There are multiple ways to solve this problem, but the easiest way IMHO is to use recursion. Below I have provided the iterative and recursive approaches to solving the problem of generating all the combinations of a set. The general idea for both approaches is to generate the set of indexes which belong to the elements you want to select.
For the recursive approach, you want to keep track of three pieces of information: A boolean array indicating whether an item was selected or not, the position you're at in your list of items and a variable r tracking the number of elements remaining to select. Then as long as r != 0 (meaning you still need to select r elements to finish this combination) you backtrack selecting an element you have not yet selected and move forward in the array.
The code shown below was taken from my github repo.
/**
* Here we present two methods (recursive and iterative) of generating all
* the combinations of a set by choosing only r of n elements.
*
* Time Complexity: O( n choose r )
*
* @author William Fiset, Micah Stairs
**/
public class Combinations {
// This method finds all the combinations of size r in a set
public static void combinationsChooseR(int[] set, int r) {
if (r < 0) return;
if (set == null) return;
boolean [] used = new boolean[set.length];
combinations(set, r, 0, used);
}
// To find all the combinations of size r we need to recurse until we have
// selected r elements (aka r = 0), otherwise if r != 0 then we need to select
// an element which is found after the position of our last selected element
private static void combinations(int[] set, int r, int at, boolean[] used) {
final int N = set.length;
// If there are more elements left to select than what are available
// This is a short circuiting optimization we can take advantage of
int elementsLeftToPick = N - at;
if (elementsLeftToPick < r) return;
// We selected 'r' elements so we found a valid subset!
if (r == 0) {
System.out.print("{ ");
for (int i = 0; i < N; i++)
if (used[i])
System.out.print(set[i] + " ");
System.out.println("}");
} else {
for (int i = at; i < N; i++) {
// Try including this element
used[i] = true;
combinations(set, r - 1, i + 1, used);
// Backtrack and try the instance where we did not include this element
used[i] = false;
}
}
}
// Use this method in combination with a do while loop to generate all the
// combinations of a set choose r in a iterative fashion. This method returns
// false once the last combination has been generated.
// NOTE: Originally the selection needs to be initialized to {0,1,2,3 ... r-1}
public static boolean nextCombination(int[] selection, int N, int r) {
if (r > N) throw new IllegalArgumentException("r must be <= N");
int i = r - 1;
while (selection[i] == N - r + i)
if (--i < 0) return false;
selection[i]++;
for (int j = i + 1; j < r; j++) selection[j] = selection[i] + j - i;
return true;
}
public static void main(String[] args) {
// Recursive approach
int R = 3;
int[] set = {1,2,3,4,5};
combinationsChooseR(set, R);
// prints:
// { 1 2 3 }
// { 1 2 4 }
// { 1 2 5 }
// { 1 3 4 }
// { 1 3 5 }
// { 1 4 5 }
// { 2 3 4 }
// { 2 3 5 }
// { 2 4 5 }
// { 3 4 5 }
// Suppose we want to select all combinations of colors where R = 3
String[] colors = {"red", "purple", "green", "yellow", "blue", "pink"};
R = 3;
// Initialize the selection to be {0, 1, ... , R-1}
int[] selection = { 0, 1, 2 };
do {
// Print each combination
for (int index : selection)
System.out.print(colors[index] + " ");
System.out.println();
} while( nextCombination(selection, colors.length, R) );
// prints:
// red purple green
// red purple yellow
// red purple blue
// red purple pink
// red green yellow
// red green blue
// red green pink
// red yellow blue
// red yellow pink
// red blue pink
// purple green yellow
// purple green blue
// purple green pink
// purple yellow blue
// purple yellow pink
// purple blue pink
// green yellow blue
// green yellow pink
// green blue pink
// yellow blue pink
}
}
