# 3146. Permutation Difference between Two Strings

## Description

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.

## Solutions

### Solution 1

• class Solution {
public int findPermutationDifference(String s, String t) {
int[] d = new int[26];
int n = s.length();
for (int i = 0; i < n; ++i) {
d[s.charAt(i) - 'a'] = i;
}
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += Math.abs(d[t.charAt(i) - 'a'] - i);
}
return ans;
}
}

• class Solution {
public:
int findPermutationDifference(string s, string t) {
int d[26]{};
int n = s.size();
for (int i = 0; i < n; ++i) {
d[s[i] - 'a'] = i;
}
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += abs(d[t[i] - 'a'] - i);
}
return ans;
}
};

• class Solution:
def findPermutationDifference(self, s: str, t: str) -> int:
d = {c: i for i, c in enumerate(s)}
return sum(abs(d[c] - i) for i, c in enumerate(t))


• func findPermutationDifference(s string, t string) (ans int) {
d := [26]int{}
for i, c := range s {
d[c-'a'] = i
}
for i, c := range t {
ans += max(d[c-'a']-i, i-d[c-'a'])
}
return
}

• function findPermutationDifference(s: string, t: string): number {
const d: number[] = Array(26).fill(0);
const n = s.length;
for (let i = 0; i < n; ++i) {
d[s.charCodeAt(i) - 'a'.charCodeAt(0)] = i;
}
let ans = 0;
for (let i = 0; i < n; ++i) {
ans += Math.abs(d[t.charCodeAt(i) - 'a'.charCodeAt(0)] - i);
}
return ans;
}