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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;
    }
    
    

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