yet another hackerrank task
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#include <iostream>
#include <bitset>
#include <vector>
#include <array>
#include <cassert>
// zero_number_ascii_offset is an ascii number of '0' which is the offset for all ascii digits as well
constexpr char zero_number_ascii_offset = 48;
// typedef for single digits, just a presence of a digit
typedef std::bitset<10> singles_t;
// typedef for counts of doubles digits
typedef std::array<size_t, 10> doubles_t;
// to_digit converts char to digit
uint8_t to_digit(char c) {
return c - zero_number_ascii_offset;
}
// to_char converts digit to char
char to_char(uint8_t n) {
return n + zero_number_ascii_offset;
}
/* solution
* let's assume the biggest palindrome has:
* 1. biggest single in the center (xxx9xxx > xxx8xxx);
* 2. biggest doubles at corners (9xxxx9 > 8xxxx8).
*
* the algorithm:
* 1. count all doubles and singles, takes O(N);
* 2. construct the result string with constant length;
* 3. if there is biggest single then put it to the center;
* 4. fill the string from the center to corners using doubles;
*
* complexity: O(N)
*/
std::string solution(std::string &S) {
std::bitset<10> singles{};
std::array<size_t, 10> doubles{};
size_t doubles_count = 0;
size_t doubles_sum = 0;
for (char c: S) {
uint8_t digit = to_digit(c);
assert(digit < 10);
if (singles.test(digit)) {
singles.reset(digit);
++doubles[digit];
++doubles_count;
doubles_sum += digit;
} else {
singles.set(digit);
}
}
char biggest_single = 0;
for (uint8_t i = 10; i > 0; i--) {
if (singles.test(i - 1)) {
biggest_single = to_char(i - 1);
break;
}
}
bool has_single = biggest_single > 0;
size_t single_offset = (has_single ? 1 : 0);
size_t result_len = doubles_count * 2 + single_offset;
if (doubles_sum == 0) {
result_len = 1;
biggest_single = biggest_single > 0 ? biggest_single : '0';
}
std::string result(result_len, biggest_single);
if (doubles_sum > 0) {
size_t iter = 0;
for (uint8_t i = 0; i < 10; i++) {
size_t count = doubles[i];
while (count-- > 0) {
result[result_len / 2 - 1 - iter] = to_char(i);
result[result_len / 2 + single_offset + iter] = to_char(i);
iter++;
}
}
}
return result;
}
int main() {
std::cout << solution("090") << std::endl;
return 0;
}