Solved
Easy
Companies
Hint
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. 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.
class Solution {
public int findPermutationDifference(String s, String t) {
int[] sCharArr= new int[26], tCharArr=new int[26];
for(int i=0 ;i<s.length(); i++){
sCharArr[s.charAt(i)-'a'] = i;
}
for(int i=0 ;i<t.length(); i++){
tCharArr[t.charAt(i)-'a'] = i;
}
int ans=0;
for(int i=0 ;i<26; i++){
ans+= Math.abs(sCharArr[i] - tCharArr[i]);
}
return ans;
}
}
/**
3146. Permutation Difference between Two Strings
Solved
Easy
Companies
Hint
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.
*/
