finding two pairs from integer array out of two elements

Question

Two pairs: If there are two pairs of dice with the same number, the player scores the sum of these dice. If not, the player scores 0. For example, 1, 1, 2, 3, 3 placed on "two pairs" gives 8.

examples: 1,1,2,3,3 results 8 1,1,2,3,4 results 0 1,1,2,2,2 results 6

How can find this efficiently?

I've been using following code to find a single pair

int max_difference = 0;
int val1 = 0 , val2 = 0;
Arrays.sort(dice);
for (int i = 0; i < dice.length - 1; i++) {
    int x = dice[i+1] - dice[i];
    if(x <= max_difference) {
        max_difference = x;
        val1 = dice[i];
        val2 = dice[i+1];

    }
}
pairScore = val1 + val2;

"two pairs" = 2 x 2 dice with the same value? So if I have 1, 1, 2, 3, 4 the result is 0? What about 1, 1, 1, 2, 2? And 1, 1, 1, 1, 2? — Mark
– Mark, Commented Jul 26, 2016 at 9:14
In which case should it be 6? 1, 1, 1, 2, 2? So it doesn#t matter that there are 3 1s it still counts as a pair? — Mark
– Mark, Commented Jul 26, 2016 at 9:16
The distinct values are limited. So use bucket sorting, or frequency map as @Thomas answered. — waltersu
– waltersu, Commented Jul 26, 2016 at 9:24

Mark · Accepted Answer · 2016-07-26 09:26:51Z

5

No need to make it that complicated, since you're only searching for the resulting number...

int prev = 0;
int result = 0;
int pairs = 0;

Arrays.sort(dice);
for (int i = 0; i < dice.length; i++) 
{
    int current = dice[i];
    if (current == prev) 
    {
        result += current*2;
        pairs++;
        prev = 0;
    } 
    else prev = current;
}

if (pairs == 2) return result;
else return 0;

answered Jul 26, 2016 at 9:26

Mark

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Comments

Thomas · Accepted Answer · 2016-07-26 09:32:04Z

3

I'd use a frequency map, i.e. the number is the key and the value is a counter (so a Map<Integer, Integer>). However, since it is used for dices you could simplify that using an array with a length equal to the maximum dice value (6 for standard dice). Then check the frequencies for each number and get the number of pairs from it.

Example:

int[] diceFrequency = new int[6];

//assuming d is in the range [1,6]
for( int d : dice ) {
  //increment the counter for the dice value
  diceFrequency[d-1]++; 
}

int numberOfPairs = 0;
int pairSum = 0;

for( int i = 0; i < diceFrequency.length; i++ ) {
  //due to integer division you get only the number of pairs, 
  //i.e. if you have 3x 1 you get 1 pair, for 5x 1 you get 2 pairs
  int num = diceFrequency[i] / 2;

  //total the number of pairs is just increases
  numberOfPairs += num; 

  //the total value of those pairs is the dice value (i+1) 
  //multiplied by the number of pairs and 2 (since it's a pair)
  pairSum += (i + 1 ) * 2 * num;
}

if( numerOfPairs >= 2 ) {
  //you have at least 2 pairs, do whatever is appropriate
}

edited Jul 26, 2016 at 9:32

answered Jul 26, 2016 at 9:22

Thomas

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5 Comments

Om Komawar Over a year ago

hey diceFrequency[d-1]++; what does this do?

Om Komawar Over a year ago

diceFrequency[d-1]++; what does this do?

Raman Sahasi Over a year ago

//increment the counter for the dice value. It's written in comments

Thomas Over a year ago

diceFrequency[d-1]++ increases the value of the element at index d-1 by 1. Since the array index is 0-based but the dice value is 1-based you need to use d-1 instead of just d.

Thomas Over a year ago

@RamanSahasi to be fair I added the comments after the question since I realized some more explanation is needed.

freeism · Accepted Answer · 2016-07-28 02:40:00Z

3

How about use hashmap as below?

public static void main(String[] args) {
    List<Integer> list = Lists.newArrayList(1, 1, 2, 3, 3, 4);
    Map<Integer, Integer> map = Maps.newHashMap();

    int result = 0;

    for (int i : list) {
        int frequency = (int) MapUtils.getObject(map, i, 0);

        if (frequency < 2) {
            map.put(i, ++frequency);
        }
    }

    for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
        if (entry.getValue() > 1) {
            result += entry.getKey() * entry.getValue();
        }
    }

    System.out.println(result);
}

edited Jul 28, 2016 at 2:40

answered Jul 26, 2016 at 9:26

freeism

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4 Comments

Thomas Over a year ago

AFAIK that kind of map is called a frequency map (just for future reference :) ). Besides that you could optimize the second loop by using entrySet() since you'd then not have to do additional lookups.

freeism Over a year ago

The keySet() is replaced by the entrySet() as @Thomas said. Thanks for advice.

Roland Illig Over a year ago

When you have a full house (e.g. 2 2 2 3 3), your code counts all the dice, it probably should only count two of the 2s.

freeism Over a year ago

@Roland Illig You have a good point. It was fixed as above. thank you :)

Raman Sahasi · Accepted Answer · 2016-07-26 09:33:43Z

1

public static void main(String[] args) {
    List<Integer> list = Lists.newArrayList(1, 1, 2, 3, 3);
    Map<Integer, Integer> map = new HashMap<>();

    int count= 0;

    for (int num : list)
        if(map.containsKey(num ))
            map.put(num , map.get(num )+1);
        else
            map.put(num , 1);

    for (int num  : map.keySet()) {
        if (map.get(num ) > 1) {
            count= count+ (num  * map.get(num ));
        }
    }

    System.out.println(count);
}

answered Jul 26, 2016 at 9:33

Raman Sahasi

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Comments

Mustafa YALNIZ · Accepted Answer · 2016-07-27 06:55:16Z

I hope, I could understand the problem. So we can use thiscode part.

    List<Integer> diceList = new ArrayList<Integer>();
    diceList.add(1);
    diceList.add(1);
    diceList.add(2);
    diceList.add(4);
    diceList.add(3);
    diceList.add(3);
    diceList.add(6);
    diceList.add(8);

    Integer prev = null, score = 0;

    boolean flag = false;
    for (Integer val : diceList) {
        if (prev == null || !prev.equals(val)) {
            if (flag) {
                score = score + prev;
                flag = false;
            }
            prev = val;

        } else if (prev == val) {
            score = score + prev;
            flag = true;
        }
    }
    System.out.println(score);

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