof(arr)/sizeof(arr[0]);
    int k = 4;
    minpoint(arr, n, k);
    return 0; 
}
// This code contributed by chitranayal

Java

// Java implementation to find the 
// maximum number of valley elements 
// in the subarrays of size K 
class GFG{
      
// Function to find the valley elements 
// in the array which contains 
// in the subarrays of the size K 
static void minpoint(int arr[], int n, int k) 
{ 
    int min_point = 0; 
    for(int i = 1; i < k - 1; i++) 
    { 
         
       // Increment min_point 
       // if element at index i 
       // is smaller than element 
       // at index i + 1 and i-1 
       if(arr[i] < arr[i - 1] && 
          arr[i] < arr[i + 1]) 
          min_point += 1; 
    } 
      
    // final_point to maintain maximum 
    // of min points of subarray 
    int final_point = min_point; 
          
    // Iterate over array 
    // from kth element 
    for(int i = k ; i < n; i++) 
    { 
         
       // Leftmost element of subarray 
       if(arr[i - ( k - 1 )] < arr[i - ( k - 1 ) + 1] && 
          arr[i - ( k - 1 )] < arr[i - ( k - 1 ) - 1]) 
          min_point -= 1; 
            
       // Rightmost element of subarray 
       if(arr[i - 1] < arr[i] && 
          arr[i - 1] < arr[i - 2]) 
          min_point += 1; 
            
       // If new subarray have greater 
       // number of min points than previous 
       // subarray, then final_point is modified 
       if(min_point > final_point) 
          final_point = min_point; 
    } 
      
    // Max minimum points in 
    // subarray of size k 
    System.out.println(final_point); 
} 
      
// Driver Code 
public static void main (String[] args)
{ 
    int arr[] = { 2, 1, 4, 2, 3, 4, 1, 2 }; 
    int n = arr.length; 
    int k = 4; 
      
    minpoint(arr, n, k); 
} 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation to find the 
# maximum number of valley elements
# in the subarrays of size K
  
# Function to find the valley elements
# in the array which contains 
# in the subarrays of the size K
def minpoint(arr, n, k):
    min_point = 0
    for i in range(1, k-1):
          
        # Increment min_point
        # if element at index i 
        # is smaller than element
        # at index i + 1 and i-1
        if(arr[i] < arr[i - 1] and arr[i] < arr[i + 1]):
            min_point += 1
  
    # final_point to maintain maximum
    # of min points of subarray
    final_point = min_point
      
    # Iterate over array
    # from kth element
    for i in range(k, n):
          
        # Leftmost element of subarray
        if(arr[i - ( k - 1 )] < arr[i - ( k - 1 ) + 1] and\
           arr[i - ( k - 1 )] < arr[i - ( k - 1 ) - 1]):
            min_point -= 1
          
        # Rightmost element of subarray
        if(arr[i - 1] < arr[i] and arr[i - 1] < arr[i - 2]):
            min_point += 1
          
        # if new subarray have greater
        # number of min points than previous 
        # subarray, then final_point is modified
        if(min_point > final_point):
            final_point = min_point
      
    # Max minimum points in 
    # subarray of size k
    print(final_point)
  
# Driver Code
if __name__ == "__main__":
    arr = [2, 1, 4, 2, 3, 4, 1, 2]
    n = len(arr)
    k = 4
    minpoint(arr, n, k)

C#

// C# implementation to find the 
// maximum number of valley elements 
// in the subarrays of size K 
using System;
  
class GFG{
      
// Function to find the valley elements 
// in the array which contains 
// in the subarrays of the size K 
static void minpoint(int []arr, int n, int k) 
{ 
    int min_point = 0; 
    for(int i = 1; i < k - 1; i++) 
    { 
  
       // Increment min_point 
       // if element at index i 
       // is smaller than element 
       // at index i + 1 and i-1 
       if(arr[i] < arr[i - 1] && 
          arr[i] < arr[i + 1]) 
          min_point += 1; 
    } 
          
    // final_point to maintain maximum 
    // of min points of subarray 
    int final_point = min_point; 
              
    // Iterate over array 
    // from kth element 
    for(int i = k ; i < n; i++) 
    { 
         
       // Leftmost element of subarray 
       if(arr[i - ( k - 1 )] < arr[i - ( k - 1 ) + 1] && 
          arr[i - ( k - 1 )] < arr[i - ( k - 1 ) - 1]) 
          min_point -= 1; 
         
       // Rightmost element of subarray 
       if(arr[i - 1] < arr[i] && 
          arr[i - 1] < arr[i - 2]) 
          min_point += 1; 
              
       // If new subarray have greater 
       // number of min points than previous 
       // subarray, then final_point is modified 
       if(min_point > final_point) 
          final_point = min_point; 
    } 
          
    // Max minimum points in 
    // subarray of size k 
    Console.WriteLine(final_point); 
} 
          
// Driver Code 
public static void Main (string[] args)
{ 
    int []arr = { 2, 1, 4, 2, 3, 4, 1, 2 }; 
    int n = arr.Length; 
    int k = 4; 
          
    minpoint(arr, n, k); 
} 
}
  
// This code is contributed by AnkitRai01

输出：

1

时间复杂度： O(N)