Java – find two pairs from an array of integers in two elements

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

Example: 1,3 result 8 1,4 result 0 1,2 result 6

How to find this effectively?

I've been using the following code to find a 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;

Solution

I use the frequency graph, that is, the number is the key and the value is the counter (so map < integer, integer >) However, since it is used for dice, you can simplify dice by using an array with a length equal to the maximum dice value (the standard dice is 6) Then check the frequency of 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
}