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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 <stdio.h>
2#include <string.h>
3
4int digitSum(char *num) {
5 int sum = 0;
6 while (*num) {
7 sum += *num - '0';
8 num++;
9 }
10 return sum;
11}
12
13int sumOfDigitsAfterConvert(char *s, int k) {
14 char num[300] = "";
15 char buffer[3];
16 for (int i = 0; s[i] != '\0'; i++) {
17 int val = s[i] - 'a' + 1;
18 sprintf(buffer, "%d", val);
19 strcat(num, buffer);
20 }
21
22 int result = 0;
23 while (k--) {
24 result = digitSum(num);
25 sprintf(num, "%d", result);
26 }
27
28 return result;
29}
30
31int main() {
32 printf("%d\n", sumOfDigitsAfterConvert("zbax", 2));
33 return 0;
34}The function sumOfDigitsAfterConvert converts the string s into a numeric representation of alphabet positions. The helper function digitSum computes the sum of digits of a given number string. The transformation step is repeated k times by repeatedly summing the digits of the number string.
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#
Here, the conversion and the first summing of digits are done simultaneously using modulo operations. This reduces the need for intermediate storage and allows for direct digit sum transformation.