// 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 }