How do you solve Edit Distance on LeetCode?

Edit Distance (LeetCode #72) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Dynamic programming is essential for optimizing recursive problems.

What is the brute force solution for Edit Distance?

The brute force approach for Edit Distance is Brute Force, with O(3^n) time complexity and O(n) space complexity. The brute force approach involves recursively calculating the minimum operations needed to convert one string into another by trying all possible operations (insert, delete, replace) at each step. This method explores every possible combination, which can be very inefficient for longer strings. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Edit Distance?

Edit Distance uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Edit Distance?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches before jumping into coding. Practice writing the dynamic programming table as it helps in visualizing the problem.

What are common mistakes when solving Edit Distance?

Not initializing the dp array correctly. Forgetting to handle base cases when one string is empty.

#72

Edit Distance

Medium

StringDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(3^n)	O(m * n)
Space	O(n)	O(m * n)

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Intuition

Time O(m * n)Space O(m * n)

The optimal solution uses dynamic programming to build a table that stores the minimum edit distances for all prefixes of the two strings. This avoids redundant calculations by reusing previously computed results.

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Algorithm

4 steps

1Step 1: Create a 2D array dp where dp[i][j] represents the edit distance between the first i characters of word1 and the first j characters of word2.
2Step 2: Initialize the first row and column of the dp array based on the lengths of the strings.
3Step 3: Fill the dp array using the recurrence relation based on the three operations (insert, delete, replace).
4Step 4: Return the value in dp[word1.length][word2.length] as the result.

solution.py12 lines

1def minDistance(word1, word2):
2    m, n = len(word1), len(word2)
3    dp = [[0] * (n + 1) for _ in range(m + 1)]
4    for i in range(m + 1): dp[i][0] = i
5    for j in range(n + 1): dp[0][j] = j
6    for i in range(1, m + 1):
7        for j in range(1, n + 1):
8            if word1[i - 1] == word2[j - 1]:
9                dp[i][j] = dp[i - 1][j - 1]
10            else:
11                dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1])
12    return dp[m][n]

ℹ

Complexity note: The time complexity is quadratic because we fill a 2D table of size m x n, where m and n are the lengths of the two strings. The space complexity is also quadratic due to the storage of the dp table.

1Dynamic programming is essential for optimizing recursive problems.
2Understanding how to build and utilize a table for storing intermediate results is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.