------
## 最终结论
```python
def min_steps_to_anagram(s, t):
    from collections import Counter
    
    freq_s = Counter(s)
    freq_t = Counter(t)
    
    # Calculate the number of steps required to make t an anagram of s
    steps = 0
    for char in freq_s:
        if char in freq_t:
            steps += max(0, freq_s[char] - freq_t[char])
        else:
            steps += freq_s[char]
    
    return steps

# Example usage:
s1 = "bab"
t1 = "aba"
print(min_steps_to_anagram(s1, t1))  # Output: 1

s2 = "leetcode"
t2 = "practice"
print(min_steps_to_anagram(s2, t2))  # Output: 5

s3 = "anagram"
t3 = "mangaar"
print(min_steps_to_anagram(s3, t3))  # Output: 0

s4 = "xxyyzz"
t4 = "xxyyzz"
print(min_steps_to_anagram(s4, t4))  # Output: 0

s5 = "friend"
t5 = "family"
print(min_steps_to_anagram(s5, t5))  # Output: 4
```