Skip to content

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 \(O(1)\).

1
2
3
class Solution:
    def scoreOfString(self, s: str) -> int:
        return sum(abs(a - b) for a, b in pairwise(map(ord, s)))

1
2
3
4
5
6
7
8
9
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
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
}

1
2
3
4
5
6
7
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;
}

Comments