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// 2375. Construct Smallest Number From DI String
#include "leetcode.h"
/*
* you are given a 0-indexed string 'pattern' of length 'n' consisting of the
* characters 'I' meaning increased and 'D' meaning decreased. a 0-indexed
* string 'num' of length 'n + 1' is created using thhe following conditions:
* 'num' consists of the digits 1 to 9, where each digit is used at most once,
* if 'pattern[i] == I', then 'num[i] < num[i + 1]', and if 'pattern[i] == D',
* thhen 'num[i] > num[i + 1]'. return the lexicographically smallest possible
* string 'num' that meets the conditions.
*/
bool find_smallest(char *pattern, char *nums, int prev, int vis) {
int new_vis, max_val, start_val;
if (*pattern == 'I') {
max_val = 10;
start_val = prev + 1;
} else {
max_val = prev;
start_val = 1;
}
for (int i = start_val; i < max_val; i++) {
if (!((vis >> (i - 1)) & 1)) {
*nums = i;
new_vis = vis | (1 << (i - 1));
if (!(*(pattern + 1)) ||
find_smallest(pattern + 1, &(nums[1]), i, new_vis))
return true;
}
}
return false;
}
char *smallestNumber(char *pattern) {
int n = strlen(pattern), vis;
char *nums = (char *)malloc((n + 1) * sizeof(char));
char *ans = (char *)malloc((n + 2) * sizeof(char));
for (int i = 1; i < 10; i++) {
vis = 1 << (i - 1);
if (find_smallest(pattern, nums, i, vis)) {
ans[0] = i + '0';
break;
}
}
for (int i = 0; i < n + 1; i++)
ans[i + 1] = nums[i] + '0';
ans[n + 1] = 0;
free(nums);
return ans;
}
int main() {
char *p1 = "IIIDIDDD", *p2 = "DDD";
printf("%s\n", smallestNumber(p1)); // expect: 123549876
printf("%s\n", smallestNumber(p2)); // expect: 4321
}
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