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:
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
).k
times in total.For example, if s = "zbax"
and k = 2
, then the resulting integer would be 8
by the following operations:
"zbax" ➝ "(26)(2)(1)(24)" ➝ "262124" ➝ 262124
262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17
17 ➝ 1 + 7 ➝ 8
Return the resulting integer after performing the operations described above.
Example:
s = "leetcode", k = 2
Output: 6
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
This problem involves converting a string into a number, then repeatedly summing its digits until a single-digit number remains. The brute force approach simulates this process exactly as described by the problem.
Here's how the algorithm would work step-by-step:
def sum_of_digits_brute_force(input_string, k_iterations):
string_sum = 0
# Convert each letter to number and sum
for character in input_string:
string_sum += ord(character) - ord('a') + 1
# Perform k iterations
for _ in range(k_iterations):
new_sum = 0
# Calculate the digit sum
while string_sum > 0:
new_sum += string_sum % 10
string_sum //= 10
string_sum = new_sum
return string_sum
The goal is to convert letters to numbers, add them up, and then repeat the adding process a few times. We efficiently calculate the sums in each round without storing large intermediary strings or doing unnecessary conversions.
Here's how the algorithm would work step-by-step:
def sum_of_digits_of_string_after_convert(input_string, k_iterations):
initial_sum = 0
for char in input_string:
# Convert each letter to its number and add to initial sum
initial_sum += ord(char) - ord('a') + 1
current_sum = initial_sum
for _ in range(k_iterations):
next_sum = 0
# Iterate through the digits of the current sum
while current_sum > 0:
next_sum += current_sum % 10
current_sum //= 10
# Update the current sum for the next iteration
current_sum = next_sum
return current_sum
Case | How to Handle |
---|---|
Null or empty string input | Return 0 immediately as there are no digits to sum. |
Input string contains non-alphabetic characters | Filter out non-alphabetic characters before processing the string to avoid incorrect conversions. |
Large k value causing multiple iterations | Ensure the solution has reasonable time complexity even with large k, preventing potential timeouts. |
String 'a' * very large number | The sum will overflow if not using long or arbitrary precision integers, so use appropriate data types to store the sum. |
k is zero | Return the initial conversion result without performing any further sum of digits operations. |
String resulting in a single-digit sum after one conversion | The while loop should terminate correctly if the sum becomes a single digit before k iterations are completed. |
Maximum string length that may cause integer overflow | Utilize long integers to handle potential overflow during the digit sum operations. |
The intermediate sum is very large but the initial string is short | Use long datatype to accommodate for potentially large sums arising from conversion and repetitive summation. |