Introduction

In this blog post, we will explore how to solve the LeetCode problem "3146" 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 such that every character occurs at most once in s and t is a permutation of s. The permutation difference between s and t is defined as the sum of the absolute difference between the index of the occurrence of each character in s and the index of the occurrence of the same character in t. Return the permutation difference between s and t. Example 1: Input: s = "abc", t = "bac" Output: 2 Explanation: For s = "abc" and t = "bac", the permutation difference of s and t is equal to the sum of: The absolute difference between the index of the occurrence of "a" in s and the index of the occurrence of "a" in t. The absolute difference between the index of the occurrence of "b" in s and the index of the occurrence of "b" in t. The absolute difference between the index of the occurrence of "c" in s and the index of the occurrence of "c" in t. That is, the permutation difference between s and t is equal to |0 - 1| + |1 - 0| + |2 - 2| = 2. Example 2: Input: s = "abcde", t = "edbac" Output: 12 Explanation: The permutation difference between s and t is equal to |0 - 3| + |1 - 2| + |2 - 4| + |3 - 1| + |4 - 0| = 12. Constraints: 1 <= s.length <= 26 Each character occurs at most once in s. t is a permutation of s. s consists only of lowercase English letters.

Explanation

Here's an efficient solution to calculate the permutation difference between two strings s and t, along with explanations:

• High-level approach:

  • Create a dictionary (hash map) to store the index of each character in string s. This allows for O(1) lookup of the index of a character in s.
  • Iterate through string t, and for each character, find its index in t and its corresponding index in s using the dictionary.
  • Calculate the absolute difference between these indices and accumulate the sum.

• Complexity Analysis:

  • Runtime Complexity: O(n), where n is the length of the strings s and t.
  • Storage Complexity: O(1), because the hash map will store at most 26 characters (since the length of s is at most 26 and each character occurs at most once). This can be considered constant.

Code

    def permutation_difference(s: str, t: str) -> int:
    """
    Calculates the permutation difference between strings s and t.

    Args:
        s: The first string.
        t: The second string, which is a permutation of s.

    Returns:
        The permutation difference between s and t.
    """

    s_index = {}
    for i in range(len(s)):
        s_index[s[i]] = i

    permutation_diff = 0
    for i in range(len(t)):
        permutation_diff += abs(i - s_index[t[i]])

    return permutation_diff