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Combinations.java
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package com.thealgorithms.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;
}
}