CODEWITHSUNDEEP
javaalgorithmcombinationsprobability
sky.flyer
sky.flyer

Reputation: 91

combination of two lists in java

Is there any algorithm to achieve this combination of output? Input :

arr1 = {x, y, z}
arr2 = {a, b}

Output :

xa, ya, za
xa, ya, zb
xa, yb, za
xa, yb, zb
xb, ya, za
xb, ya, zb
xb, yb, za
xb, yb, zb

Upvotes: 2

Views: 743

Answers (3)

shmosel
shmosel

Reputation: 50746

static char[] arr1 = {'x', 'y', 'z'};
static char[] arr2 = {'a', 'b'};

public static void main(String[] args) {
    print(new char[arr1.length - 1], 0);
}

static void print(char[] store, int depth) {
    for(char c : arr2) {
        if(depth < store.length) {
            store[depth] = c;
            print(store, depth + 1);
        } else {
            for(int i = 0; i < store.length; i++) {
                System.out.print(arr1[i] + "" + store[i] + ", ");
            }
            System.out.println(arr1[depth] + "" + c);
        }
    }
}

EDIT: Couldn't resist trying out @DenisKulagin's method, so here goes:

public static void main(String[] args) {
    char[] arr1 = {'x', 'y', 'z'};
    char[] arr2 = {'a', 'b'};

    for(int i = 0; i < 1 << arr1.length; i++) {
        for(int j = 0; j < arr1.length; j++) {
            int inverted = arr1.length - 1 - j;
            int index = (i & (1 << inverted)) >>> inverted;
            System.out.print(arr1[j] + "" + arr2[index] + " ");
        }
        System.out.println();
    }
}

Not quite as flexible as my version, since arr2 can only contain 2 elements, but definitely a clever approach.

Upvotes: 3

aw-think
aw-think

Reputation: 4803

It maybe a 2 power 3 (arr2.length power arr1.length). This should give the num of rows.

So you need an Algorithm that calculates every single row. The shorter one gives the num and the longer one the exponent.

Upvotes: 0

Denis Kulagin
Denis Kulagin

Reputation: 8937

  1. Encode {a, a, a} as {0, 0, 0} - 0/binary.
  2. Encode {b, b, b} as {1, 1, 1} - 7/binary.
  3. Loop through 0..7, joining {x, y, z} & resulting 3-tuple.
  4. Profit!

Upvotes: 1

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