package Maths;

/** @see <a href="https://en.wikipedia.org/wiki/Combination">Combination</a> */
public class Combinations {
 public static void main(String[] args) {
 assert combinations(1, 1) == 1;
 assert combinations(10, 5) == 252;
 assert combinations(6, 3) == 20;
 assert combinations(20, 5) == 15504;

 // Since, 200 is a big number its factorial will go beyond limits of long even when 200C5 can be saved in a long
 // variable. So below will fail
 // assert combinations(200, 5) == 2535650040l;

 assert combinationsOptimized(100, 0) == 1;
 assert combinationsOptimized(1, 1) == 1;
 assert combinationsOptimized(10, 5) == 252;
 assert combinationsOptimized(6, 3) == 20;
 assert combinationsOptimized(20, 5) == 15504;
 assert combinationsOptimized(200, 5) == 2535650040l;
 }

 /**
 * Calculate of factorial
 *
 * @param n the number
 * @return factorial of given number
 */
 public static long factorial(int n) {
 if (n < 0) {
 throw new IllegalArgumentException("number is negative");
 }
 return n == 0 || n == 1 ? 1 : n * factorial(n - 1);
 }

 /**
 * Calculate combinations
 *
 * @param n first number
 * @param k second number
 * @return combinations of given {@code n} and {@code k}
 */
 public static long combinations(int n, int k) {
 return factorial(n) / (factorial(k) * factorial(n - k));
 }

 /**
 * The above method can exceed limit of long (overflow) when factorial(n) is larger than limits of long variable.
 * Thus even if nCk is within range of long variable above reason can lead to incorrect result.
 * This is an optimized version of computing combinations.
 * Observations:
 * nC(k + 1) = (n - k) * nCk / (k + 1)
 * We know the value of nCk when k = 1 which is nCk = n
 * Using this base value and above formula we can compute the next term nC(k+1)
 * @param n
 * @param k
 * @return nCk
 */
 public static long combinationsOptimized(int n, int k) {
 if (n < 0 || k < 0) {
 throw new IllegalArgumentException("n or k can't be negative");
 }
 if (n < k) {
 throw new IllegalArgumentException("n can't be smaller than k");
 }
 // nC0 is always 1
 long solution = 1;
 for(int i = 0; i < k; i++) {
 long next = (n - i) * solution / (i + 1);
 solution = next;
 }
 return solution;
 }
}