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3498. Reverse Degree of a String

Description

Given a string s, calculate its reverse degree.

The reverse degree is calculated as follows:

  1. For each character, multiply its position in the reversed alphabet ('a' = 26, 'b' = 25, ..., 'z' = 1) with its position in the string (1-indexed).
  2. Sum these products for all characters in the string.

Return the reverse degree of s.

 

Example 1:

Input: s = "abc"

Output: 148

Explanation:

Letter Index in Reversed Alphabet Index in String Product
'a' 26 1 26
'b' 25 2 50
'c' 24 3 72

The reversed degree is 26 + 50 + 72 = 148.

Example 2:

Input: s = "zaza"

Output: 160

Explanation:

Letter Index in Reversed Alphabet Index in String Product
'z' 1 1 1
'a' 26 2 52
'z' 1 3 3
'a' 26 4 104

The reverse degree is 1 + 52 + 3 + 104 = 160.

 

Constraints:

  • 1 <= s.length <= 1000
  • s contains only lowercase English letters.

Solutions

Solution 1: Simulation

We can simulate the reverse degree of each character in the string. For each character, calculate its position in the reverse alphabet, multiply it by its position in the string, and then sum up all the results.

Time complexity is \(O(n)\), where \(n\) is the length of the string. Space complexity is \(O(1)\).

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class Solution:
    def reverseDegree(self, s: str) -> int:
        ans = 0
        for i, c in enumerate(s, 1):
            x = 26 - (ord(c) - ord("a"))
            ans += i * x
        return ans

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class Solution {
    public int reverseDegree(String s) {
        int n = s.length();
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            int x = 26 - (s.charAt(i - 1) - 'a');
            ans += i * x;
        }
        return ans;
    }
}

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class Solution {
public:
    int reverseDegree(string s) {
        int n = s.length();
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            int x = 26 - (s[i - 1] - 'a');
            ans += i * x;
        }
        return ans;
    }
};

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func reverseDegree(s string) (ans int) {
    for i, c := range s {
        x := 26 - int(c-'a')
        ans += (i + 1) * x
    }
    return
}

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function reverseDegree(s: string): number {
    let ans = 0;
    for (let i = 1; i <= s.length; ++i) {
        const x = 26 - (s.charCodeAt(i - 1) - 'a'.charCodeAt(0));
        ans += i * x;
    }
    return ans;
}

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