# 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;
}