Edit Distance

Intuition

Define dp[i][j] as the minimum number of operations to convert the first i characters of word1 into the first j characters of word2. Turning any prefix into the empty string costs one deletion per character, and building a prefix from the empty string costs one insertion per character, which fixes the base row and column as their own indices. When the current characters word1[i-1] and word2[j-1] are equal, no work is needed at this position, so dp[i][j] inherits dp[i-1][j-1]. When they differ, we take one operation plus the cheapest of three moves: an insert from dp[i][j-1], a delete from dp[i-1][j], or a replace from dp[i-1][j-1]. Filling the table in increasing order of i and j ensures each dependency is already computed.

Algorithm walkthrough

1. 1Allocate a table of size (m + 1) by (n + 1) where m and n are the two string lengths.
2. 2Initialize the first column with dp[i][0] = i and the first row with dp[0][j] = j to encode pure deletes and inserts.
3. 3Scan i from 1 to m and j from 1 to n, filling each cell from its neighbors.
4. 4If word1[i-1] equals word2[j-1], copy the diagonal value dp[i-1][j-1] with no added cost.
5. 5Otherwise set dp[i][j] to 1 plus the minimum of the left, top, and diagonal cells for insert, delete, and replace.
6. 6The bottom-right cell dp[m][n] is the minimum edit distance for the full strings.

Implementation

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(m + 1):
            dp[i][0] = i
        for j in range(n + 1):
            dp[0][j] = j
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    dp[i][j] = 1 + min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1])
        return dp[m][n]