数组中的 K 倒计时计数

给定一个长度为N的数组arr[]和一个数字K ，任务是计算数组中 K 倒计时的数量。

A contiguous subarray is said to be a K-countdown if it is of length K and contains the integers K, K-1, K-2, …, 2, 1 in that order. For example, [4, 3, 2, 1] is 4-countdown and [6, 5, 4, 3, 2, 1] is a 6-countdown.

例子：

Input: K = 2, arr[] = {3 2 1 2 2 1} 
Output: 2 
Explanation: Here, K=2 so the array has 2 2-Countdowns(2, 1). One countdown is from index 1 to 2 and the other is from index 4 to 5.
Input: K = 3, arr[] = {4 3 2 1 5 3 2 1} 
Output: 2 
Explanation: Here, K=3 so the array has 2 3-Countdowns(3, 2, 1) 
 

方法：遍历给定的数组，每次遇到数字 K 时，检查是否所有数字 K、K-1、K-2、……直到 1 都按顺序出现在数组中。如果是，则计数加 1。如果下一个数字不按顺序排列，则查找下一次出现的 K。
下面是上述方法的实现：

C++

// C++ code for the above program.
 
#include 
using namespace std;
 
// Function to to count the
// number of K-countdowns for
// multiple queries
int countKCountdown(int arr[],
                    int N,
                    int K)
{
 
    // flag which stores the
    // current value of value
    // in the countdown
    int flag = -1;
 
    // count of K-countdowns
    int count = 0;
 
    // Loop to iterate over the
    // elements of the array
    for (int i = 0; i < N; i++) {
 
        // condition check if
        // the elements
        // of the array is
        // equal to K
        if (arr[i] == K)
            flag = K;
 
        // condition check if
        // the elements
        // of the array is in
        // continuous order
        if (arr[i] == flag)
            flag--;
 
        // condition check if
        // the elements
        // of the array are not
        // in continuous order
        else
            flag = -1;
 
        // condition check to
        // increment the counter
        // if the there is a
        // K-countdown present
        // in the array
        if (flag == 0)
            count++;
    }
 
    // returning the count of
    // K-countdowns
    return count;
}
 
// Driver Code
int main()
{
    int N = 8;
    int K = 3;
    int arr[N] = { 4, 3, 2, 1,
                   5, 3, 2, 1 };
 
    // Function Call
    cout << countKCountdown(arr, N, K);
}

Java

// Java code for the above program.
class GFG{
     
// Function to to count the
// number of K-countdowns for
// multiple queries
public static int countKCountdown(int arr[],
                                  int N, int K)
{
     
    // Flag which stores the
    // current value of value
    // in the countdown
    int flag = -1;
     
    // Count of K-countdowns
    int count = 0;
     
    // Loop to iterate over the
    // elements of the array
    for(int i = 0; i < N; i++)
    {
        
       // Condition check if the
       // elements of the array is
       // equal to K
       if (arr[i] == K)
           flag = K;
            
       // Condition check if the
       // elements of the array is
       // in continuous order
       if (arr[i] == flag)
           flag--;
        
       // Condition check if the
       // elements of the array are
       // not in continuous order
       else
           flag = -1;
        
       // Condition check to increment
       // the counter if the there is a
       // K-countdown present in the array
       if (flag == 0)
           count++;
    }
     
    // Returning the count of
    // K-countdowns
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 8;
    int K = 3;
    int arr[] = { 4, 3, 2, 1, 5, 3, 2, 1 };
     
    System.out.print(countKCountdown(arr, N, K));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 code for the above program.
 
# Function to to count the
# number of K-countdowns for
# multiple queries
def countKCountdown(arr, N, K):
 
    # flag which stores the
    # current value of value
    # in the countdown
    flag = -1;
 
    # count of K-countdowns
    count = 0;
 
    # Loop to iterate over the
    # elements of the array
    for i in range(0, N):
 
        # condition check if
        # the elements
        # of the array is
        # equal to K
        if (arr[i] == K):
            flag = K;
 
        # condition check if
        # the elements
        # of the array is in
        # continuous order
        if (arr[i] == flag):
            flag -= 1;
 
        # condition check if
        # the elements
        # of the array are not
        # in continuous order
        else:
            flag = -1;
 
        # condition check to
        # increment the counter
        # if the there is a
        # K-countdown present
        # in the array
        if (flag == 0):
            count += 1;
     
    # returning the count of
    # K-countdowns
    return count;
 
# Driver Code
N = 8;
K = 3;
arr = [ 4, 3, 2, 1,
        5, 3, 2, 1 ];
 
# Function Call
print(countKCountdown(arr, N, K))
 
# This code is contributed by Akanksha_Rai

C#

// C# code for the above program.
using System;
class GFG{
     
// Function to to count the
// number of K-countdowns for
// multiple queries
public static int countKCountdown(int []arr,
                                  int N, int K)
{
     
    // Flag which stores the
    // current value of value
    // in the countdown
    int flag = -1;
     
    // Count of K-countdowns
    int count = 0;
     
    // Loop to iterate over the
    // elements of the array
    for(int i = 0; i < N; i++)
    {
         
    // Condition check if the
    // elements of the array is
    // equal to K
    if (arr[i] == K)
        flag = K;
             
    // Condition check if the
    // elements of the array is
    // in continuous order
    if (arr[i] == flag)
        flag--;
         
    // Condition check if the
    // elements of the array are
    // not in continuous order
    else
        flag = -1;
         
    // Condition check to increment
    // the counter if the there is a
    // K-countdown present in the array
    if (flag == 0)
        count++;
    }
     
    // Returning the count of
    // K-countdowns
    return count;
}
 
// Driver code
public static void Main()
{
    int N = 8;
    int K = 3;
    int []arr = { 4, 3, 2, 1, 5, 3, 2, 1 };
     
    Console.Write(countKCountdown(arr, N, K));
}
}
 
// This code is contributed by Akanksha_Rai

Javascript

//Javascript code for the above program.
 
// Function to to count the
// number of K-countdowns for
// multiple queries
function countKCountdown( arr, N, K)
{
 
    // flag which stores the
    // current value of value
    // in the countdown
    var flag = -1;
 
    // count of K-countdowns
    var count = 0;
 
    // Loop to iterate over the
    // elements of the array
    for (var i = 0; i < N; i++) {
 
        // condition check if
        // the elements
        // of the array is
        // equal to K
        if (arr[i] == K)
            flag = K;
 
        // condition check if
        // the elements
        // of the array is in
        // continuous order
        if (arr[i] == flag)
            flag--;
 
        // condition check if
        // the elements
        // of the array are not
        // in continuous order
        else
            flag = -1;
 
        // condition check to
        // increment the counter
        // if the there is a
        // K-countdown present
        // in the array
        if (flag == 0)
            count++;
    }
 
    // returning the count of
    // K-countdowns
    return count;
}
 
var N = 8;
var K = 3;
var arr = [ 4, 3, 2, 1, 5, 3, 2, 1 ];
// Function Call
document.write( countKCountdown(arr, N, K));
 
// This code is contributed by SoumikMondal

输出：

2

时间复杂度： O(N)
辅助空间复杂度： O(1)

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