长度小于或等于K的非递减子数组的数量

给定N个元素的数组arr []和整数K ，任务是查找长度小于或等于K的非递减子数组的数量。
例子：

Input: arr[] = {1, 2, 3}, K = 2 
Output: 5 
{1}, {2}, {3}, {1, 2} and {2, 3} are the valid subarrays.
Input: arr[] = {3, 2, 1}, K = 1 
Output: 3 
 

天真的方法：一种简单的方法是生成所有长度小于或等于K的子数组，然后检查子数组是否满足条件。因此，该方法的时间复杂度将为O(N ³ ) 。
高效的方法：更好的方法是使用两指针技术。

• 对于任何索引i ，找到最大索引j ，以使子数组arr [i…j]不变。这可以通过简单地从i + 1开始增加j的值并检查arr [j]是否大于arr [j – 1]来实现。
• 假设在上一步中找到的子数组的长度为L。计算X = max(0，L – K)和(L *(L + 1))/ 2 –(X *(X + 1))/ 2将添加到最终答案中。这是因为对于长度为L的数组，长度为K的子数组的数目。
  • 从第一个元素= L – K = X开始的此类子数组的数量。
  • 从第二个元素开始的此类子数组的数量= L – K – 1 = X – 1 。
  • 从第三个元素= L – K – 2 = X – 2开始的子数组的数量。
  • 依此类推，直到0，即1 + 2 + 3 + .. + X =(X *(X + 1))/ 2 。如果从总递增子数组中减去此值，则结果将是长度小于或等于K的递增子数组的计数

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the required count
int findCnt(int* arr, int n, int k)
{
    // To store the final result
    int ret = 0;
 
    // Two pointer loop
    int i = 0;
    while (i < n) {
 
        // Initialising j
        int j = i + 1;
 
        // Looping till the subarray increases
        while (j < n and arr[j] >= arr[j - 1])
            j++;
        int x = max(0, j - i - k);
 
        // Update ret
        ret += ((j - i) * (j - i + 1)) / 2 - (x * (x + 1)) / 2;
 
        // Update i
        i = j;
    }
 
    // Return ret
    return ret;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(int);
    int k = 2;
 
    cout << findCnt(arr, n, k);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the required count
static int findCnt(int[] arr, int n, int k)
{
    // To store the final result
    int ret = 0;
 
    // Two pointer loop
    int i = 0;
    while (i < n)
    {
 
        // Initialising j
        int j = i + 1;
 
        // Looping till the subarray increases
        while (j < n && arr[j] >= arr[j - 1])
            j++;
        int x = Math.max(0, j - i - k);
 
        // Update ret
        ret += ((j - i) * (j - i + 1)) / 2 -
                          (x * (x + 1)) / 2;
 
        // Update i
        i = j;
    }
 
    // Return ret
    return ret;
}
 
// Driver code
public static void main(String []args)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
    int k = 2;
 
    System.out.println(findCnt(arr, n, k));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the required count
def findCnt(arr, n, k) :
 
    # To store the final result
    ret = 0;
 
    # Two pointer loop
    i = 0;
    while (i < n) :
 
        # Initialising j
        j = i + 1;
 
        # Looping till the subarray increases
        while (j < n and arr[j] >= arr[j - 1]) :
            j += 1;
             
        x = max(0, j - i - k);
 
        # Update ret
        ret += ((j - i) * (j - i + 1)) // 2 - \
                          (x * (x + 1)) / 2;
 
        # Update i
        i = j;
 
    # Return ret
    return ret;
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 2, 3 ];
    n = len(arr);
    k = 2;
 
    print(findCnt(arr, n, k));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
                     
class GFG
{
 
// Function to return the required count
static int findCnt(int[] arr, int n, int k)
{
    // To store the final result
    int ret = 0;
 
    // Two pointer loop
    int i = 0;
    while (i < n)
    {
 
        // Initialising j
        int j = i + 1;
 
        // Looping till the subarray increases
        while (j < n && arr[j] >= arr[j - 1])
            j++;
        int x = Math.Max(0, j - i - k);
 
        // Update ret
        ret += ((j - i) * (j - i + 1)) / 2 -
                        (x * (x + 1)) / 2;
 
        // Update i
        i = j;
    }
 
    // Return ret
    return ret;
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
    int k = 2;
 
    Console.WriteLine(findCnt(arr, n, k));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript implementation of the approach
 
// Function to return the required count
function findCnt(arr, n, k)
{
    // To store the final result
    var ret = 0;
 
    // Two pointer loop
    var i = 0;
    while (i < n) {
 
        // Initialising j
        var j = i + 1;
 
        // Looping till the subarray increases
        while (j < n && arr[j] >= arr[j - 1])
            j++;
        var x = Math.max(0, j - i - k);
 
        // Update ret
        ret += ((j - i) * (j - i + 1)) / 2 - (x * (x + 1)) / 2;
 
        // Update i
        i = j;
    }
 
    // Return ret
    return ret;
}
 
// Driver code
var arr = [1, 2, 3 ];
var n = arr.length;
var k = 2;
document.write( findCnt(arr, n, k));
 

输出：

5

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