Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.
You have the following three operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
Example 1:
Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution" Output: 5 Explanation: intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')
0 <= word1.length, word2.length <= 500word1andword2consist of lowercase English letters.
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)]
# Base cases initialization
for i in range(1, m + 1):
dp[i][0] = i # Cost of deleting all characters from word1[0:i]
for j in range(1, n + 1):
dp[0][j] = j # Cost of inserting all characters to form word2[0:j]
# Fill the dp table
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] # No operation needed
else:
dp[i][j] = min(dp[i - 1][j] + 1, # Delete
dp[i][j - 1] + 1, # Insert
dp[i - 1][j - 1] + 1) # Replace
return dp[m][n]O(m×n), where m is the length of word1 and n is the length of word2. This complexity arises from the nested loops required to fill the DP table.
O(m×n) for the DP table. This can be optimized to O(n) using a rolling array technique if minimizing space is critical, by maintaining only the current and the previous row of the DP table.