Number of sub-sequence such that it has one consecutive element with difference less than or equal to 1

Given an array arr[] of N elements. The task is to find the number of sub-sequences which have at least two consecutive elements such that absolute difference between them is ≤ 1.

Examples:

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

Naive approach: The idea is to find all the possible sub-sequences and check if there exist sub-sequence with any consecutive pair with difference ≤1 and increase the count.

Efficient approach: The idea is to iterate over the given array and for each ith-element, try to find the required sub-sequence ending with ith element as its last element.
For every i, we want to use arr[i], arr[i] -1, arr[i] + 1, so we will define 2D array, dp[][], where dp[i][0] will contain the number of sub sequence that do not have any consecutive pair with difference less than 1 and dp[i][1] contain the number of sub sequence having any consecutive pair with difference ≤1.
Also, we will maintain two variables required_subsequence and not_required_subsdequence to maintain the count of sub sequences which have at least one consecutive element with difference ≤1 and count of sub-sequences which do not contain any consecutive element pair with difference ≤1.

Now, considering the sub-array arr[1] …. arr[i], we will perform the following steps:

  1. Compute the number of sub sequences which do not have any consecutive pair with difference less than 1 but will have by adding the ith element in the sub sequence. These are basically sum of dp[arr[i] + 1][0], dp[arr[i] – 1][0] and dp[arr[i]][0].
  2. Total number of sub sequences have at least one consecutive pair with difference at least 1 and ending at i is equal to total sub-sequences found till i (just append arr[i] at the last) + sub sequences which turns into sub sequence have at least consecutive pair with difference less than 1 on adding arr[i].
  3. Total sub sequence which do not have any consecutive pair with difference less than 1 and ending at i = total sub-sequence which do not have any consecutive pair with difference less than 1 before i + 1 (just the current element as a sub sequence).
  4. Update required_sub-sequence, not_required_subsequence and dp[arr[i][0]] and the final answer will be required_subsequence.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

12


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