LeetCode - Edit Distance

Edit Distance

2014.2.25 23:07

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

Solution1:

　　This is a typical problem of dynamic programming. Let f[i][j] be the edit distance of w1[1:i] and w2[1:j]:

　　　　1. f[0][0] = 0, all empty

　　　　2. f[i][0] = i, delete i letters

　　　　3. f[0][j] = j, insert j letters

　　　　4. f[i][j] could be f[i][j - 1] + 1, insert a letter.

　　　　5. f[i][j] could be f[i - 1][j] + 1, delete a letter.

　　　　6. If w1[i] == w2[j], f[i][j] could be f[i - 1][j - 1].

　　　　7. If w1[i] != w2[j], f[i][j] could be f[i - 1][j - 1] + 1, replace a letter.

　　f[i][j] will be the minimum of them all. The result is f[len1][len2].

　　The solution uses O(n^2) time and space.

Accepted code:

// 2CE, 2WA, 1AC, standard DP problem.
#include <algorithm>
#include <string>
using namespace std;

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1, n2;
        
        if (n1 < n2) {
            return minDistance(word2, word1);
        }
        
        n1 = (int)word1.length();
        n2 = (int)word2.length();
        if (n1 == 0) {
            return n2;
        } else if (n2 == 0) {
            return n1;
        }
        
        int **dp;

        dp = new int*[n1 + 1];
        dp[0] = new int[(n1 + 1) * (n2 + 1)];
        
        int i, j;
        for (i = 1; i <= n1; ++i) {
            dp[i] = dp[0] + (i * (n2 + 1));
        }
        
        dp[0][0] = 0;
        for (i = 1; i <= n1; ++i) {
            dp[i][0] = i;
        }
        for (j = 1; j <= n2; ++j) {
            dp[0][j] = j;
        }
        
        for (i = 1; i <= n1; ++i) {
            for (j = 1; j <= n2; ++j) {
                dp[i][j] = min(dp[i - 1][j] , dp[i][j - 1]) + 1;
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
                } else {
                    dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + 1);
                }
            }
        }
        int result = dp[n1][n2];
        
        delete[] dp[0];
        delete[] dp;
        
        return result;
    }
};

Solution2:

　　Obviously the space usage can be reduced to O(n) with proper adjustment.

　　See the code below.

Accepted code:

// 1AC, space-optimized version using DP.
#include <algorithm>
#include <string>
using namespace std;

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1, n2;
        
        if (n1 < n2) {
            return minDistance(word2, word1);
        }
        
        n1 = (int)word1.length();
        n2 = (int)word2.length();
        if (n1 == 0) {
            return n2;
        } else if (n2 == 0) {
            return n1;
        }
        
        int **dp;

        dp = new int*[2];
        dp[0] = new int[2 * (n2 + 1)];
        dp[1] = dp[0] + (n2 + 1);
        
        int i, j;
        
        for (j = 0; j <= n2; ++j) {
            dp[0][j] = j;
        }
        
        int flag = 1;
        for (i = 1; i <= n1; ++i) {
            dp[flag][0] = i;
            for (j = 1; j <= n2; ++j) {
                dp[flag][j] = min(dp[!flag][j] , dp[flag][j - 1]) + 1;
                if (word1[i - 1] == word2[j - 1]) {
                    dp[flag][j] = min(dp[flag][j], dp[!flag][j - 1]);
                } else {
                    dp[flag][j] = min(dp[flag][j], dp[!flag][j - 1] + 1);
                }
            }
            flag = !flag;
        }
        int result = dp[!flag][n2];
        
        delete[] dp[0];
        delete[] dp;
        
        return result;
    }
};