1945. Sum of Digits of String After Convert LeetCode Solution
In this guide, you will get 1945. Sum of Digits of String After Convert LeetCode Solution with the best time and space complexity. The solution to Sum of Digits of String After Convert problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Sum of Digits of String After Convert
You are given a string s consisting of lowercase English letters, and an integer k. Your task is to convert the string into an integer by a special process, and then transform it by summing its digits repeatedly k times. More specifically, perform the following steps:
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).
Transform the integer by replacing it with the sum of its digits.
Repeat the transform operation (step 2) k times in total.
For example, if s = “zbax” and k = 2, then the resulting integer would be 8 by the following operations:
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.
Complexity Analysis
Time Complexity:
Space Complexity:
1945. Sum of Digits of String After Convert LeetCode Solution in C++
class Solution {
public:
int getLucky(string s, int k) {
int ans = convert(s);
for (int i = 1; i < k; ++i)
ans = getDigitSum(ans);
return ans;
}
private:
int convert(string s) {
int sum = 0;
for (const char c : s) {
const int val = c - 'a' + 1;
// Do one transform to prevent integer overflow.
sum += val < 10 ? val : (val % 10 + val / 10);
}
return sum;
}
int getDigitSum(int num) {
int digitSum = 0;
while (num > 0) {
digitSum += num % 10;
num /= 10;
}
return digitSum;
}
}; /* code provided by PROGIEZ */
1945. Sum of Digits of String After Convert LeetCode Solution in Java
class Solution {
public int getLucky(String s, int k) {
int ans = convert(s);
for (int i = 1; i < k; ++i)
ans = getDigitSum(ans);
return ans;
}
private int convert(final String s) {
int sum = 0;
for (final char c : s.toCharArray()) {
final int val = c - 'a' + 1;
// Do one transform to prevent integer overflow.
sum += val < 10 ? val : (val % 10 + val / 10);
}
return sum;
}
private int getDigitSum(int num) {
int digitSum = 0;
while (num > 0) {
digitSum += num % 10;
num /= 10;
}
return digitSum;
}
} // code provided by PROGIEZ
1945. Sum of Digits of String After Convert LeetCode Solution in Python
class Solution:
def getLucky(self, s: str, k: int) -> int:
ans = self._convert(s)
for _ in range(k):
ans = self._getDigitSum(ans)
return ans
def _convert(self, s: str) -> int:
return int(''.join(str(ord(c) - ord('a') + 1) for c in s))
def _getDigitSum(self, num: int) -> int:
return sum(int(digit) for digit in str(num)) # code by PROGIEZ
