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This approach constructs the smallest lexicographical string by starting with 'a' for all characters initially. Then, it tries to modify the string starting from the end towards the beginning so that the sum of all character values becomes equal to k. We iterate backwards and replace 'a's with whatever character provides exactly the needed additional sum up to 25 (since 26 - 1 = 25, where 1 is the value of 'a').
Time Complexity: O(n), as we iterate over the string once.
Space Complexity: O(n), for the storage of the result string.
1#include <iostream>
2#include <string>
3
4std::string getSmallestString(int n, int k) {
5 std::string result(n, 'a');
6 k -= n; // Reduce k by the sum of 'a's
7 for (int i = n - 1; i >= 0 && k > 0; --i) {
8 int add = std::min(k, 25);
9 result[i] += add;
10 k -= add;
11 }
12 return result;
13}
14
15int main() {
16 int n = 3, k = 27;
17 std::cout << getSmallestString(n, k) << std::endl;
18 return 0;
19}In C++, we utilize a string initialized to 'a' characters. We adjust k by decreasing it by n, the sum of initial 'a's, and iteratively update the string characters from the end to suit the required numeric value without breaching the lexicographical order.
This approach generates the solution by starting with the smallest lexicographical letter 'a' and incrementally modifying the last possible position where it can introduce a remainder of the sum. We start assigning 'a' to fill the leftovers systematically from the beginning and increment positions forward when no more space is left towards the end.
Time Complexity: O(n)
Space Complexity: O(n) for storing the resulting string.
Character arrays provide a powerful method to manipulate data within Java. The Java method iterates from the back to meet the calculated sum without reorganizing or modifying dict entries.