LeetCode 72. Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character

b) Delete a character

c) Replace a character

As indicated in the question. There are three ways to adjust two strings.

We can adjust the words either bottom up or top down.

1: Bottom Up: dp[i][j] = min(min(dp[i-1][j], dp[i][j-1])), dp[i-1][j-1]) + 1; 

dp[i-1][j-1] is to replace ith character to jth character. dp[i-1][j] is to delete ith character, dp[i][j-1] is to insert jth character.

#include <vector>
#include <string>
#include <iostream>
using namespace std;

int minDistance(string word1, string word2) {
    int m = word1.size();
    int n = word2.size();
    vector< vector<int> > dis(m + 1, vector<int>(n + 1, 0));
    for(int i = 0; i <= m; ++i) {
        dis[i][0] = i;
    }

    for(int i = 0; i <= n; ++i) {
        dis[0][i] = i;
    }

    for(int i = 1; i <= m; ++i) {
        for(int j = 1; j <= n; ++j) {
            if(word1[i-1] == word2[j-1]) {
                dis[i][j] = dis[i-1][j-1];
            } else {
                // in this case word[i-1] != word[j-1];
                // in this case, we have three choices. 1: delete, 2: insert, 3: replace.
                // dis[i-1][j-1] : replace
                // dis[i-1][j] : delete
                // dis[i][j-1] : insert
                dis[i][j] = min(min(dis[i-1][j], dis[i][j-1]), dis[i-1][j-1]) + 1;
            }
        }
    }
    return dis[m][n];
}

int main(void) {
    string word1 = "ab";
    string word2 = "abcd";
    int distance = minDistance(word1, word2);
    cout << distance << endl;
}

I honestly dont think I understand this question every well.... even though there is no problem with coding it out. It is better to practise multiple solutions!