1945. Sum of Digits of String After Convert

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.

Solutions

Solution 1

Python Code
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class Solution:
    def getLucky(self, s: str, k: int) -> int:
        s = ''.join(str(ord(c) - ord('a') + 1) for c in s)
        for _ in range(k):
            t = sum(int(c) for c in s)
            s = str(t)
        return int(s)

Java Code
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class Solution {
    public int getLucky(String s, int k) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            sb.append(c - 'a' + 1);
        }
        s = sb.toString();
        while (k-- > 0) {
            int t = 0;
            for (char c : s.toCharArray()) {
                t += c - '0';
            }
            s = String.valueOf(t);
        }
        return Integer.parseInt(s);
    }
}

C++ Code
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class Solution {
public:
    int getLucky(string s, int k) {
        string t;
        for (char c : s) t += to_string(c - 'a' + 1);
        s = t;
        while (k--) {
            int t = 0;
            for (char c : s) t += c - '0';
            s = to_string(t);
        }
        return stoi(s);
    }
};

Go Code
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func getLucky(s string, k int) int {
	var t strings.Builder
	for _, c := range s {
		t.WriteString(strconv.Itoa(int(c - 'a' + 1)))
	}
	s = t.String()
	for k > 0 {
		k--
		t := 0
		for _, c := range s {
			t += int(c - '0')
		}
		s = strconv.Itoa(t)
	}
	ans, _ := strconv.Atoi(s)
	return ans
}

TypeScript Code
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function getLucky(s: string, k: number): number {
    let ans = '';
    for (const c of s) {
        ans += c.charCodeAt(0) - 'a'.charCodeAt(0) + 1;
    }
    for (let i = 0; i < k; i++) {
        let t = 0;
        for (const v of ans) {
            t += Number(v);
        }
        ans = `${t}`;
    }
    return Number(ans);
}

Rust Code
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impl Solution {
    pub fn get_lucky(s: String, k: i32) -> i32 {
        let mut ans = String::new();
        for c in s.as_bytes() {
            ans.push_str(&(c - b'a' + 1).to_string());
        }
        for _ in 0..k {
            let mut t = 0;
            for c in ans.as_bytes() {
                t += (c - b'0') as i32;
            }
            ans = t.to_string();
        }
        ans.parse().unwrap()
    }
}

PHP Code
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class Solution {
    /**
     * @param String $s
     * @param Integer $k
     * @return Integer
     */
    function getLucky($s, $k) {
        $rs = '';
        for ($i = 0; $i < strlen($s); $i++) {
            $num = ord($s[$i]) - 96;
            $rs = $rs . strval($num);
        }
        while ($k != 0) {
            $sum = 0;
            for ($j = 0; $j < strlen($rs); $j++) {
                $sum += intval($rs[$j]);
            }
            $rs = strval($sum);
            $k--;
        }
        return intval($rs);
    }
}