Leetcode #72: Edit Distance
In this guide, we solve Leetcode #72 Edit Distance in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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') Constraints: 0 <= word1.length, word2.length <= 500 word1 and word2 consist of lowercase English letters.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
Python Solution
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
m, n = len(word1), len(word2)
f = [[0] * (n + 1) for _ in range(m + 1)]
for j in range(1, n + 1):
f[0][j] = j
for i, a in enumerate(word1, 1):
f[i][0] = i
for j, b in enumerate(word2, 1):
if a == b:
f[i][j] = f[i - 1][j - 1]
else:
f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1
return f[m][n]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.