Product array puzzle -searching|sorting

Given an array a[i] of size n, construct a Product Array p (size n) such that p[i] is equal to the product of all the elements of a except a[i].

Example 1:

Input: a[] = {10, 3, 5, 6, 2}
Output: p[] = {180, 600, 360, 300, 900}
The elements of output array are 
{3*5*6*2, 10*5*6*2, 10*3*6*2, 
10*3*5*2, 10*3*5*6}

 

Example 2:

Input: a[] = {1, 2, 1, 3, 4}
Output: p[] = {24, 12, 24, 8, 6}
The elements of output array are 
{3*4*1*2, 1*1*3*4, 4*3*2*1, 
1*1*4*2, 1*1*3*2}

Example 3:

input a[] = {12,0,1}
Output p[]: 0 12
{0*1, 12*1 , 12*0}

approach:

step 1: first traverse through the array and find the product of all the elements in the array and count the number of zeros in the array

step 2: if the number of zero's is equal to 1, than the index in the array a where zero was present should be filled with the product of all elements in the p array.

step 3: else if the number of zero's is greater than 1, all the elements in the p array will be zero.

step 4: else ,if there are no zero's , than all the elements in the p array will be filled with product / a[i]

implementation: in c++

long long int productExceptSelf(long long int a[], int n) 
{
   long long int pro=1;
   long long int zero=0;
   long long int p[n];

   for (int i=0;i<n;i++)
   {   
       if(i==0)
       zero++;
       else
       pro=pro*a[i];
   }

  if(zero==1)
  {
   for (int i=0;i<n;i++)
   {   
       if(i==0)
       p[i]=pro;
       else
       {
       p[i]=0;
       }
   }
  }
  
   else if(zero>1)
   { 
       for (int i=0;i<n;i++)
       p[i]=0;
   }

   else
   {
     for (int i=0;i<n;i++)
     p[i]=(pro/a[i]);
   }
}

time complexity: O(n),because of using a single loop

space complexity:O(n) , because of using an extra array p[]