README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3110.Score%20of%20a%20String/README_EN.md	1152	Biweekly Contest 128 Q1	String

3110. Score of a String

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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 $O(1)$.

Python3

class Solution:
    def scoreOfString(self, s: str) -> int:
        return sum(abs(a - b) for a, b in pairwise(map(ord, s)))

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}