Package g1901_2000.s1945_sum_of_digits_of_string_after_convert

Class Solution

java.lang.Object
g1901_2000.s1945_sum_of_digits_of_string_after_convert.Solution
public class Solution extends Object
1945 - Sum of Digits of String After Convert.

Easy

You are given a string s consisting of lowercase English letters, and an integer k.

First, convert s into an integer by replacing each letter with its position in the alphabet (i.e., replace 'a' with 1, 'b' with 2, …, 'z' with 26). Then, transform the integer by replacing it with the sum of its digits. Repeat the transform operation k times in total.

For example, if s = "zbax" and k = 2, then the resulting integer would be 8 by the following operations:

  • Convert: "zbax" \u279d "(26)(2)(1)(24)" \u279d "262124" \u279d 262124
  • Transform #1: 262124 \u279d 2 + 6 + 2 + 1 + 2 + 4 \u279d 17
  • Transform #2: 17 \u279d 1 + 7 \u279d 8

Return the resulting integer after performing the operations described above.

Example 1:

Input: s = “iiii”, k = 1

Output: 36

Explanation: The operations are as follows:

  • Convert: “iiii” \u279d “(9)(9)(9)(9)” \u279d “9999” \u279d 9999

  • Transform #1: 9999 \u279d 9 + 9 + 9 + 9 \u279d 36 Thus the resulting integer is 36.

Example 2:

Input: s = “leetcode”, k = 2

Output: 6

Explanation: The operations are as follows:

  • Convert: “leetcode” \u279d “(12)(5)(5)(20)(3)(15)(4)(5)” \u279d “12552031545” \u279d 12552031545

  • Transform #1: 12552031545 \u279d 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 \u279d 33

  • Transform #2: 33 \u279d 3 + 3 \u279d 6 Thus the resulting integer is 6.

Example 3:

Input: s = “zbax”, k = 2

Output: 8

Constraints:

  • 1 <= s.length <= 100
  • 1 <= k <= 10
  • s consists of lowercase English letters.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • getLucky

      public int getLucky(String s, int k)