0072-edit-distance.cpp

/*
    Given 2 strings, return minimum number of operations to convert word1 to word2

    Naive: check all possible edit sequences & choose shortest one
    Optimal: DP, if chars at i & j same, no operations needed, else 3 cases:
    (1) replace (i - 1, j - 1), (2) delete (i - 1, j), (3) insert (i, j - 1)

    Time: O(m x n)
    Space: O(m x n)
*/

class Solution {
public:
    int minDistance(string word1, string word2) {
        if (word1.empty() && word2.empty()) {
            return 0;
        }
        if (word1.empty() || word2.empty()) {
            return 1;
        }
        
        int m = word1.size();
        int n = word2.size();
        
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        
        // base cases (convert to empty string w/ deletions), dist is just length
        for (int i = 1; i <= m; i++) {
            dp[i][0] = i;
        }
        for (int j = 1; 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]) {
                    // no operation needed, same char
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    // min(replace, delete, insert) + 1 <-- since an op was needed
                    dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        
        return dp[m][n];
    }
};

// Since we only need at most dp[i - 1][j - 1], can space optimize to O(n)
// class Solution {
// public:
//     int minDistance(string word1, string word2) {
//         int m = word1.size();
//         int n = word2.size();
//         int prev = 0;
//         vector<int> curr(n + 1);
//         for (int j = 1; j <= n; j++) {
//             curr[j] = j;
//         }
//         for (int i = 1; i <= m; i++) {
//             prev = curr[0];
//             curr[0] = i;
//             for (int j = 1; j <= n; j++) {
//                 int temp = curr[j];
//                 if (word1[i - 1] == word2[j - 1]) {
//                     curr[j] = prev;
//                 } else {
//                     curr[j] = min(prev, min(curr[j - 1], curr[j])) + 1;
//                 }
//                 prev = temp;
//             }
//         }
//         return curr[n];
//     }
// };