**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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