Count Reverse Pairs

class Solution {
    public int countRevPairs(int[] arr) {
        int n = arr.length;
        int temp[] = new int[n];
        return divide(arr, temp, 0, n - 1);
    }

    static int divide(int arr[], int temp[], int low, int high) {
        int c = 0;
        if (low < high) {
            int mid = (low + high) / 2;
            c = c + divide(arr, temp, low, mid);
            c = c + divide(arr, temp, mid + 1, high);
            c = c + count(arr, temp, low, mid, high);
            mergeSort(arr, temp, low, mid, high);
        }

        return c;
    }

    static void mergeSort(int arr[], int temp[], int low, int mid, int high) {
        int left = low;
        int right = mid + 1;
        int k = low;
        while (left <= mid && right <= high) {
            if (arr[left] <= arr[right]) {
                temp[k] = arr[left];
                left++;
                k++;
            } else {
                temp[k] = arr[right];
                right++;
                k++;
            }
        }

        while (left <= mid) {
            temp[k] = arr[left];
            left++;
            k++;
        }

        while (right <= high) {
            temp[k] = arr[right];
            right++;
            k++;
        }

        for (int i = low; i <= high; i++)
            arr[i] = temp[i];
    }

    static int count(int arr[], int temp[], int low, int mid, int high) {
        int c = 0;
        int h = mid + 1;
        for (int i = low; i <= mid; i++) {
            while (h <= high && arr[i] > (2 * arr[h]))
                h++;
            c = c + h - (mid + 1);
        }

        return c;
    }
}