Algorithms – Grouping Numbers to Minimize Group-Means

Here's a quick implementation in java. The idea is to find how many cards can be picked each step by counting the number of cards with a number below this step and subtracting the number of cards already picked. Of course a lot of work is unnecessarily repeated, but this should give a good idea of how to approach this problem.

public class PickCards {
public static int countLessOrEqualThanMax(int[] cards, int max) {
    int count = 0;
    for (int number : cards) {
        if(number <= max) count++;
    }
    return count;
}

public static void main(String[] args) {
    int[] cards = {0, 1, 2, 4, 3};
    int numberOfWays = 1;
    for(int i = 0; i < cards.length; i++) {
        int ways = Math.max(0, countLessOrEqualThanMax(cards, i) - i);
        numberOfWays *= ways;
    }
    System.out.println(numberOfWays);
}

}

Take a subset, add each element into a hash table. If the hash table already contains a value - print out saying there is a duplicate.

A hash table will only allocate 1 memory block per entry. Hence its easy to find if the number already exists. Since these are just regular entries your table will be something like this:

struct Hashtable{
   int number;
};

static struct Hashtable hashTable[10];

int getHash(int x){ return x; }

void addHash(int number)
{
    if(hashTable[number].number == -1)
    { /*OK*/
        hashTable[number].number = number;
    }else{
        printf("DUPLICATE FOUND! BAILING OUT!");
    }
}

void initTable(){ for(int i = 0; i < 10; i++){hashTable[i].number = -1;}}