Number of Permutations such that no Three Terms forms Increasing Subsequence

Given a number N. The task is to find the number of permutations of 1 to N such that no three terms of the permutation form an increasing subsequence.

Examples:

Input : N = 3
Output : 5
Valid permutations : 132, 213, 231, 312 and 321 and not 123

Input : N = 4
Output : 14

The above problem is an application of Catalan numbers. So, the task is to only find the n’th Catalan Number. First few Catalan numbers are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, … (considered from 0th number)

Below is the program to find Nth Catalan Number:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the
// nth catalan number
#include <bits/stdc++.h>
using namespace std;
  
// Returns value of Binomial Coefficient C(n, k)
unsigned long int binomialCoeff(unsigned int n,
                                unsigned int k)
{
    unsigned long int res = 1;
  
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
  
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] / [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// A Binomial coefficient based function
// to find nth catalan
// number in O(n) time
unsigned long int catalan(unsigned int n)
{
    // Calculate value of 2nCn
    unsigned long int c = binomialCoeff(2 * n, n);
  
    // return 2nCn/(n+1)
    return c / (n + 1);
}
  
// Driver code
int main()
{
    int n = 3;
  
    cout << catalan(n) << endl;
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the
// nth catalan number
import java.io.*;
  
class GFG
{
  
// Returns value of Binomial
// Coefficient C(n, k)
static long binomialCoeff(long n, long k)
{
    long res = 1;
  
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
  
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] / 
    // [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i) 
    {
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// A Binomial coefficient based 
// function to find nth catalan
// number in O(n) time
static long catalan(long n)
{
    // Calculate value of 2nCn
    long c = binomialCoeff(2 * n, n);
  
    // return 2nCn/(n+1)
    return c / (n + 1);
}
  
// Driver code
public static void main (String[] args) 
{
    int n = 3;
      
    System.out.println(catalan(n));
}
}
  
// This code has been contributed
// by inder_verma.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python program to find the 
# nth catalan number 
  
# Returns value of Binomial 
# Coefficient C(n, k) 
def binomialCoeff(n, k):
    res = 1
      
    # Since C(n, k) = C(n, n-k)
    if k > n - k:
        k=n-k
    # Calculate value of 
    # [n*(n-1)*---*(n-k+1)] // 
    # [k*(k-1)*---*1]
  
    for i in range(k):
        res = res * (n - i)
        res = res // (i + 1)
    return res
      
# A Binomial coefficient based 
# function to find nth catalan 
# number in O(n) time 
def catalan(n):
      
    # Calculate value of 2nCn 
    c = binomialCoeff(2 * n, n)
      
    # return 2nCn/(n+1) 
    return c // (n + 1)
      
# Driver code 
n = 3
print(catalan(n))
  
# This code is contributed 
# by sahil shelangia

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the
// nth catalan number
using System;
  
class GFG
{
  
// Returns value of Binomial
// Coefficient C(n, k)
static long binomialCoeff(long n, 
                          long k)
{
    long res = 1;
  
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
  
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] / 
    // [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i) 
    {
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// A Binomial coefficient based 
// function to find nth catalan
// number in O(n) time
static long catalan(long n)
{
    // Calculate value of 2nCn
    long c = binomialCoeff(2 * n, n);
  
    // return 2nCn/(n+1)
    return c / (n + 1);
}
  
// Driver code
public static void Main (String[] args) 
{
    int n = 3;
      
    Console.WriteLine(catalan(n));
}
}
  
// This code is contributed
// by Kirti_Mangal

chevron_right


PHP

$n – $k)
$k = $n – $k;

// Calculate value of
// [n*(n-1)*—*(n-k+1)] //
// [k*(k-1)*—*1]
for ($i = 0; $i < $k; $i++) { $res = $res * ($n - $i); $res = $res / ($i + 1); } return $res; } // A Binomial coefficient based // function to find nth catalan // number in O(n) time function catalan($n) { // Calculate value of 2nCn $c = binomialCoeff(2 * $n, $n); // return 2nCn/(n+1) return $c / ($n + 1); } // Driver code $n = 3; print(catalan($n)); // This code is contributed // by mits ?>

Output:

5


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.