#1945

Easy

LeetCode Problem Workspace

Sum of Digits of String After Convert

Convert a lowercase string into digits and repeatedly sum them k times using a direct string plus simulation approach.

Problem Statement

Given a lowercase string s and an integer k, first convert each character in s to its alphabetical position ('a' = 1, 'b' = 2, ..., 'z' = 26) and concatenate the numbers to form a new integer. Then, perform a transformation k times where in each step you sum all digits of the current number.

Return the final integer obtained after performing these k transformations. For example, if s = "zbax" and k = 2, convert to 26 2 1 24 forming 262124, sum digits for the first transform, then sum digits again for the second transform to get the result.

Examples

Example 1

Input: s = "iiii", k = 1

Output: 36

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

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

Example details omitted.

Constraints

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

Solution Approach

Direct String Conversion

Map each character to its numeric position and concatenate to form a string representation of the integer. This avoids integer overflow and keeps the process clear for simulation.

Iterative Digit Summing

Iterate k times: in each iteration, sum all digits of the current number string, then replace the number string with this sum. This simulates the repeated transformation without recursion or complex math.

Optimization via Early Sum

After the first transformation, the number is at most 1000 (since 1 <= s.length <= 100 and max char = 26). For subsequent k-1 transformations, summing digits is fast, allowing the solution to run in linear time.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

Time complexity is O(n) for the initial string conversion and digit summing, where n is the length of s. Space complexity is O(1) since we reuse variables and the digit sum strings are short and bounded.

What Interviewers Usually Probe

Looking for correct mapping of letters to numbers without skipping steps.
Checking if the candidate handles repeated digit summation efficiently.
Verifying understanding of string plus simulation pattern to avoid unnecessary integer parsing.

Common Pitfalls or Variants

Common pitfalls

Trying to convert the entire concatenated string to integer at once, risking overflow.
Performing k transformations recursively without realizing iterative summing suffices.
Ignoring that after the first transform, the number is small enough for simple summing.

Follow-up variants

Change k to a large number and optimize by observing digital root properties.
Use uppercase letters and handle case-insensitive conversion to positions.
Instead of sum, perform product of digits repeatedly k times to modify the simulation pattern.

FAQ

How do I start converting the string in Sum of Digits of String After Convert?

Map each lowercase letter to its alphabetical position and join them as a string to form the initial number.

Can I perform the k transformations recursively?

Yes, but iterative summing is safer and faster since the number quickly becomes small after the first transform.

What is the maximum size of the number after initial conversion?

At most 2600 when s has length 100 and all letters are 'z', which is still manageable for string-based summing.

Does the problem pattern involve simulation?

Yes, it follows the String plus Simulation pattern by converting characters and simulating k transformations.

Why not convert the entire string to an integer immediately?

Direct integer conversion may overflow; using a string to sum digits prevents errors and aligns with simulation logic.

terminal

Solution

Solution 1: Simulation

We can simulate the process described in the problem.

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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)

Continue Practicing

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