Leetcode 1945: Sum of Digits of String After Convert
You are given a string consisting of lowercase English letters and an integer k. Your task is to convert the string into an integer by replacing each letter with its corresponding position in the alphabet (where ‘a’ = 1, ‘b’ = 2, …, ‘z’ = 26), and then repeatedly sum its digits k times. The final result is the integer obtained after performing the digit sum operation k times.
Problem
Approach
Steps
Complexity
Input: You are given a string s of lowercase English letters and an integer k. The string s represents the word to be converted into an integer, and k represents the number of times the digit sum operation should be performed.
Example: s = "abcd", k = 2
Constraints:
• 1 <= s.length <= 100
• 1 <= k <= 10
• s consists of lowercase English letters
Output: Return the integer obtained after performing the digit sum operation on the string's converted integer representation k times.
Example: Output: 7
Constraints:
• The result should be a single integer.
Goal: The goal is to convert each character in the string to its respective position in the alphabet, then repeatedly sum the digits of the resulting integer k times to get the final result.
Steps:
• Step 1: Convert each character in the string to its corresponding number in the alphabet.
• Step 2: Concatenate the numbers to form a large integer.
• Step 3: Perform the digit sum operation on the integer k times.
• Step 4: Return the resulting integer after k digit sum operations.
Goal: Constraints ensure that the input string is of manageable length, and the number of transformations (k) is small.
Steps:
• 1 <= s.length <= 100
• 1 <= k <= 10
• s consists of lowercase English letters
Assumptions:
• The input string will contain only lowercase English letters.
• The value of k will be at least 1.
• Input: Input: s = "abcd", k = 2
• Explanation: The string 'abcd' is converted to '1234'. First, we sum the digits: 1+2+3+4 = 10. Then, we sum the digits of 10: 1+0 = 1. After 2 transformations, the result is 1.
• Input: Input: s = "aaa", k = 1
• Explanation: The string 'aaa' becomes '111'. Summing the digits gives: 1+1+1 = 3. After 1 transformation, the result is 3.
Approach: The approach is to first map the letters of the string to their respective numeric values based on their position in the alphabet, and then perform digit sum transformations k times.
Observations:
• We need to map each letter to its respective position in the alphabet.
• After forming the integer from these positions, we will repeatedly sum the digits.
• This is a simple string manipulation and digit sum problem that can be solved in a straightforward manner using basic operations.
Steps:
• Step 1: Convert each character to its corresponding position in the alphabet.
• Step 2: Concatenate the numeric values to form a large integer.
• Step 3: Perform the digit sum transformation k times.
• Step 4: Return the final integer after k transformations.
Empty Inputs:
• Empty string input is not valid as per the constraints.
Large Inputs:
• If the string has a length close to the upper limit (100 characters), the solution should efficiently handle the large number formed.
Special Values:
• Single character strings should return a simple transformation result.
Constraints:
• Ensure that k is within the valid range of 1 to 10.
int getLucky(string s, int k) {
string tmp;
for (char c : s) tmp += to_string(c - 'a' + 1);
int n = 0;
for (char c : tmp) n += c - '0';
for (int i = 1; i < k; ++i) {
int next = 0;
while (n) {
next += n % 10;
n /= 10;
}
n = next;
}
return n;
}
1 : Function Definition
int getLucky(string s, int k) {
The function definition begins with the input parameters: 's' (a string) and 'k' (an integer) representing the number of iterations for the digit reduction process.
2 : Variable Initialization
string tmp;
A string 'tmp' is initialized to store the numeric string derived from the input string 's'.
3 : Loop over String
for (char c : s) tmp += to_string(c - 'a' + 1);
Each character of the input string 's' is converted to a corresponding number ('a' -> 1, 'b' -> 2, ..., 'z' -> 26) and appended to 'tmp'.
4 : Variable Initialization
int n = 0;
The variable 'n' is initialized to 0. It will hold the sum of the digits in the string 'tmp'.
5 : Summing Digits
for (char c : tmp) n += c - '0';
The digits in the string 'tmp' are summed up and stored in the variable 'n'. Each character is converted to its numeric value using 'c - '0'.
6 : Loop for Reduction
for (int i = 1; i < k; ++i) {
A loop is initiated that will run 'k' times, reducing the sum 'n' by repeatedly summing its digits.
7 : Variable Initialization
int next = 0;
A temporary variable 'next' is initialized to accumulate the sum of the digits of 'n'.
8 : While Loop for Digit Summation
while (n) {
A while loop is used to reduce 'n' by summing its digits. The loop continues until 'n' becomes 0.
9 : Digit Summation
next += n % 10;
The last digit of 'n' is added to 'next'.
10 : Update n
n /= 10;
The last digit of 'n' is removed by performing integer division by 10.
11 : Update n
n = next;
'n' is updated to 'next', which is the sum of the digits of the previous 'n'.
12 : Return Statement
return n;
The final reduced value of 'n' is returned after 'k' iterations.
Best Case: O(n) where n is the length of the string.
Average Case: O(n * k) where n is the length of the string and k is the number of digit sum operations.
Worst Case: O(n * k), the worst-case time complexity occurs when the input string is large and k is at its maximum value.
Description: The time complexity depends on both the string length and the number of times the transformation is performed.
Best Case: O(1), if we use constant space for digit summing operations.
Worst Case: O(n), as we need space for storing the converted string and the intermediate results.
Description: The space complexity depends on the length of the input string and the storage required for intermediate results.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus