# Minimum Edit Distance

Hi geeks! In this article, we going to see about Minimum Edit Distance problem with Dynamic Programming Approach.

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

1. Insert a character
2. Delete a character
3. 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')```

## C# Solution

```public class Solution {
public int MinDistance(string word1, string word2) {

//Boundary cases
if(string.IsNullOrEmpty(word1) && string.IsNullOrEmpty(word2))
return 0;
else if(string.IsNullOrEmpty(word1))
return word2.Length;
else if(string.IsNullOrEmpty(word2))
return word1.Length;

int m = word1.Length+1;
int n =  word2.Length+1;

int[,] dp = new int[m,n];

for(int i=0;i<m;i++)
{
//Row wise
dp[i,0] = i;
}

for(int j=0;j<n;j++)
{
dp[0,j] = j;
}

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] = Math.Min(Math.Min(dp[i,j-1],dp[i-1,j]),dp[i-1,j-1]) + 1;
}
}
}
return dp[m-1,n-1];

}
}```

Time Complexity: O(m*n)

Space Complexity: O(m*n)