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LeetCode 72 | Difficulty: Medium

Medium

Problem Description

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.

Topics: String, Dynamic Programming

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Approach

Dynamic Programming

Break the problem into overlapping subproblems. Define a state (what information do you need?), a recurrence (how does state[i] depend on smaller states?), and a base case. Consider both top-down (memoization) and bottom-up (tabulation) approaches.

When to use

Optimal substructure + overlapping subproblems (counting ways, min/max cost, feasibility).

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

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Solutions

Solution 1: C# (Best: 104 ms)

Metric	Value
Runtime	104 ms
Memory	N/A
Date	2018-03-05

Solution

public class Solution {
    public int MinDistance(string word1, string word2) {
        int m=word1.Length, n=word2.Length;
    int[,] dp = new int[m+1,n+1];
    for (int i = 0; i <= m; i++)
    {
        dp[i,0] = i;
    }
    for (int j = 0; j <= n; j++)
    {
        dp[0,j] = j;
    }
    for (int i = 1; i <= m; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if(word1[i-1]==word2[j-1])
            {
                dp[i,j] = dp[i-1,j-1];
            }
            else
            {
                dp[i, j] = Math.Min(Math.Min(dp[i - 1, j - 1] + 1,          dp[i, j - 1]+1 ),           dp[i - 1, j]+1);
            }
            
            
        }
    }
    return dp[m,n];
}
}

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Complexity Analysis

Approach	Time	Space
DP (2D)	$O(n × m)$	$O(n × m)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Define the DP state clearly. Ask: "What is the minimum information I need to make a decision at each step?"
• Consider if you can reduce space by only keeping the last row/few values.