permutation-difference-between-two-strings_3146.py

# 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| + |2 - 2| + |1 - 0| = 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.

# ---------------------------------------Runtime 37 ms Beats 100.00% Memory 16.58 MB Beats 100.00%---------------------------------------


class Solution:
    def findPermutationDifference(self, s: str, t: str) -> int:
        r = list(range(len(s)))
        s_map = dict(zip(s, r))
        t_map = dict(zip(t, r))
        return sum([abs(s_map[char] - t_map[char]) for char in s])