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This approach involves converting each character to its corresponding position in the alphabet and forming a numeric string. We then sum the digits of this number 'k' times to get the resulting integer.
Time Complexity: O(N + kM), where N is the length of the string and M is the number of digits in the transformed number.
Space Complexity: O(M), as we store the transformed number as a string.
1#include <iostream>
2#include <string>
3using namespace std;
4
5int digitSum(string num) {
6 int sum = 0;
7 for(char ch : num)
8 sum += ch - '0';
9 return sum;
10}
11
12int sumOfDigitsAfterConvert(string s, int k) {
13 string num = "";
14 for (char c : s)
15 num += to_string(c - 'a' + 1);
16
17 int result = 0;
18 while (k--) {
19 result = digitSum(num);
20 num = to_string(result);
21 }
22
23 return result;
24}
25
26int main() {
27 cout << sumOfDigitsAfterConvert("zbax", 2) << endl;
28 return 0;
29}In C++, we convert each character of s to its corresponding alphabetic position using c - 'a' + 1, and concatenate it to form a number string. The digit sum is calculated using digitSum function, and this process is repeated k times.
This approach directly calculates the digit sum during the conversion of the string to a number by aggregating the sum of each digit representation, reducing the need for intermediate large integer string handling.
Time Complexity: O(N + k*logM), where M is the intermediate sum value.
Space Complexity: O(1).
1
public class Solution {
public static int SumOfDigitsAfterConvert(string s, int k) {
int sum = 0;
foreach (var c in s) {
int val = c - 'a' + 1;
while (val > 0) {
sum += val % 10;
val /= 10;
}
}
while (--k > 0) {
int newSum = 0;
while (sum > 0) {
newSum += sum % 10;
sum /= 10;
}
sum = newSum;
}
return sum;
}
public static void Main() {
Console.WriteLine(SumOfDigitsAfterConvert("zbax", 2));
}
}C# provides an iterative approach for summing digits, minimizing interim storage and focusing on direct in-place transformations through loop constructs.