计算具有严格递减连续元素的子数组

给定一个包含整数的数组arr[] 。任务是找到差为1的递减子数组的数量。

例子：

Input: arr[] = {3, 2, 1, 4}
Output: 7
Explanation: Following are the possible decreasing subarrays with difference 1. 
[3], [2], [1], [4], [3,2], [2,1], and [3,2,1]
Therefore, the answer is 7. 

Input: arr[] = {5, 4, 3, 2, 1, 6}
Output: 16

朴素的方法：这个问题可以通过使用动态规划来解决。请按照以下步骤解决给定的问题。

1. 对于每个索引 i 的任务是计算以i结尾的子数组的数量，它遵循这种模式arr[i-2]==arr[i-1]+1 , arr[i-1]==arr[i]+1 .
2. 初始化一个变量ans = 0 ，以存储差值为1的递减子数组的数量。
3. 我们可以创建一个dp[]数组来存储每个索引的这些连续元素的计数。
4. dp[i] 是以i结尾且遵循此模式的子数组的数量。
5. 遍历dp[]并将 ans 中的每个值相加。
6. 返回ans作为最终结果。

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to count number of
// decreasing subarrays with difference 1
long long getcount(vector& p)
{
    int size = p.size(), cnt = 0;
    long long ans = 0;
    vector dp(size, cnt);
    for (int i = 0; i < size; i++) {
        if (i == 0)
            cnt = 1;
        else if (p[i] + 1 == p[i - 1])
            cnt++;
        else
            cnt = 1;
        dp[i] = cnt;
    }
    for (int i = 0; i < size; i++)
        ans += dp[i];
    return ans;
}
 
// Driver Code
int main()
{
    vector arr{ 5, 4, 3, 2, 1, 6 };
 
    // Function Call
    cout << getcount(arr);
 
    return 0;
}

Java

// Java code to implement the above approach
import java.util.*;
public class GFG
{
 
  // Function to count number of
  // decreasing subarrays with difference 1
  static long getcount(int p[])
  {
    int size = p.length, cnt = 0;
    long ans = 0;
 
    int dp[] = new int[size];
    for(int i = 0; i < size; i++) {
      dp[i] = cnt;
    }
 
    for (int i = 0; i < size; i++) {
      if (i == 0)
        cnt = 1;
      else if (p[i] + 1 == p[i - 1])
        cnt++;
      else
        cnt = 1;
      dp[i] = cnt;
    }
    for (int i = 0; i < size; i++)
      ans += dp[i];
    return ans;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int arr[] = { 5, 4, 3, 2, 1, 6 };
 
    // Function Call
    System.out.println(getcount(arr));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code to implement the above approach
 
# Function to count number of
# decreasing subarrays with difference 1
def getcount(p):
    size = len(p)
    cnt = 0
    ans = 0
    dp = [cnt for i in range(size)]
    for i in range(size):
        if (i == 0):
            cnt = 1
        elif (p[i] + 1 == p[i - 1]):
            cnt += 1
        else:
            cnt = 1
        dp[i] = cnt
         
    for i in range(size):
        ans += dp[i]
    return ans
 
# Driver Code
arr = [5, 4, 3, 2, 1, 6]
 
# Function Call
print(getcount(arr))
 
# This code is contributed by Shubham Singh

C#

// C# code to implement the above approach
using System;
class GFG
{
   
  // Function to count number of
  // decreasing subarrays with difference 1
  static long getcount(int []p)
  {
    int size = p.Length, cnt = 0;
    long ans = 0;
 
    int []dp = new int[size];
    for(int i = 0; i < size; i++) {
      dp[i] = cnt;
    }
 
    for (int i = 0; i < size; i++) {
      if (i == 0)
        cnt = 1;
      else if (p[i] + 1 == p[i - 1])
        cnt++;
      else
        cnt = 1;
      dp[i] = cnt;
    }
    for (int i = 0; i < size; i++)
      ans += dp[i];
    return ans;
  }
 
  // Driver code
  public static void Main()
  {
    int []arr = { 5, 4, 3, 2, 1, 6 };
 
    // Function Call
    Console.Write(getcount(arr));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

 
// JavaScript program for above approach
 
// Function to count number of decreasing
// subarrays with difference 1
function getcount(p)
{
    let size = p.length, cnt = 0;
    let ans = 0;
    let dp = new Array(size).fill(cnt);
     
    for(let i = 0; i < size; i++)
    {
        if (i == 0)
            cnt = 1;
        else if (p[i] + 1 == p[i - 1])
            cnt++;
        else
            cnt = 1;
             
        dp[i] = cnt;
    }
    for(let i = 0; i < size; i++)
        ans += dp[i];
         
    return ans;
}
 
// Driver Code
let arr = [ 5, 4, 3, 2, 1, 6 ];
 
// Function Call
document.write(getcount(arr));
 
// This code is contributed by Potta Lokesh
 

输出

16

时间复杂度： O(N)
辅助空间： O(N)

高效方法：在上述方法中，辅助空间复杂度可以通过用变量替换dp[]数组来进一步优化到恒定空间，以跟踪当前子数组的数量。请按照以下步骤解决给定的问题。

• 初始化一个变量说count = 0 。
• 当arr[i]-arr[i-1 ]==1开始遍历数组，生成一个递减1的数字链，然后count++ 。
• 将计数添加到 ans。
• 当链中断时， arr[i]-arr[i-1] !=1然后重置计数。
• 返回ans作为最终结果。

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to count the number of
// decreasing subarrays with difference 1
long long getcount(vector& arr)
{
    long long int ans = arr.size();
    long long int count = 0;
    for (int i = 1; i < arr.size(); i++) {
        if (arr[i - 1] - arr[i] == 1)
            count++;
        else
            count = 0;
        ans = ans + count;
    }
    return ans;
}
 
// Driver Code
int main()
{
    vector arr{ 5, 4, 3, 2, 1, 6 };
 
    // Function Call
    cout << getcount(arr);
 
    return 0;
}

Java

// Java program for above approach
class GFG
{
 
  // Function to count the number of
  // decreasing subarrays with difference 1
  static long getcount(int[] arr)
  {
    int ans = arr.length;
    int count = 0;
    for (int i = 1; i < arr.length; i++) {
      if (arr[i - 1] - arr[i] == 1)
        count++;
      else
        count = 0;
      ans = ans + count;
    }
    return ans;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int[] arr = { 5, 4, 3, 2, 1, 6 };
 
    // Function Call
    System.out.print(getcount(arr));
 
  }
}
 
// This code is contributed by 29AjayKumar

Python3

#Python program for the above approach
 
# Function to count the number of
# decreasing subarrays with difference 1
def getcount(arr):
    ans = len(arr)
    count = 0
    for i in range(1, len(arr)):
        if (arr[i - 1] - arr[i] == 1):
            count+=1
        else:
            count = 0
        ans = ans + count
         
    return ans
 
  # Driver Code
arr = [ 5, 4, 3, 2, 1, 6 ]
 
# Function Call
print(getcount(arr))
 
# This code is contributed by Shubham Singh

C＃

// C# program for above approach
using System;
 
public class GFG
{
 
  // Function to count the number of
  // decreasing subarrays with difference 1
  static long getcount(int[] arr)
  {
    int ans = arr.Length;
    int count = 0;
    for (int i = 1; i < arr.Length; i++) {
      if (arr[i - 1] - arr[i] == 1)
        count++;
      else
        count = 0;
      ans = ans + count;
    }
    return ans;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int[] arr = { 5, 4, 3, 2, 1, 6 };
 
    // Function Call
    Console.Write(getcount(arr));
 
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript program for above approach
 
// Function to count the number of
  // decreasing subarrays with difference 1
  function getcount(arr)
  {
    var ans = arr.length;
    var count = 0;
    for (var i = 1; i < arr.length; i++) {
      if (arr[i - 1] - arr[i] == 1)
        count++;
      else
        count = 0;
      ans = ans + count;
    }
    return ans;
  }
 
  // Driver Code
var arr = [ 5, 4, 3, 2, 1, 6 ];
 
// Function Call
document.write(getcount(arr));
 
// This code is contributed by 29AjayKumar

输出

16

时间复杂度： O(N)
辅助空间： O(1)