# Solving Leetcode Interviews in Seconds with AI: Minimum Number of Steps to Make Two Strings Anagram


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1347" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given two strings of the same length s and t. In one step you can choose any character of t and replace it with another character. Return the minimum number of steps to make t an anagram of s. An Anagram of a string is a string that contains the same characters with a different (or the same) ordering.   Example 1:  Input: s = "bab", t = "aba" Output: 1 Explanation: Replace the first 'a' in t with b, t = "bba" which is anagram of s.  Example 2:  Input: s = "leetcode", t = "practice" Output: 5 Explanation: Replace 'p', 'r', 'a', 'i' and 'c' from t with proper characters to make t anagram of s.  Example 3:  Input: s = "anagram", t = "mangaar" Output: 0 Explanation: "anagram" and "mangaar" are anagrams.     Constraints:  1 <= s.length <= 5 * 104 s.length == t.length s and t consist of lowercase English letters only.  

	# Explanation
	Here's the breakdown of the solution:

*   **Character Frequency Counting:** Count the frequency of each character in both strings `s` and `t`.
*   **Identify Differences:** Iterate through the character frequencies. If `t` has more of a certain character than `s`, it needs to be replaced.
*   **Calculate Replacements:** Sum the differences between character counts where `t` has more occurrences than `s`. This sum represents the minimum number of replacements needed.

*   **Runtime Complexity:** O(N), where N is the length of the strings `s` and `t`.
*   **Storage Complexity:** O(1), as we use a fixed-size array (26 elements) to store character frequencies.

	
	# Code
	```python
	def min_steps_to_anagram(s: str, t: str) -> int:
    """
    Calculates the minimum number of steps to make t an anagram of s.

    Args:
        s: The first string.
        t: The second string.

    Returns:
        The minimum number of steps to make t an anagram of s.
    """

    s_counts = [0] * 26
    t_counts = [0] * 26

    for char in s:
        s_counts[ord(char) - ord('a')] += 1
    for char in t:
        t_counts[ord(char) - ord('a')] += 1

    steps = 0
    for i in range(26):
        if t_counts[i] > s_counts[i]:
            steps += t_counts[i] - s_counts[i]

    return steps
	```