# Solving Leetcode Interviews in Seconds with AI: Reverse Degree of a String


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3498" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given a string s, calculate its reverse degree. The reverse degree is calculated as follows:  For each character, multiply its position in the reversed alphabet ('a' = 26, 'b' = 25, ..., 'z' = 1) with its position in the string (1-indexed). 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.  

	# Explanation
	Here's the approach, complexity, and code for calculating the reverse degree of a string:

*   **Iterate through the string:** Process each character of the input string from left to right.
*   **Calculate the reversed alphabet position:** Determine the reversed alphabet position of each character (a=26, b=25, ..., z=1).  This can be efficiently done using the character's ASCII value.
*   **Accumulate the product:** Multiply the reversed alphabet position by the character's position in the string (1-indexed) and add it to the running sum, which represents the reverse degree.

*   **Runtime Complexity:** O(n), where n is the length of the string.
*   **Storage Complexity:** O(1), as we only use a few constant extra variables.

	
	# Code
	```python
	def reverse_degree(s: str) -> int:
    """
    Calculates the reverse degree of a string.

    The reverse degree is calculated as follows:
    For each character, multiply its position in the reversed alphabet ('a' = 26, 'b' = 25, ..., 'z' = 1)
    with its position in the string (1-indexed).
    Sum these products for all characters in the string.

    Example:
    reverse_degree("abc") == 148
    reverse_degree("zaza") == 160
    """
    reverse_degree_sum = 0
    for i, char in enumerate(s):
        reversed_alphabet_position = ord('z') - ord(char) + 1
        reverse_degree_sum += reversed_alphabet_position * (i + 1)
    return reverse_degree_sum
	```