Edit Distance


Problem Statement :


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.


Solution :



title-img 


                            Solution in C :

#define min(a,b,c) ((a)<(b)?((a)<(c)?(a):(c)):((b)<(c)?(b):(c)))

int minDistance(char * word1, char * word2){    
    int len1 = strlen(word1);
    int len2 = strlen(word2);
	
    if(len1 == 0)
        return len2;
    if(len2 == 0)
        return len1;
    
    int i, j;
    int** table = (int**)malloc(len1 * sizeof(int*));
    for(i = 0; i < len1; i++){
        table[i] = (int*)malloc(len2 * sizeof(int));
    }
    
    if(word1[0] == word2[0])
        table[0][0] = 0;
    else
        table[0][0] = 1;
    // 第一列
        for(i = 1; i < len2; i++){
            if(word1[0] == word2[i])
                table[0][i] = i;
            else
                table[0][i] = table[0][i-1] + 1;
        }
    //第一行
        for(i = 1; i < len1; i++){
            if(word2[0] == word1[i])
                table[i][0] = i;
            else
                table[i][0] = table[i-1][0] + 1;
        }
    //處理其他
    for(i = 1; i < len1; i++){
        for(j = 1; j < len2; j++){
            if(word1[i] == word2[j])
                table[i][j] = table[i-1][j-1];
            
            else 
                 table[i][j] = 1 + min(table[i-1][j], table[i][j-1], table[i-1][j-1]); 
        }
    }
    
    int ans = table[len1-1][len2-1];
    for(i = 0; i < len1; i++){
        free(table[i]);
    }
    free(table);
    
    return ans;

}
                        


                        Solution in C++ :

class Solution {
public:
     int minDistance(string word1, string word2) {
        
        int m=word1.length(),n=word2.length();
        
        int dp[m+1][n+1];
        
        for(int i=0;i<=m;i++)
            dp[i][0]=i;
        
        for(int i=0;i<=n;i++)
            dp[0][i]=i;
        
        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] = 1+ min(dp[i][j - 1], min(dp[i - 1][j], dp[i - 1][j - 1]));
                }
            }
        }
        return dp[m][n];
    }
};
                    


                        Solution in Java :

class Solution {
    public int minDistance(String word1, String word2) {
        if (word1 == null || word2 == null) {
            throw new IllegalArgumentException("Input strings are null");
        }

        int insertCost = 1;
        int deleteCost = 1;
        int replaceCost = 1;
        int l1 = word1.length();
        int l2 = word2.length();

        if (l1 == 0) {
            return l2 * insertCost;
        }
        if (l2 == 0) {
            return l1 * deleteCost;
        }
        // Bellow condition can be used only if all three costs are same.
        if (l1 > l2) {
            return minDistance(word2, word1);
        }

        int[] dp = new int[l1 + 1];
        // Setting DP array for 0th column of 2D DP array. Here l2 is blank, thus we
        // have to delete everything in l1.
        for (int i = 1; i <= l1; i++) {
            dp[i] = dp[i - 1] + deleteCost;
        }

        for (int j = 1; j <= l2; j++) {
            int prev = dp[0];
            dp[0] += insertCost; // l1 is blank, Inserting l2 chars in l1.
            char c2 = word2.charAt(j - 1);
            for (int i = 1; i <= l1; i++) {
                char c1 = word1.charAt(i - 1);
                int temp = dp[i];
                // Both chars are same, so the distance will also remain same as dp[i-1][j-1]
                if (c1 == c2) {
                    dp[i] = prev;
                } else {
                    // Replace l1[i - 1] by l2[j - 1] ==> dp[i][j] = dp[i - 1][j - 1] + 1
                    // Delete l1[i-1] from l1[0..i-1] ==> dp[i-1][j] + 1
                    // Insert l2[j-1] into l1[0..i-1] ==> dp[i][j-1] + 1
                    dp[i] = Math.min(prev + replaceCost, Math.min(dp[i - 1] + deleteCost, dp[i] + insertCost));
                }
                prev = temp;
            }
        }

        return dp[l1];
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def minDistance(self, s1: str, s2: str) -> int:

        @lru_cache(maxsize=None)
        def f(i, j):
            if i == 0 and j == 0: return 0
            if i == 0 or j == 0: return i or j
            if s1[i - 1] == s2[j - 1]:
                return f(i - 1, j - 1)
            return min(1 + f(i, j - 1), 1 + f(i - 1, j), 1 + f(i - 1, j - 1))

        m, n = len(s1), len(s2)
        return f(m, n)


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