Given two strings s and t of equal length, you may perform the following operation on t any number of times: choose an index in t and replace the character at that index with any lowercase English letter.

Return the minimum number of operations required to make t an anagram of s.

Note: An anagram is a rearrangement of characters that results in the same character counts for each letter.

Constraints:

• $1 \leq$ s.length $\leq 5 \cdot 10^4$

• s.length $==$ t.length

• s and t consist of lowercase English letters only.

Input: s = "bab"-str, t = "aba"-str
Output: 1-int
Explanation: Counting letters in both strings, s needs two b’s and one a, while t has one b and two a’s. t is missing one b, so changing one extra a in t to b takes 1 operation.

Input: s = "leetcode"-str, t = "practice"-str
Output: 5-int
Explanation: Comparing character frequencies, for each letter we count how many more times it appears in s than in t (only positive deficits matter). The total deficit across all letters is 5, so t must change 5 characters to match the letters of s.

Input: s = "anagram"-str, t = "mangaar"-str
Output: 0-int
Explanation: Both strings already contain the same count of each letter (they are anagrams), so no replacements are needed.

Input: s = "xxyyzz"-str, t = "xxxxxx"-str
Output: 4-int
Explanation: s requires x:2, y:2, and z:2, while t has x:6. t is missing 2 y’s and 2 z’s, so 4 replacements are required (changing 4 of the extra x’s into y’s and z’s).

Now, let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps you check if you’re solving the correct problem:

We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.

Implement your solution in the following coding playground.