I wrote the following code to answer this, but I am wondering if there is any better approach.For exampleIf the following array is given:{10,21,34,45,56}we should find the differentiate of each number with other values of the array and in case it is equals to 11 should increment the counter.10-21=-1110-34=-2410-45=-3510-56=-4621-10=11     <21-34=-1321-45=-2421-56=-3534-10=2434-21=1334-45=-1134-56=-2245-10=3545-21=2445-34=11   <<45-56=-1156-10=4656-21=3556-34=2256-45=11   <<<CodeList<Integer> a = new ArrayList<Integer>();       a.add(10);       a.add(21);       a.add(34);       a.add(45);       a.add(56);      int number = 11;      int counter = 0;       for(int i=0;i<a.size();i++){           for(int j=0;j<a.size();j++){               if(j!=i){                   int t = a.get(i) - a.get(j);                   if(t == number){                       System.err.println(a.get(i) +-+ a.get(j) + = + t);                       counter++;                   }               }           }       }       System.err.println(counter);    }OutputThe application is correctly showing the results but I am wondering if there is any other approach to the above solution.21-10=1145-34=1156-45=113
Find the number of times that the difference of array values are equal to the number
java;array
I think you should look at the problem in a different way and come up with a simpler solution.Try thinking at having a set of elements. What you want to know is if there is any pair of elements such that a - b = x.You're given x, so for a given value of a in your set you should check whether it also contains x + a.Your implementation should first move the content of the list to a Set. Then you have to iterate through the set and for each element a check if the set contains also x + a. If it is true then you should add (a,b) to your solution.In your example, x = 11. If you consider a = 10 you'll check if it contains 11 + 10 = 21. It does so you can the pair (21,10) is a valid solution. Conversely, when you consider a = 21, you obtain b = 32, which is not part of the set and therefore you have to discard that pair.int checkDifferences(Set<Integer> numbers, int difference) {    int occurrences = 0;    for (Integer number : numbers) {        if (numbers.contains(number + difference)) {            occurrences++;        }    }    return occurrences;}