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int temp=1;
for(int i=0;i<size;i++)
temp*=b[i];
for(i=0;i<size;i++)
c[i]=a[i]*temp/b[i];
...............................O(n)

int[] getArrayC(int[] a, int[] b) {

      int i;
      int product = 1;
      int length = a.length();
      for ( i = 0 ; i < n ; i++ ) {
           product = product*b[i];
      }

      for ( i = 0 ; i < n ; i++ ) {
           c[i] = a[i]*product/b[i]; 
      }

      return c;

}

hi there. if the length of the array is 3, then how can we get b[3]?

Have two more arrays of size n.
X(i) = product of elements of B will i-1
Y(i) = product of elements of B from i+1 to n

these can be calulated in O(n)
now C(i) = A(i) * X(i) * Y(i)

Simple loop just once from i =0 to 2 and keep maintaining values as
j= i+1 % 3 and k = i+2 % 3

You can't get O(log n). It takes O(n) to just look at all the elements, which is necessary for answering this problem.