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3110. Score of a String
Description
You are given a string s
. The score of a string is defined as the sum of the absolute difference between the ASCII values of adjacent characters.
Return the score of s
.
Example 1:
Input: s = "hello"
Output: 13
Explanation:
The ASCII values of the characters in s
are: 'h' = 104
, 'e' = 101
, 'l' = 108
, 'o' = 111
. So, the score of s
would be \|104 - 101\| + \|101 - 108\| + \|108 - 108\| + \|108 - 111\| = 3 + 7 + 0 + 3 = 13
.
Example 2:
Input: s = "zaz"
Output: 50
Explanation:
The ASCII values of the characters in s
are: 'z' = 122
, 'a' = 97
. So, the score of s
would be \|122 - 97\| + \|97 - 122\| = 25 + 25 = 50
.
Constraints:
2 <= s.length <= 100
s
consists only of lowercase English letters.
Solutions
Solution 1: Simulation
We directly traverse the string $s$, calculating the sum of the absolute differences of the ASCII codes of adjacent characters.
The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
-
class Solution { public int scoreOfString(String s) { int ans = 0; for (int i = 1; i < s.length(); ++i) { ans += Math.abs(s.charAt(i - 1) - s.charAt(i)); } return ans; } }
-
class Solution { public: int scoreOfString(string s) { int ans = 0; for (int i = 1; i < s.size(); ++i) { ans += abs(s[i] - s[i - 1]); } return ans; } };
-
class Solution: def scoreOfString(self, s: str) -> int: return sum(abs(a - b) for a, b in pairwise(map(ord, s)))
-
func scoreOfString(s string) (ans int) { for i := 1; i < len(s); i++ { ans += abs(int(s[i-1]) - int(s[i])) } return } func abs(x int) int { if x < 0 { return -x } return x }
-
function scoreOfString(s: string): number { let ans = 0; for (let i = 1; i < s.length; ++i) { ans += Math.abs(s.charCodeAt(i) - s.charCodeAt(i - 1)); } return ans; }