
Problem Statement
In this task, you receive a string s consisting solely of lowercase English letters, along with an integer k. The primary goal is to convert this string into a numeric value uniquely and then modify this numerical representation by summing its digits repeatedly for k iterations. The conversion and transformation process involves:
- Convert characters to numbers: Replace each character with its corresponding position in the alphabet (i.e.,
'a'becomes 1,'b'becomes 2, ...,'z'becomes 26). Concatenate these values into a single string of digits. - Transform the result
ktimes: Take the string of digits and repeatedly sum all the digitsktimes. After each sum, replace the string with the new sum’s digits.
The final result is the numeric outcome after applying the above transformation k times.
Examples
Example 1
Input:
s = "iiii", k = 1
Output:
36
Explanation:
'i' → 9 So "iiii" becomes "9999" Sum of digits (once): 9 + 9 + 9 + 9 = 36
Example 2
Input:
s = "vultrcode", k = 2
Output:
6
Explanation:
v → 22 u → 21 l → 12 t → 20 r → 18 c → 3 o → 15 d → 4 e → 5 → Concatenated string = "22211220183 1545" Actual sequence: "vultrcode" → "222112201831545" First transformation (sum of digits): 2+2+2+1+1+2+2+0+1+8+3+1+5+4+5 = 39 Second transformation: 3 + 9 = 12 Final result = 12
(Note: You may recompute the exact digits above; this is illustrative. Real calculation should follow literal character position replacement.)
Example 3
Input:
s = "zbax", k = 2
Output:
8
Explanation:
z → 26 b → 2 a → 1 x → 24 → Concatenated string: "262124" First transformation: 2 + 6 + 2 + 1 + 2 + 4 = 17 Second transformation: 1 + 7 = 8 Final result = 8
Constraints
1 <= s.length <= 1001 <= k <= 10sconsists of lowercase English letters only
Approach and Intuition
Here's a step-by-step breakdown of the approach:
Convert each character to its alphabetical index:
- Use
ord(char) - ord('a') + 1to convert each character insto its corresponding number. - Concatenate all these numbers into a single string (not added, just glued together).
- Example:
'abc' → "123"not1+2+3.
- Use
Transform the string into a number:
- The concatenated string of numbers becomes a large digit string like
"262124".
- The concatenated string of numbers becomes a large digit string like
Apply digit-sum
ktimes:- On each iteration, convert the string to a list of digits and compute their sum.
- Replace the number with this sum and repeat
ktimes.
Return the final result after k transformations.
This method ensures that the transformation from characters to digits and the subsequent digit summing are handled efficiently in linear time.
Solutions
- C++
- Java
- Python
class Solution {
public:
int transformString(string strInput, int iterations) {
int tempResult = 0;
for (char c : strInput) {
int alphaIndex = c - 'a' + 1;
while (alphaIndex > 0) {
tempResult += alphaIndex % 10;
alphaIndex /= 10;
}
}
for (int j = 1; j < iterations; ++j) {
int sumDigits = 0;
while (tempResult > 0) {
sumDigits += tempResult % 10;
tempResult /= 10;
}
tempResult = sumDigits;
if (tempResult < 10) {
break;
}
}
return tempResult;
}
};
This solution in C++ features a function, transformString, designed to sum the digits of string characters after conversion and potentially multiple iterations of summing digits. Here's a concise breakdown of how it works:
- The function accepts a string
strInputand an integeriterationsrepresenting the number of times the summing of digits is performed. - The string characters are converted to their respective positions in the alphabet (e.g., 'a' = 1, 'b' = 2, etc.).
- Each character's position value is then added cumulatively after breaking it down into individual digits.
- After the initial conversion and addition, the result stored in
tempResultis then re-evaluated over several iterations (specified byiterationsparameter). - During each iteration, digits of
tempResultare summed up untiltempResultis a single-digit number or until all iterations are completed.
- A notable implementation detail is that the function exits early if the resulting
tempResultbecomes a single-digit number before reaching the maximum specified iterations, optimizing processing time.
This approach meticulously handles the transformation and reduction of digital values through straightforward mathematical manipulations, ensuring efficiency and correctness in obtaining the final sum.
class Solution {
public int computeLuck(String str, int transformations) {
int sum = 0;
for (char c : str.toCharArray()) {
int val = c - 'a' + 1;
while (val > 0) {
sum += val % 10;
val /= 10;
}
}
for (int j = 1; j < transformations; ++j) {
int tempSum = 0;
while (sum > 0) {
tempSum += sum % 10;
sum /= 10;
}
sum = tempSum;
if (sum < 10) {
break;
}
}
return sum;
}
}
This solution provides a method to compute the resultant sum of digits of a string after converting each character to its respective position in the alphabet and applying transformation reductions. The function, computeLuck, takes two parameters: a string str and an integer transformations indicating how many times the sum should be transformed.
First, iterate over each character of the string to convert it into a numerical value corresponding to its position in the alphabet (
'a' = 1,'b' = 2, ...,'z' = 26).For each character, convert it to its alphabetical position and compute the sum of its digits.
Continue to transform the sum for the specified number of transformations:
- In each transformation, reduce the current sum by summing its individual digits.
- Check after each transformation if the sum has reduced to a single digit. If it has, exit the loop early as further transformations will not change the sum.
Finally, return the result, which is a single-digit sum after the specified number of transformations.
This method effectively handles the reduction of numbers to their component digits and can dynamically control the depth of transformation based on input, making it flexible for varied use cases.
class Solution:
def transformString(self, input_str: str, num_transforms: int) -> int:
result = 0
for character in input_str:
ascii_position = ord(character) - ord('a') + 1
while ascii_position > 0:
result += ascii_position % 10
ascii_position //= 10
for transform in range(1, num_transforms):
sum_digits = 0
while result > 0:
sum_digits += result % 10
result //= 10
result = sum_digits
if result < 10:
break
return result
This Python code defines a class Solution with a method transformString that calculates the sum of the digits of a string after performing several specified transformations. Here’s how the method works:
- Initialize
resultto 0 which will hold the cumulative sum. - Convert each character in the input string to a number by finding its position in the alphabet (i.e., 'a' = 1, 'b' = 2, ..., 'z' = 26) using the formula
ord(character) - ord('a') + 1. - For each character’s numeric value, add the sum of its digits to
result. - After processing all characters in the input string, proceed to transform the resulting number as many times as specified by
num_transforms:- For each transformation, calculate the sum of the digits of the current
result. - Update
resultwith this new sum. - If
resultbecomes a single digit before reaching the total number of specified transformations, exit the loop early.
- For each transformation, calculate the sum of the digits of the current
The final result, after all transformations are complete or terminated early, is returned as the output of the transformString method.
This approach effectively applies a digit-root-like operation multiple times, which can be useful for tasks that involve digit manipulation and transformation in strings within computational contexts.