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:
"a"
in s
and the index of the occurrence of "a"
in t
."b"
in s
and the index of the occurrence of "b"
in t
."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
s
.t
is a permutation of s
.s
consists only of lowercase English letters.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
}
}