CareerCup

left_partial = 1;
right_partial = 1;
for (i = 0 ; i < n; i++)
{
prod[i] = left_partial;
left_partial *= arr[i];
}

for (i = n - 1; i >= 0; i--)
{
prod[i] *= right_partial;
right_partial *= arr[i];
}

O(n) time complexity
O(1) space complexity

if element stored in int e[];
go from begin to tail, h[i]=h[i-1]*e[i]
go from tail to head, t[i]=t[i+1]*e[i]
then the result is prod[i]=h[i-1]*t[i+1]

int main()
{
int i,j,pro=1;
int prod[5],arr[10]={1,2,3,4,5};
for(i=0;i<5;++i)
{
for(j=0;j<5;++j)
{
if(i==j)
continue;
else
pro=pro*arr[j];
}
prod[i]=pro;
pro=1;
}
for(i=0;i<5;++i)
printf("%d\t",prod[i]);
printf("\n");
return 0;
}

Multiply all the numbers in one variable and than traverse the array and divide the product by the i-th to store i-th product value using division without using the division operator.
Division can be done using bit shifting method.

int prod =1;
for(int i=0;i<n;i++)
prod = prod*arr[i];

for(int i=0;i<n;i++)
prod[i] = prod/a[i];



is this solution acceptable?

no its not. problem says you should not use / operator

no its not. problem says you should not use / operator

What if the integer array contains a zreo? You cannot use divide in this case.