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You are given a string s consisting of lowercase English letters, and an integer k. First, you need to convert each character of s into its position in the alphabet (i.e., convert ‘a’ to 1, ‘b’ to 2, …, ‘z’ to 26). Then, transform the resulting number by summing its digits repeatedly, for a total of k times. Return the resulting integer after performing the transformations.

Clarifying Questions

  1. Example Clarification:
    • For example, if s = "leetcode" and k = 2, the steps would be:
      • Convert characters to positions: l=12, e=5, e=5, t=20, c=3, o=15, d=4, e=5.
      • Concatenate these numbers: 12552031545.
      • Sum the digits of this number once: 1+2+5+5+2+0+3+1+5+4+5 = 33.
      • Sum the digits again: 3+3 = 6.
      • Result is 6.
  2. Edge Cases:
    • What if s is an empty string? (Although not explicit, we’ll assume s is non-empty based on typical constraints.)
    • What is the minimum and maximum length of s and value of k?

Code

Here’s the Python code to solve this problem:

def getLucky(s: str, k: int) -> int:
    # Convert the string to a number formed by the position in the alphabet
    num_str = ''.join(str(ord(char) - ord('a') + 1) for char in s)
    
    # Function to sum digits of a number in string form
    def sum_of_digits(n: str) -> int:
        return sum(int(char) for char in n)
    
    # Start with the massive number string
    curr_str = num_str
    
    # Perform the sum of digits process k times
    for _ in range(k):
        curr_str = str(sum_of_digits(curr_str))
        
    return int(curr_str)

# Example usage
s = "leetcode"
k = 2
print(getLucky(s, k)) # Output: 6

Strategy

  1. Convert Characters to Positions:
    • Iterate through each character in the string s.
    • Convert each character to its corresponding position (e.g., ‘a’ to 1, ‘b’ to 2, etc.).
    • Concatenate these positions to form a large number in string format.
  2. Perform Digit Sum Transformations:
    • Define a helper function that takes a string of digits and computes the sum of its digits.
    • Apply this helper function k times to the initial large number string.
    • After k transformations, convert the final string back to an integer and return it.

Time Complexity

  1. Character to Position Conversion:
    • This step runs in O(n) time where n is the length of the string s.
  2. Sum of Digits:
    • Each transformation in sum of digits iterates over the digits of the number formed. The length of number formed is approximately in order of magnitude O(n) where n is the original string length.
    • Performing this operation k times introduces an additional factor.

Thus, the total time complexity is about O(kn), where n is the length of the input string and k is the number of repetitions.

This ensures efficiency even for reasonable input sizes.

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