# Edit Distance - Amazon Top Interview Questions

### Problem Statement :

```Given two strings a and b, find the minimum edit distance between the two. One edit distance is defined as

Deleting a character or
Inserting a character or
Replacing a character
Constraints

n ≤ 1,000 where n is the length of a
m ≤ 1,000 where m is the length of b

Example 1

Input

a = "zhello"

b = "helli"

Output

2

Explanation

"z" is removed and the "o" is replaced with "i"

Example 2

Input

a = "dycare"

b = "daycare"

Output

1

Explanation

"a" is inserted into the first string to get "daycare".```

### Solution :

```                        ```Solution in C++ :

vector<vector<int>> dp;
int get(string &a, string &b, int i, int j) {
if (i < 0 or j < 0) return max(i, j) + 1;
if (dp[i][j] != -1) return dp[i][j];
if (a[i] != b[j])
return dp[i][j] =
1 + min({get(a, b, i - 1, j), get(a, b, i, j - 1), get(a, b, i - 1, j - 1)});
return dp[i][j] = get(a, b, i - 1, j - 1);
}
int solve(string a, string b) {
dp.assign(a.size(), vector<int>(b.size(), -1));
return get(a, b, a.size() - 1, b.size() - 1);
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(String a, String b) {
int[][] edits = new int[b.length() + 1][a.length() + 1];
for (int i = 0; i < edits.length; i++) {
for (int j = 0; j < edits[0].length; j++) {
edits[0][j] = j;
}
edits[i][0] = i;
}
for (int i = 1; i < b.length() + 1; i++) {
for (int j = 1; j < a.length() + 1; j++) {
if (b.charAt(i - 1) == a.charAt(j - 1)) {
edits[i][j] = edits[i - 1][j - 1];
} else {
edits[i][j] = 1
+ Math.min(edits[i - 1][j - 1], Math.min(edits[i - 1][j], edits[i][j - 1]));
}
}
}
return edits[b.length()][a.length()];
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, a, b):
n, m = len(a), len(b)
dp = [[1e9 for _ in range(m + 1)] for p in range(n + 1)]
dp[0][0] = 0
for i in range(1, n):
dp[i][0] = i

for i in range(1, m):
dp[0][i] = i

for i in range(1, n + 1):
for j in range(1, m + 1):
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1
if a[i - 1] == b[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
return dp[n][m]```
```

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