Why Levenshtein distance from "abcd" to "badc" is 3?

Question

Starting from "abcd", I want to go to "badc" :

• I remove "a" -> "bcd";
• I insert "a" at the right position -> "bacd";
• I remove "c" -> "bad";
• I insert "c" at the right position -> "badc".

It's 4 operations. I can't find out a shorter way to do it. However, Levenshtein distance returns me a cost of 3. Why is that?

Thanks for your response.

Answer 1

You can go from abcd to badc in 3 operations. Remember you can do insertions, deletions and substitutions:

1. Remove a -> bcd.
2. Substitute c with a -> bad.
3. Add c at the end -> badc.

More generally, an insert + remove pair of operations can be coalesced into a substitution if the insertion and the removal are adjacent.