Leetcode #583: Delete Operation for Two Strings
In this guide, we solve Leetcode #583 Delete Operation for Two Strings 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 steps required to make word1 and word2 the same. In one step, you can delete exactly one character in either string.
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 = "sea", word2 = "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
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 i in range(1, m + 1):
f[i][0] = i
for j in range(1, n + 1):
f[0][j] = j
for i, a in enumerate(word1, 1):
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]) + 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.