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Diffstat (limited to 'min_num_opp_arr_equal_1.c')
-rw-r--r-- | min_num_opp_arr_equal_1.c | 48 |
1 files changed, 48 insertions, 0 deletions
diff --git a/min_num_opp_arr_equal_1.c b/min_num_opp_arr_equal_1.c new file mode 100644 index 0000000..c6b5dc3 --- /dev/null +++ b/min_num_opp_arr_equal_1.c @@ -0,0 +1,48 @@ +// 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 +} |