Maximum contiguous product in an Array

Given an array of natural numbers (+ve, 0, -ve) find the maximum product of continuous elements.


input  : a = { 11, 4, 5, -2, 0, -3, -4, 4, 2, 3, 5, -7 };
output : 3360(From index positions 7 to 12)

class MaximumContiguousProduct {

 public static void main(String[] args) {
  int[] a = { 11, 4, 5, -2, 0, -3, -4, 4, 2, 3, 5, -7 };
  int product = getMaximumProduct(a);
  System.out.println(product);
 }

 /**
  * Track zero positions and calculate maximum product till zero
  * 
  * @param a
  * @return
  */
 public static int getMaximumProduct(int[] a) {

  if (a == null || a.length == 0)
   return 0;

  int start = 0;
  int maxSubArrayProduct = 0;
  int maxProduct = -3421545;
  for (int i = 0; i < a.length; i++) {
   if (a[i] == 0) {// if ith element is zero then send i-1 element as
    // end position
    maxSubArrayProduct = getMaxProductInSubArray(a, start, i - 1);
    start = i + 1;// skip the ith element.
   } else if (i == a.length - 1) {
    maxSubArrayProduct = getMaxProductInSubArray(a, start,
      a.length - 1);
   }
   if (maxSubArrayProduct > maxProduct)
    maxProduct = maxSubArrayProduct;
  }
  return maxProduct;
 }

 // in this sub array check if array contains even number of -ve numbers or
 // odd.
 // If it is even number,max product is product of all numbers
 // else we will get two products.
 private static int getMaxProductInSubArray(int[] a, int start, int end) {

  if (a == null || a.length == 0 || start > end)
   return 0;
  else if (start == end)
   return a[start];

  int numberOfNegativeNumbers = 0;
  int totalProduct = 1;
  for (int i = start; i <= end; i++) {
   if (a[i] < 0)
    numberOfNegativeNumbers++;
   totalProduct = totalProduct * a[i];
  }

  if (numberOfNegativeNumbers % 2 == 0)// handling case for even number of
   // negative numbers.
   return totalProduct;

  // for odd number of -ve numbers,total prod value is -ve.calculate prod1
  // and prod2
  int prod1 = 1;
  int i = start;
  while (a[i] > 0) {// calculating till we trace first -ve number
   prod1 = prod1 * a[i];
   i++;
  }
  int prodRem1 = totalProduct / (prod1 * a[i]);

  int prod2 = 1;
  i = end;
  while (a[i] > 0) {// calculating till we trace first -ve number
   prod2 = prod2 * a[i];
   i--;
  }

  int prodRem2 = totalProduct / (prod2 * a[i]);

  return prodRem1 > prodRem2 ? prodRem1 : prodRem2;
 }
}