LeetCode in Kotlin

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.

Solution

import kotlin.math.abs

class Solution {
    fun findPermutationDifference(s: String, t: String): Int {
        val res = IntArray(26)
        res.fill(-1)
        var sum = 0
        for (i in s.indices) {
            res[s[i].code - 'a'.code] = i
        }
        for (i in t.indices) {
            sum += abs((res[t[i].code - 'a'.code] - i))
        }
        return sum
    }
}

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