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Java-based LeetCode algorithm problem solutions, regularly updated
3146\. Permutation Difference between Two Strings
Easy
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.