A Product Array Puzzle

Given an array arr[] of n integers, construct a Product Array prod[] (of same size) such that prod[i] is equal to the product of all the elements of arr[] except arr[i]. Solve it without division operator and in O(n).

The above method can be optimized to work in space complexity O(1). Thanks to Dileep for suggesting the below solution.
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void productArray(int arr[], int n)
{
int i, temp = 1;

/* Allocate memory for the product array */
int *prod = (int *)malloc(sizeof(int)*n);

/* Initialize the product array as 1 */
memset(prod,1,n);

/* In this loop, temp variable contains product of
elements on left side excluding arr[i] */
for(i=0; i {
prod[i] = temp;
temp *= arr[i];
}

/* Initialize temp to 1 for product on right side */
temp = 1;

/* In this loop, temp variable contains product of
elements on right side excluding arr[i] */
for(i= n-1; i>=0; i--)
{
prod[i] *= temp;
temp *= arr[i];
}

/* print the constructed prod array */
for (i=0; i printf("%d ", prod[i]);

return;
}

Time Complexity: O(n)
Space Complexity: O(n)
Auxiliary Space: O(1)


Optimization :

void ArrayMul(int buf[],int res[],size_t len) /*res[] is assumed to be of size len & all members initialized to 1 */
{
  size_t i;
  int left = 1,
		right = 1;

  for(i = 0; i<len;i++)
  {
	  res[i] *= left;
	  res[lenn-i-1] *= right;

	  left *= buf[i];
	  right *= buf[len-i-1];
  }
}