# Delete Characters to Equalize Strings - Google Top Interview Questions

### Problem Statement :

```Given two lowercase alphabet strings a and b, consider an operation where we delete any character in either string.
Return the minimum number of operations required such that both strings are equal.

Constraints

0 ≤ n ≤ 1,000 where n is the length of a

0 ≤ m ≤ 1,000 where m is the length of b

Example 1

Input

a = "zyyx"

b = "yfyx"

Output

2

Explanation

We delete the "z" in a and delete "f" in b```

### Solution :

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

int solve(string a, string b) {
int m = a.size();
int n = b.size();

int dp[2][n + 1];
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0)
dp[i % 2][j] = j;
else if (j == 0)
dp[i % 2][j] = i;
else if (a[i - 1] == b[j - 1])
dp[i % 2][j] = dp[1 - i % 2][j - 1];
else
dp[i % 2][j] = 1 + min(dp[1 - i % 2][j], dp[i % 2][j - 1]);
}
}

return dp[m % 2][n];
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int dp(String a, String b, int i, int j, int[][] memo) {
if (i == a.length() && j == b.length()) {
return 0;
}
if (i == a.length()) {
return b.length() - j;
}
if (j == b.length()) {
return a.length() - i;
}

if (memo[i][j] != -1) {
return memo[i][j];
}
int ans = Integer.MAX_VALUE;
if (a.charAt(i) == b.charAt(j)) {
ans = Math.min(ans, dp(a, b, i + 1, j + 1, memo));
} else {
ans = Math.min(ans, 1 + dp(a, b, i + 1, j, memo));
ans = Math.min(ans, 1 + dp(a, b, i, j + 1, memo));
}
return memo[i][j] = ans;
}
public int solve(String a, String b) {
int[][] memo = new int[a.length() + 5][b.length() + 5];
for (int[] x : memo) {
Arrays.fill(x, -1);
}
return dp(a, b, 0, 0, memo);
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, a, b):

m = len(a)
n = len(b)

@functools.lru_cache(None)
def dp(i, j):
if i == m:
return n - j
elif j == n:
return m - i
else:
if a[i] == b[j]:
return dp(i + 1, j + 1)
else:
return 1 + min(dp(i + 1, j), dp(i, j + 1))

return dp(0, 0)```
```

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