This approach required only four states, whereas the first approach required eight. The first approach uses three flip flops, whereas the second uses only two. This is not much of a difference. Consider, however, what happens if the problem required a 1 output iff the input has been 1 for 25 or more consecutive clock times. For the first approach, we would need to save the last 25 inputs, using 25 flip flops. The state table would have 225 rows. The second approach requires 26 states (no 1’s through 25 or more 1’s). They could be coded with just five flip flops.