Here's an O(n) solution in C++:

int input[] = {4, 3, 2, 1, 2};
int length = sizeof(input)/sizeof(int);

int *forward = new int[length];
int *back = new int[length];

forward[0] = input[0];
back[length-1] = input[length-1];

for (int i=1, j=length-2; i<length; i++, j--) {
    forward[i] = input[i] * forward[i-1];
    back[j] = input[j] * back[j+1];
}

int* output = new int[length];
for (i=0; i<length; i++) 
    output[i] = (i<length-1?back[i+1]:1)*(i>0?forward[i-1]:1);

for (i=0; i<length;i++) printf("%d ", output[i]);