// Complete the nonDivisibleSubset function below.
    static int nonDivisibleSubset(int k, int[] S) {
        // In maths. if (a + b) % k = 0 => then ((a % k) + (b % k)) % k = 0
        // Example: (5 + 7) % 6 = 0 => then (5 % 6) + (7 % 6) > (5 + 1) % 6 = 0

        // Solution: Find remainder of each element in the array.
        // then, choose max element from the pair which together can able to be divided by k. If one pair is "i" then other pair will be "k-i"
        // For example: S = {2, 3, 7, 8, 12} and k = 5.
        // Now we have 3 numbers whose remainder 2 => ( 2 % 5 = 2, 7 % 5 = 2, 12 % 5 = 2)
        // and also we have 2 numbers whose remainder 3 => (3, 8)
        // Right now we have to choose one of the element from that pair (3, 2) (where 3 > numbers 2, 7, 12 && 2 > numbers 3, 8)
        // Because of the problem, we will choose the max which is 3.

        int[] remainderArr = new int[k];
        // find remainder of each element in the array S
        // For example k = 4, S = {0, 5, 7, 10} => remainderArr will be: {0, 1, 1, 1}
        // where each index represents remainder. For example remainderArr[2] = 1 means
        // that there is 1 number whose remainder 2 after divided 4. (10 % 4 = 2)
        for (Integer each : S) {
            remainderArr[each % k]++;
        }

        // After getting each remainder, index 0 (actually remainder 0) is a special case
        // Think of it like this:
        //      1. There will be no element such as k - 0 = k. (remainderArr[k] will give us ArrayIndexOutOfBoundsException)
        //      2. If there are 2 elements in remainderArr[0], we have to choose only 1, otherwise, we can sum up 2 or more
        //         zeros, then non-sub divisible set could be divisible by k.
        int zeroRemainder = remainderArr[0];

        // That's why, our initial subset size is 1, if there is a zero remainder,
        //                                           otherwise it is 0
        int maxNumberOfDivisibleSet = zeroRemainder > 0 ? 1 : 0;

        // Another thing is that pair which is itself. That's means, let's say k = 4, therefore pair of remainderArr[2]
        // will also be remainderArr[2]( i = 2 then, k - i = 2). Thus, we have to choose only 1 element from that pair (or we should increment
        // the result number just 1)
        // if condition "i != k - i" will handle this situation.
        for (int i = 1; i <= (k / 2); i++) {
            if (i != k - i)
                maxNumberOfDivisibleSet += Math.max(remainderArr[i], remainderArr[k - i]);
            else
                maxNumberOfDivisibleSet ++;
        }

        return maxNumberOfDivisibleSet;
    }