A string `s` is called **good** if there are no two different characters in `s` that have the same **frequency**.

Given a string `s`, return _the **minimum** number of characters you need to delete to make_ `s` _**good**._

The **frequency** of a character in a string is the number of times it appears in the string. For example, in the string `"aab "`, the **frequency** of `'a'` is `2`, while the **frequency** of `'b'` is `1`.

**Example 1:**

**Input:** s =  "aab "
**Output:** 0
**Explanation:** `s` is already good.

**Example 2:**

**Input:** s =  "aaabbbcc "
**Output:** 2
**Explanation:** You can delete two 'b's resulting in the good string  "aaabcc ".
Another way it to delete one 'b' and one 'c' resulting in the good string  "aaabbc ".

**Example 3:**

**Input:** s =  "ceabaacb "
**Output:** 2
**Explanation:** You can delete both 'c's resulting in the good string  "eabaab ".
Note that we only care about characters that are still in the string at the end (i.e. frequency of 0 is ignored).

**Constraints:**

*   `1 <= s.length <= 105`
*   `s` contains only lowercase English letters.


def can_convert(s, t, k):
    diff = 0
    for i in range(len(s)):
        curr_diff = (ord(t[i]) - ord(s[i]) + 26) % 26
        if curr_diff > 0:
            diff += curr_diff
            k -= curr_diff
    return k >= 0 and k >= diff - len(s) and k % 26 == 0
