# Count ways to divide circle using N non-intersecting chords

Given a number N, find the number of ways you can draw N chords in a circle with 2*N points such that no 2 chords intersect.
Two ways are different if there exists a chord which is present in one way and not in other.

Examples:

```Input : N = 2
Output : 2
Explanation: If points are numbered 1 to 4 in
clockwise direction, then different ways to
draw chords are:
{(1-2), (3-4)} and {(1-4), (2-3)}

Input : N = 1
Output : 1
Explanation: Draw a chord between points 1 and 2.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

If we draw a chord between any two points, can you observe the current set of points getting broken into two smaller sets S_1 and S_2. If we draw a chord from a point in S_1 to a point in S_2, it will surely intersect the chord we’ve just drawn.
So, we can arrive at a recurrence that Ways(n) = sum[i = 0 to n-1] { Ways(i)*Ways(n-i-1) }.
Here we iterate over i, assuming that size of one of the sets is i and size of another set automatically is (n-i-1) since we’ve already used a pair of points and i pair of points in one set.

## C++

 `// cpp code to count ways  ` `// to divide circle using ` `// N non-intersecting chords. ` `#include ` `using` `namespace` `std; ` ` `  `int` `chordCnt( ``int` `A){ ` ` `  `    ``// n = no of points required ` `    ``int` `n = 2 * A; ` `     `  `    ``// dp array containing the sum ` `    ``int` `dpArray[n + 1]={ 0 }; ` `    ``dpArray = 1; ` `    ``dpArray = 1; ` `    ``for` `(``int` `i=4;i<=n;i+=2){ ` `        ``for` `(``int` `j=0;j

## Java

 `// Java code to count ways ` `// to divide circle using ` `// N non-intersecting chords. ` `import` `java.io.*; ` ` `  `class` `GFG { ` `    ``static` `int` `chordCnt(``int` `A) ` `    ``{ ` ` `  `        ``// n = no of points required ` `        ``int` `n = ``2` `* A; ` ` `  `        ``// dp array containing the sum ` `        ``int``[] dpArray = ``new` `int``[n + ``1``]; ` `        ``dpArray[``0``] = ``1``; ` `        ``dpArray[``2``] = ``1``; ` `        ``for` `(``int` `i = ``4``; i <= n; i += ``2``) { ` `            ``for` `(``int` `j = ``0``; j < i - ``1``; j += ``2``)  ` `            ``{ ` `                ``dpArray[i] += (dpArray[j] *  ` `                              ``dpArray[i - ``2` `- j]); ` `            ``} ` `        ``} ` ` `  `        ``// returning the required number ` `        ``return` `dpArray[n]; ` `    ``} ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `N; ` `        ``N = ``2``; ` `        ``System.out.println(chordCnt(N)); ` `        ``N = ``1``; ` `        ``System.out.println(chordCnt(N)); ` `        ``N = ``4``; ` `        ``System.out.println(chordCnt(N)); ` `    ``} ` `} ` ` `  `// This code is contributed by Gitanjali. `

## Python 3

 `# python code to count ways to divide ` `# circle using N non-intersecting chords. ` `def` `chordCnt( A): ` ` `  `    ``# n = no of points required ` `    ``n ``=` `2` `*` `A ` ` `  `    ``# dp array containing the sum ` `    ``dpArray ``=` `[``0``]``*``(n ``+` `1``) ` `    ``dpArray[``0``] ``=` `1` `    ``dpArray[``2``] ``=` `1` `    ``for` `i ``in` `range``(``4``, n ``+` `1``, ``2``): ` `        ``for` `j ``in` `range``(``0``, i``-``1``, ``2``): ` `            ``dpArray[i] ``+``=` `(dpArray[j]``*``dpArray[i``-``2``-``j]) ` ` `  `    ``# returning the required number ` `    ``return` `int``(dpArray[n]) ` ` `  `# driver code ` `N ``=` `2` `print``(chordCnt( N)) ` `N ``=` `1` `print``(chordCnt( N)) ` `N ``=` `4` `print``(chordCnt( N)) `

## C#

 `// C# code to count ways to divide  ` `// circle using N non-intersecting chords. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `int` `chordCnt(``int` `A) ` `    ``{ ` `        ``// n = no of points required ` `        ``int` `n = 2 * A; ` ` `  `        ``// dp array containing the sum ` `        ``int``[] dpArray = ``new` `int``[n + 1]; ` `        ``dpArray = 1; ` `        ``dpArray = 1; ` `         `  `        ``for` `(``int` `i = 4; i <= n; i += 2)  ` `        ``{ ` `            ``for` `(``int` `j = 0; j < i - 1; j += 2) ` `            ``{ ` `                ``dpArray[i] += (dpArray[j] * dpArray[i - 2 - j]); ` `            ``} ` `        ``} ` ` `  `        ``// returning the required number ` `        ``return` `dpArray[n]; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `N; ` `        ``N = 2; ` `        ``Console.WriteLine(chordCnt(N)); ` `        ``N = 1; ` `        ``Console.WriteLine(chordCnt(N)); ` `        ``N = 4; ` `        ``Console.WriteLine(chordCnt(N)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```2
1
14
```

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