Water collected between towers

Question

I recently came across an interview question asked by Amazon and I am not able to find an optimized algorithm to solve this question:

You are given an input array wh

Answer 1

Here is my take to the problem, I use a loop to see if the previous towers is bigger than the actual one. If it is then I create another loop to check if the towers coming after the actual one are bigger or equal to the previous tower. If that's the case then I just add all the differences in height between the previous tower and all other towers. If not and if my loop reaches my last object then I simply reverse the array so that the previous tower becomes my last tower and call my method recursively on it. That way I'm certain to find a tower bigger than my new previous tower and will find the correct amount of water collected.

public class towers {
    public static int waterLevel(int[] i) {
        int totalLevel = 0;

        for (int j = 1; j < i.length - 1; j++) {
            if (i[j - 1] > i[j]) {
                for (int k = j; k < i.length; k++) {
                    if (i[k] >= i[j  - 1]) {
                        for (int l = j; l < k; l++) { 
                            totalLevel += (i[j - 1] - i[l]);
                        }

                        j = k;
                        break;
                    }  

                    if (k == i.length - 1) {
                        int[] copy = Arrays.copyOfRange(i, j - 1, k + 1);
                        int[] revcopy = reverse(copy);
                        totalLevel += waterLevel(revcopy);
                    }
                }
            }
        }

        return totalLevel;
    }

    public static int[] reverse(int[] i) {
        for (int j = 0; j < i.length / 2; j++) {
            int temp = i[j];
            i[j] = i[i.length - j - 1];
            i[i.length - j - 1] = temp;
        }

        return i;
    }

    public static void main(String[] args) {
        System.out.println(waterLevel(new int[] {1, 6, 3, 2, 2, 6}));
    }
}