子序列数，使其具有一个连续的元素，其差值小于或等于1

给定N个元素的数组arr [] 。任务是找到具有至少两个连续元素的子序列的数量，以使它们之间的绝对差≤1 。

例子：

Input: arr[] = {1, 6, 2, 1}
Output: 6
{1, 2}, {1, 2, 1}, {2, 1}, {6, 2, 1}, {1, 1} and {1, 6, 2, 1}
are the sub-sequences that have at least one consecutive pair
with difference less than or equal to 1.

Input: arr[] = {1, 6, 2, 1, 9}
Output: 12

幼稚的方法：想法是找到所有可能的子序列，并检查是否存在相差≤1的任何连续对的子序列，并增加计数。

高效的方法：想法是遍历给定的数组，对于每个ith元素，尝试找到所需的以ith元素作为最后一个元素结尾的子序列。
对于每个i，我们要使用arr [i]，arr [i] -1，arr [i] + 1，因此我们将定义2D数组dp [] [] ，其中dp [i] [0]将包含不具有相差小于1的任何连续对的子序列数和dp [i] [1]包含具有相差≤1的任何连续对的子序列数。
另外，我们将维护两个变量required_subsequence和not_required_subsdequence，以维护具有至少一个连续元素差≤1的子序列的计数和不包含任何具有差值≤1的连续元素对的子序列的计数。

现在，考虑子数组arr [1]…。 arr [i]，我们将执行以下步骤：

1. 通过在子序列中添加第ith个元素，可以计算出不具有相差小于1的连续对的子序列的数目。这些基本上是dp [arr [i] + 1] [0]，dp [arr [i] – 1] [0]和dp [arr [i]] [0]的总和。
2. 子序列的总数具有至少一个连续的对，其差值至少为1，并且在i处终止，直到找到i为止的总子序列(仅在最后追加arr [i])+变成子序列的子序列具有加arr [i]时，至少有一对小于1的连续对。
3. 在i + 1之前不具有相差小于1且不小于1的连续对的总子序列=在i + 1之前不具有相差小于1的任何连续对的总子序列(仅将当前元素作为子序列) 。
4. 更新required_sub-sequence，not_required_subsequence和dp [arr [i] [0]]，最终答案将为required_subsequence。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
const int N = 10000;
  
// Function to return the number of subsequences
// which have at least one consecutive pair
// with difference less than or equal to 1
int count_required_sequence(int n, int arr[])
{
    int total_required_subsequence = 0;
    int total_n_required_subsequence = 0;
    int dp[N][2];
    for (int i = 0; i < n; i++) {
  
        // Not required sub-sequences which
        // turn required on adding i
        int turn_required = 0;
        for (int j = -1; j <= 1; j++)
            turn_required += dp[arr[i] + j][0];
  
        // Required sub-sequence till now will be
        // required sequence plus sub-sequence
        // which turns required
        int required_end_i = (total_required_subsequence
                              + turn_required);
  
        // Similarly for not required
        int n_required_end_i = (1 + total_n_required_subsequence
                                - turn_required);
  
        // Also updating total required and
        // not required sub-sequences
        total_required_subsequence += required_end_i;
        total_n_required_subsequence += n_required_end_i;
  
        // Also, storing values in dp
        dp[arr[i]][1] += required_end_i;
        dp[arr[i]][0] += n_required_end_i;
    }
  
    return total_required_subsequence;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 6, 2, 1, 9 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << count_required_sequence(n, arr) << "\n";
  
    return 0;
}

Java

// Java implemenation of above approach
public class GFG
{
      
static int N = 10000;
  
// Function to return the number of subsequences
// which have at least one consecutive pair
// with difference less than or equal to 1
static int count_required_sequence(int n, int arr[])
{
    int total_required_subsequence = 0;
    int total_n_required_subsequence = 0;
    int [][]dp = new int[N][2];
    for (int i = 0; i < n; i++) 
    {
  
        // Not required sub-sequences which
        // turn required on adding i
        int turn_required = 0;
        for (int j = -1; j <= 1; j++)
            turn_required += dp[arr[i] + j][0];
  
        // Required sub-sequence till now will be
        // required sequence plus sub-sequence
        // which turns required
        int required_end_i = (total_required_subsequence
                            + turn_required);
  
        // Similarly for not required
        int n_required_end_i = (1 + total_n_required_subsequence
                                - turn_required);
  
        // Also updating total required and
        // not required sub-sequences
        total_required_subsequence += required_end_i;
        total_n_required_subsequence += n_required_end_i;
  
        // Also, storing values in dp
        dp[arr[i]][1] += required_end_i;
        dp[arr[i]][0] += n_required_end_i;
    }
  
    return total_required_subsequence;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = { 1, 6, 2, 1, 9 };
    int n = arr.length;
  
    System.out.println(count_required_sequence(n, arr));
}
}
  
// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
import numpy as np;
  
N = 10000; 
  
# Function to return the number of subsequences 
# which have at least one consecutive pair 
# with difference less than or equal to 1 
def count_required_sequence(n, arr) :
      
    total_required_subsequence = 0; 
    total_n_required_subsequence = 0; 
    dp = np.zeros((N,2)); 
      
    for i in range(n) :
  
        # Not required sub-sequences which 
        # turn required on adding i 
        turn_required = 0; 
        for j in range(-1, 2,1) :
            turn_required += dp[arr[i] + j][0]; 
  
        # Required sub-sequence till now will be 
        # required sequence plus sub-sequence 
        # which turns required 
        required_end_i = (total_required_subsequence 
                            + turn_required); 
  
        # Similarly for not required 
        n_required_end_i = (1 + total_n_required_subsequence 
                                - turn_required); 
  
        # Also updating total required and 
        # not required sub-sequences 
        total_required_subsequence += required_end_i; 
        total_n_required_subsequence += n_required_end_i; 
  
        # Also, storing values in dp 
        dp[arr[i]][1] += required_end_i; 
        dp[arr[i]][0] += n_required_end_i; 
          
    return total_required_subsequence; 
  
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 1, 6, 2, 1, 9 ]; 
    n = len(arr); 
  
    print(count_required_sequence(n, arr)) ; 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
  
class GFG
{
      
    static int N = 10000;
      
    // Function to return the number of subsequences
    // which have at least one consecutive pair
    // with difference less than or equal to 1
    static int count_required_sequence(int n, int []arr)
    {
        int total_required_subsequence = 0;
        int total_n_required_subsequence = 0;
        int [, ]dp = new int[N, 2];
        for (int i = 0; i < n; i++) 
        {
      
            // Not required sub-sequences which
            // turn required on adding i
            int turn_required = 0;
            for (int j = -1; j <= 1; j++)
                turn_required += dp[arr[i] + j, 0];
      
            // Required sub-sequence till now will be
            // required sequence plus sub-sequence
            // which turns required
            int required_end_i = (total_required_subsequence
                                + turn_required);
      
            // Similarly for not required
            int n_required_end_i = (1 + total_n_required_subsequence
                                    - turn_required);
      
            // Also updating total required and
            // not required sub-sequences
            total_required_subsequence += required_end_i;
            total_n_required_subsequence += n_required_end_i;
      
            // Also, storing values in dp
            dp[arr[i], 1] += required_end_i;
            dp[arr[i], 0] += n_required_end_i;
        }
      
        return total_required_subsequence;
    }
      
    // Driver code
    public static void Main() 
    {
        int [] arr = new int [] { 1, 6, 2, 1, 9 };
        int n = arr.Length;
      
        Console.WriteLine(count_required_sequence(n, arr));
    }
}
  
// This code has been contributed by ihritik

输出：

12