To solve the Edit Distance problem, we need to think about how one string can be gradually transformed into another. We are allowed three operations: insert, delete, or replace a character. Each of these costs 1 unit, and the goal is to reach the target string in the fewest steps.
Let's say we are trying to convert word1
to word2
:
- If both strings are empty, then no operations are needed — the distance is 0.
- If one string is empty and the other is not, then the only option is to insert or delete characters one-by-one. For example, converting "" to "abc" takes 3 insertions.
- If the characters at the current position in both strings are the same, we don’t need to do anything — we just move to the next characters.
- But if the characters are different, we have three choices:
- Insert: Imagine we insert the next character from
word2
into word1
.
- Delete: Imagine we delete the current character from
word1
.
- Replace: Imagine we replace the current character of
word1
with the character from word2
.
We pick the operation with the smallest cost and add 1 to it.
To make this efficient, we store the results of already-solved subproblems in a 2D table. Each cell in the table dp[i][j]
tells us the edit distance between the first i
characters of word1
and the first j
characters of word2
.
By building this table from the base cases (like when one of the strings is empty), we ensure we don’t compute the same values again and again. The final answer will be at dp[m][n]
, where m
and n
are the lengths of word1
and word2
respectively.
This approach ensures we consider all possibilities and always take the minimum number of operations needed, making it suitable even for large strings.