Introduction

In this blog post, we will explore how to solve the LeetCode problem "2186" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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 s and t. In one step, you can append any character to either s or t. Return the minimum number of steps to make s and t anagrams of each other. An anagram of a string is a string that contains the same characters with a different (or the same) ordering. Example 1: Input: s = "leetcode", t = "coats" Output: 7 Explanation: - In 2 steps, we can append the letters in "as" onto s = "leetcode", forming s = "leetcodeas". - In 5 steps, we can append the letters in "leede" onto t = "coats", forming t = "coatsleede". "leetcodeas" and "coatsleede" are now anagrams of each other. We used a total of 2 + 5 = 7 steps. It can be shown that there is no way to make them anagrams of each other with less than 7 steps. Example 2: Input: s = "night", t = "thing" Output: 0 Explanation: The given strings are already anagrams of each other. Thus, we do not need any further steps. Constraints: 1 <= s.length, t.length <= 2 * 105 s and t consist of lowercase English letters.

Explanation

Here's a breakdown of the solution and the code:

• High-Level Approach:

  • Count character frequencies in both strings.
  • Calculate the difference in frequencies for each character (a-z).
  • Sum the absolute differences to find the total characters that need to be added to either string to make them anagrams.

• Complexity:

  • Runtime: O(n + m), where n is the length of string s and m is the length of string t.
  • Storage: O(1), as we only store the counts for the 26 lowercase English letters, regardless of input string size.

Code

    def minSteps(s: str, t: str) -> int:
    """
    Calculates the minimum number of steps to make two strings anagrams of each other.
    """

    s_counts = {}
    t_counts = {}

    # Count character frequencies in string s
    for char in s:
        s_counts[char] = s_counts.get(char, 0) + 1

    # Count character frequencies in string t
    for char in t:
        t_counts[char] = t_counts.get(char, 0) + 1

    steps = 0
    alphabet = "abcdefghijklmnopqrstuvwxyz"

    # Calculate the absolute difference in frequencies for each character
    for char in alphabet:
        s_count = s_counts.get(char, 0)
        t_count = t_counts.get(char, 0)
        steps += abs(s_count - t_count)

    return steps // 2  # Divide by 2 because each difference represents two strings needing adjustment

# Optimized solution with single pass and character code
def minStepsOptimized(s: str, t: str) -> int:
    counts = [0] * 26
    for char in s:
        counts[ord(char) - ord('a')] += 1
    for char in t:
        counts[ord(char) - ord('a')] -= 1

    steps = 0
    for count in counts:
        steps += abs(count)
    return steps // 2 if steps % 2 == 0 else steps