Number of sub-sequence such that it has one consecutive element with difference less than or equal to 1
Last Updated :
25 May, 2022
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 exists a 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 subsequences 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:
- 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].
- Total number of subsequences 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) + subsequences which turns into subsequence have at least consecutive pair with difference less than 1 on adding arr[i].
- Total subsequence 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 subsequence).
- 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++
#include <bits/stdc++.h>
using namespace std;
const int N = 10000;
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++) {
int turn_required = 0;
for ( int j = -1; j <= 1; j++)
turn_required += dp[arr[i] + j][0];
int required_end_i = (total_required_subsequence
+ turn_required);
int n_required_end_i = (1 + total_n_required_subsequence
- turn_required);
total_required_subsequence += required_end_i;
total_n_required_subsequence += n_required_end_i;
dp[arr[i]][1] += required_end_i;
dp[arr[i]][0] += n_required_end_i;
}
return total_required_subsequence;
}
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
public class GFG
{
static int N = 10000 ;
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++)
{
int turn_required = 0 ;
for ( int j = - 1 ; j <= 1 ; j++)
turn_required += dp[arr[i] + j][ 0 ];
int required_end_i = (total_required_subsequence
+ turn_required);
int n_required_end_i = ( 1 + total_n_required_subsequence
- turn_required);
total_required_subsequence += required_end_i;
total_n_required_subsequence += n_required_end_i;
dp[arr[i]][ 1 ] += required_end_i;
dp[arr[i]][ 0 ] += n_required_end_i;
}
return total_required_subsequence;
}
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));
}
}
|
Python3
import numpy as np;
N = 10000 ;
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) :
turn_required = 0 ;
for j in range ( - 1 , 2 , 1 ) :
turn_required + = dp[arr[i] + j][ 0 ];
required_end_i = (total_required_subsequence
+ turn_required);
n_required_end_i = ( 1 + total_n_required_subsequence
- turn_required);
total_required_subsequence + = required_end_i;
total_n_required_subsequence + = n_required_end_i;
dp[arr[i]][ 1 ] + = required_end_i;
dp[arr[i]][ 0 ] + = n_required_end_i;
return total_required_subsequence;
if __name__ = = "__main__" :
arr = [ 1 , 6 , 2 , 1 , 9 ];
n = len (arr);
print (count_required_sequence(n, arr)) ;
|
C#
using System;
public class GFG{
static int N = 10000;
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++)
{
int turn_required = 0;
for ( int j = -1; j <= 1; j++)
turn_required += dp[arr[i] + j,0];
int required_end_i = (total_required_subsequence
+ turn_required);
int n_required_end_i = (1 + total_n_required_subsequence
- turn_required);
total_required_subsequence += required_end_i;
total_n_required_subsequence += n_required_end_i;
dp[arr[i],1] += required_end_i;
dp[arr[i],0] += n_required_end_i;
}
return total_required_subsequence;
}
static public void Main ()
{
int [] arr = { 1, 6, 2, 1, 9 };
int n = arr.Length;
Console.WriteLine(count_required_sequence(n, arr));
}
}
|
Javascript
<script>
var N = 10000;
function count_required_sequence(n, arr)
{
var total_required_subsequence = 0;
var total_n_required_subsequence = 0;
var dp = Array.from(Array(N), ()=> Array(2).fill(0));
for ( var i = 0; i < n; i++) {
var turn_required = 0;
for ( var j = -1; j <= 1; j++)
turn_required += dp[arr[i] + j][0];
var required_end_i = (total_required_subsequence
+ turn_required);
var n_required_end_i = (1 + total_n_required_subsequence
- turn_required);
total_required_subsequence += required_end_i;
total_n_required_subsequence += n_required_end_i;
dp[arr[i]][1] += required_end_i;
dp[arr[i]][0] += n_required_end_i;
}
return total_required_subsequence;
}
var arr = [ 1, 6, 2, 1, 9 ];
var n = arr.length;
document.write( count_required_sequence(n, arr) + "<br>" );
</script>
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Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(N), where N represents the size of the given array.
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