// 2654. Minimum Number of Operations to Make All Array Elements Equal to 1 #include "leetcode.h" /* * given a 0-indexed array 'nums' consisting of positive integers. you can do * the following operation on the array any number of time: select an index 'i' * such that '0 <= i < n - 1' and replace either of 'nums[i]' or 'nums[i + 1]' * with their gcd values. return the minimum number of operations to make all * elements of 'nums' equal to 1. if this is impossible, return -1. the gcd of * two integers is the greatest common divisor of the two integers. */ int gcd(int a, int b) { int rem = a % b; if (!rem) return b; return gcd(b, rem); } int minOperations(int *nums, int numsSize) { int one_cnt = 0; for (int i = 0; i < numsSize; i++) if (nums[i] == 1) one_cnt++; if (one_cnt >= 1) return numsSize - one_cnt; int min_subarr_len = numsSize + 1; for (int i = 0; i < numsSize; i++) { int curr = nums[i]; for (int j = i + 1; j < numsSize; j++) { curr = gcd(curr, nums[j]); if (curr == 1) { int curr_n = (j + 1) - i; if (curr_n < min_subarr_len) min_subarr_len = curr_n; } break; } } return min_subarr_len <= numsSize ? (min_subarr_len - 1) + (numsSize - 1) : -1; } int main() { int n1[] = {2, 6, 3, 4}, n2[] = {2, 10, 6, 14}; printf("%d\n", minOperations(n1, ARRAY_SIZE(n1))); // expect: 4 printf("%d\n", minOperations(n2, ARRAY_SIZE(n2))); // expect: -1 }