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.
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 <bits/stdc++.h>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[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];}// Driver functionint main(){ int N; N = 2;cout<<chordCnt( N)<<'\n'; N = 1;cout<<chordCnt( N)<<'\n'; N = 4;cout<<chordCnt( N)<<'\n'; return 0;}// This code is contributed by Gitanjali. |
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 codeN = 2print(chordCnt( N))N = 1print(chordCnt( N))N = 4print(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[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]; } // 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
<?php// PHP code to count ways // to divide circle using // N non-intersecting chords. function chordCnt( $A){ // n = no of points required $n = 2 * $A; // dp array containing the sum $dpArray = array_fill(0, $n + 1, 0); $dpArray[0] = 1; $dpArray[2] = 1; for ($i = 4; $i <= $n; $i += 2) { for ($j = 0; $j < $i - 1; $j += 2) { $dpArray[$i] += ($dpArray[$j] * $dpArray[$i - 2 - $j]); } } // returning the required number return $dpArray[$n]; } // Driver Code $N = 2; echo chordCnt($N), "\n"; $N = 1; echo chordCnt($N), "\n"; $N = 4; echo chordCnt($N), "\n"; // This code is contributed by Ryuga ?> |
Javascript
<script>// JavaScript code to count ways // to divide circle using// N non-intersecting chords.function chordCnt( A){ // n = no of points required var n = 2 * A; // dp array containing the sum var dpArray = Array(n+1).fill(0); dpArray[0] = 1; dpArray[2] = 1; for (var i=4;i<=n;i+=2){ for (var j=0;j<i-1;j+=2){ dpArray[i] += (dpArray[j]*dpArray[i-2-j]); } } // returning the required number return dpArray[n];}// Driver functionvar N;N = 2;document.write( chordCnt( N) + '<br>');N = 1;document.write( chordCnt( N) + '<br>');N = 4;document.write( chordCnt( N) + '<br>');</script> |
Output:
2 1 14
Time Complexity: O(n2)
Auxiliary Space: O(n)
Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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