C/C++ Program for nth Catalan Number
Last Updated :
09 Nov, 2023
Catalan numbers are defined as a mathematical sequence that consists of positive integers, which can be used to find the number of possibilities of various combinations.
The nth term in the sequence denoted Cn, is found in the following formula:
The first few Catalan numbers for n = 0, 1, 2, 3, … are : 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …
Catalan numbers occur in many interesting counting problems like the following.
- Count the number of expressions containing n pairs of parentheses that are correctly matched. For n = 3, possible expressions are ((())), ()(()), ()()(), (())(), (()()).
- Count the number of possible Binary Search Trees with n keys (See this)
- Count the number of full binary trees (A rooted binary tree is full if every vertex has either two children or no children) with n+1 leaves.
- Given a number n, return the number of ways you can draw n chords in a circle with 2 x n points such that no 2 chords intersect.
Examples:
Input: n = 6
Output: 132
Input: n = 8
Output: 1430
C/C++ Program for nth Catalan Number using Recursive Solution:
Catalan numbers satisfy the following recursive formula:
Follow the steps below to implement the above recursive formula
- Base condition for the recursive approach, when n <= 1, return 1
- Iterate from i = 0 to i < n
- Make a recursive call catalan(i) and catalan(n – i – 1) and keep adding the product of both into res.
- Return the res.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
unsigned long int catalan(unsigned int n)
{
if (n <= 1)
return 1;
unsigned long int res = 0;
for ( int i = 0; i < n; i++)
res += catalan(i) * catalan(n - i - 1);
return res;
}
int main()
{
for ( int i = 0; i < 10; i++)
cout << catalan(i) << " " ;
return 0;
}
|
Output1 1 2 5 14 42 132 429 1430 4862
Time Complexity: The above implementation is equivalent to nth Catalan number.
The value of nth Catalan number is exponential which makes the time complexity exponential.
Auxiliary Space: O(n)
C/C++ Program for nth Catalan Number using Dynamic Programming:
We can observe that the above recursive implementation does a lot of repeated work. Since there are overlapping subproblems, we can use dynamic programming for this.
Below is the implementation of the above idea:
- Create an array catalan[] for storing ith Catalan number.
- Initialize, catalan[0] and catalan[1] = 1
- Loop through i = 2 to the given Catalan number n.
- Loop through j = 0 to j < i and Keep adding value of catalan[j] * catalan[i – j – 1] into catalan[i].
- Finally, return catalan[n]
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
unsigned long int catalanDP(unsigned int n)
{
unsigned long int catalan[n + 1];
catalan[0] = catalan[1] = 1;
for ( int i = 2; i <= n; i++) {
catalan[i] = 0;
for ( int j = 0; j < i; j++)
catalan[i] += catalan[j] * catalan[i - j - 1];
}
return catalan[n];
}
int main()
{
for ( int i = 0; i < 10; i++)
cout << catalanDP(i) << " " ;
return 0;
}
|
Output1 1 2 5 14 42 132 429 1430 4862
Time Complexity: O(n2)
Auxiliary Space: O(n)
C/C++ Program for nth Catalan Number using Binomial Coefficient:
We can also use the below formula to find nth Catalan number in O(n) time.
Below are the steps for calculating nCr.
- Create a variable to store the answer and change r to n – r if r is greater than n – r because we know that C(n, r) = C(n, n-r) if r > n – r
- Run a loop from 0 to r-1
- In every iteration update ans as (ans*(n-i))/(i+1), where i is the loop counter.
- So the answer will be equal to ((n/1)*((n-1)/2)*…*((n-r+1)/r), which is equal to nCr.
Below are steps to calculate Catalan numbers using the formula: 2nCn/(n+1)
- Calculate 2nCn using the similar steps that we use to calculate nCr
- Return the value 2nCn/ (n + 1)
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
unsigned long int binomialCoeff(unsigned int n,
unsigned int k)
{
unsigned long int res = 1;
if (k > n - k)
k = n - k;
for ( int i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
unsigned long int catalan(unsigned int n)
{
unsigned long int c = binomialCoeff(2 * n, n);
return c / (n + 1);
}
int main()
{
for ( int i = 0; i < 10; i++)
cout << catalan(i) << " " ;
return 0;
}
|
Output1 1 2 5 14 42 132 429 1430 4862
Time Complexity: O(n).
Auxiliary Space: O(1)
We can also use the below formulas to find nth Catalan number in O(n) time.
C/C++ Program for nth Catalan Number using the previous Catalan Number:
Below are steps to calculate Catalan numbers using the above formula:
- Initialize a variable res = 1
- Print 1 as the first Catalan Number
- Iterate from i = 1 to i < n
- Update res with res = (res * (4 * i – 2)) / (i + 1)
- print res
Below is the implementation for the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void catalan( int n)
{
int res = 1;
cout << res << " " ;
for ( int i = 1; i < n; i++) {
res = (res * (4 * i - 2)) / (i + 1);
cout << res << " " ;
}
}
int main()
{
int n = 10;
catalan(n);
return 0;
}
|
Output1 1 2 5 14 42 132 429 1430 4862
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
Please refer complete article on Program for nth Catalan Number for more details!
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