Given clock hands positions for different points of time as pairs A[I][0] and A[I][1] where the order of the hands does not matter but their angle enclosed, count the number of pairs of points of time where the angles are the same

    public static int[] getClockHandsDelta(int[][] A) {

        int[] angles = new int[A.length];

        for (int i = 0; i < A.length; i++){

            angles[i] = Math.max(A[i][0], A[i][1]) - Math.min(A[i][0],A[i][1]);

        }

        return angles;

    }

    public static int NChooseK(int n, int k)

    {

        if (k < 0 || k > n || n == 0) return 0;

        if ( k == 0 || k == n) return 1;

        return Factorial(n) / (Factorial(n-k) * Factorial(k));

    }

 

    public static int Factorial(int n) {

        if (n <= 1) return 1;

        return n * Factorial(n-1);

    }

 

    public static int countPairsWithIdenticalAnglesDelta(int[] angles){

        Arrays.sort(angles);

        int count = 1;

        int result = 0;

        for (int i = 1; i < angles.length; i++) {

            if (angles[i] == angles[i-1]) {

                count += 1;

            } else {

                if (count > 0) {

                    result += NChooseK(count, 2);

                }

                count = 1;

            }

        }

        if (count > 0) {

            result += NChooseK(count, 2);

            count = 0;

        }

        return result;

    }

 

        int [][] A = new int[5][2];
         A[0][0] = 1;    A[0][1] = 2;
         A[1][0] = 2;    A[1][1] = 4;
         A[2][0] = 4;    A[2][1] = 3;
         A[3][0] = 2;    A[3][1] = 3;
         A[4][0] = 1;    A[4][1] = 3;
 1 2 1 1 2
1 1 1 2 2
4