给定长度为N且数字为K的数组arr [] ，任务是计算数组中K个倒数的数量。
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，都要检查数组中是否依次存在所有的数字K，K-1，K-2，…(最多1个)。如果是，则将计数加1。如果下一个数字使它不按顺序排列，则将查找下一个出现的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 divyeshrabadiya07Python3
# 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_RaiC#
// 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输出：
2
时间复杂度： O(N)
辅助空间复杂度： O(1)