Formatted question description: https://leetcode.ca/all/1945.html
1945. Sum of Digits of String After Convert
Level
Easy
Description
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" ➝ "(26)(2)(1)(24)" ➝ "262124" ➝ 262124
- Transform #1:
262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17
- Transform #2:
17 ➝ 1 + 7 ➝ 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” ➝ “(9)(9)(9)(9)” ➝ “9999” ➝ 9999
- Transform #1: 9999 ➝ 9 + 9 + 9 + 9 ➝ 36
Thus the resulting integer is 36.
Example 2:
Input: s = “leetcode”, k = 2
Output: 6
Explanation: The operations are as follows:
- Convert: “leetcode” ➝ “(12)(5)(5)(20)(3)(15)(4)(5)” ➝ “12552031545” ➝ 12552031545
- Transform #1: 12552031545 ➝ 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 ➝ 33
- Transform #2: 33 ➝ 3 + 3 ➝ 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.
Solution
Since k >= 1
, at least one transform operation is needed. Therefore, transform s
into an integer, which is guaranteed to fit in a 32-bit integer. Then do the transform for another k - 1
times.
class Solution {
public int getLucky(String s, int k) {
StringBuffer sb = new StringBuffer();
int length = s.length();
for (int i = 0; i < length; i++) {
char c = s.charAt(i);
int num = c - 'a' + 1;
sb.append(num);
}
int sum = 0;
int digitLength = sb.length();
for (int i = 0; i < digitLength; i++) {
int digit = sb.charAt(i) - '0';
sum += digit;
}
for (int i = 2; i <= k; i++)
sum = convert(sum);
return sum;
}
public int convert(int num) {
int sum = 0;
while (num > 0) {
sum += num % 10;
num /= 10;
}
return sum;
}
}