72. Edit Distance *HARD*

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

* Dynamic Programming
* Definitaion
* m[i][j] is minimal distance from word1[0..i] to word2[0..j]
* So,
* 1) if word1[i] == word2[j], then m[i][j] == m[i-1][j-1].
* 2) if word1[i] != word2[j], then we need to find which one below is minimal:
* min( m[i-1][j-1], m[i-1][j], m[i][j-1] ) and +1 - current char need be changed.
* Let's take a look m[1][2] : "a" => "ab"
* +---+ +---+
* ''=> a | 1 | | 2 | '' => ab
* +---+ +---+
* +---+ +---+
* a => a | 0 | | 1 | a => ab
* +---+ +---+
*
* To know the minimal distance `a => ab`, we can get it from one of the following cases:
* 1) delete the last char in word1, minDistance( '' => ab ) + 1
* 2) delete the last char in word2, minDistance( a => a ) + 1
* 3) change the last char, minDistance( '' => a ) + 1

* For Example:
* word1="abb", word2="abccb"
* 1) Initialize the DP matrix as below:
* "" a b c c b
* "" 0 1 2 3 4 5
* a 1
* b 2
* b 3
* 2) Dynamic Programming
* "" a b c c b
* "" 0 1 2 3 4 5
* a 1 0 1 2 3 4
* b 2 1 0 1 2 3
* b 3 2 1 1 1 2

int min(int x, int y, int z) {
    return std::min(x, std::min(y,z));
}

int minDistance(string word1, string word2) {
    int n1 = word1.size();
    int n2 = word2.size();
    if (n1==) return n2;
    if (n2==) return n1;
    vector< vector<int> > m(n1+, vector<int>(n2+));
    for(int i=; i<m.size(); i++){
        m[i][] = i;
    }
    for (int i=; i<m[].size(); i++) {
        m[][i]=i;
    }

    //Dynamic Programming
    int row, col;
    for (row=; row<m.size(); row++) {
        for(col=; col<m[row].size(); col++){
            if (word1[row-] == word2[col-] ){
                m[row][col] = m[row-][col-];
            }else{
                int minValue = min(m[row-][col-], m[row-][col],  m[row][col-]);
                m[row][col] = minValue + ;
            }
        }
    }

    return m[row-][col-];
}