3498. Reverse Degree of a String

String
Simulation

TIME

O(n)

SPACE

O(1)

Intuition

The definition is already the algorithm. The only translation needed is from a character to its reversed-alphabet position: 'a' is offset 0 from 'a' and should map to 26, 'z' is offset 25 and should map to 1, so the value is 26 - offset.

Approach

Walk the string with a 0-based index i.
For character c, compute weight = 26 - (ord(c) - ord('a')) and position = i + 1.
Accumulate weight * position.

Complexity

Time: O(n)
Space: O(1)

Edge Cases

Single character "a" → 26, "z" → 1
All 'z' → sum of positions n(n+1)/2
Maximum value: n = 1000, all 'a' → 26 * 1000 * 1001 / 2 = 13,013,000, fits in a 32-bit int

Code

class Solution:
    def reverseDegree(self, s: str) -> int:
        """
        3498. Reverse Degree of a String
        Time: O(n)
        Space: O(1)
        """
        total = 0
        for i, ch in enumerate(s):
            weight = 26 - (ord(ch) - ord("a"))
            total += weight * (i + 1)
        return total

/**
 * 3498. Reverse Degree of a String
 * Time: O(n)
 * Space: O(1)
 */
class Solution {
    public int reverseDegree(String s) {
        int total = 0;
        for (int i = 0; i < s.length(); i++) {
            int weight = 26 - (s.charAt(i) - 'a');
            total += weight * (i + 1);
        }
        return total;
    }
}