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.

Examples

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.

Approaches

1️⃣ Iteration

Code:

int findPermutationDifference(string s, string t) {
        int permutation = 0;
        for(int sIndex = 0; sIndex < s.size(); sIndex++){
            int tIndex = t.find(s[sIndex]);
            permutation = permutation + abs(sIndex - tIndex);
        }
        return permutation;
    }

Complexity Analysis:

• Time Complexity: O(n)
• Space Complexity: O(1)