/*

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

*/

// Levenshtein distance

// a string metric for measuring the difference

// between two sequences. Informally, the Levenshtein distance between two

// words is the minimum number of single-character edits (i.e. insertions,

// deletions or substitutions) required to change one word into the other. The

// phrase edit distance is often used to refer specifically to Levenshtein

// distance

// Another string distance measurement is Hamming distance, only allows substitions

// DP

// word1.substring(0, i) can be converted into word2.substring(0, j) using mem[i][j] steps

// mem[i][j] = mem[i][j-1], denotes a insertion operation in word1.substring

// mem[i][j] = mem[i-1][j], denotes an deletion operation in word1.substring

// mem[i][j] = mem[i-1][j-1], denotes an replace operation in word1.substring

// time: O(m*n); space: O(m*n)

public class Solution {

public int minDistance(String word1, String word2) {

if (word1==null || word2==null){

if (word1==null && word2==null) return 0;

return word1==null ? word2.length() : word1.length();

}

int M = word1.length(), N = word2.length();

int[][] dp = new int[M+1][N+1];

for (int i=1; i<=M; i++)

dp[i][0] = i;

for (int i=1; i<=N; i++)

dp[0][i] = i;

for (int i=1; i<=M; i++){

for (int j=1; j<=N; j++){

if (word1.charAt(i-1)==word2.charAt(j-1))

dp[i][j] = dp[i-1][j-1];

else

dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1])+1;

}

}

return dp[M][N];

}

}