Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.

Example 1:

```
Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
```

Note:

The length of given words won't exceed 500.

Characters in given words can only be lower-case letters.

Solution:

Approach 1: Using Longest Common subsequence

```
class Solution {
public int minDistance(String word1, String word2) {
int[][] dp = new int[word1.length() + 1][word2.length() + 1];
for (int i = 0; i <= word1.length(); i++) {
for (int j = 0; j <= word2.length(); j++) {
if (i == 0 || j == 0)
continue;
if (word1.charAt(i - 1) == word2.charAt(j - 1))
dp[i][j] = 1 + dp[i - 1][j - 1];
else
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
return word1.length() + word2.length() - 2 * dp[word1.length()][word2.length()];
}
}
```

Approach 2: Without LCS (dynammic programming)

```
public class Solution {
public int minDistance(String s1, String s2) {
int[][] dp = new int[s1.length() + 1][s2.length() + 1];
for (int i = 0; i <= s1.length(); i++) {
for (int j = 0; j <= s2.length(); j++) {
if (i == 0 || j == 0)
dp[i][j] = i + j;
else if (s1.charAt(i - 1) == s2.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[s1.length()][s2.length()];
}
}
```

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