carloscn
commented
10 months ago Analysis
static int32_t min_number(int32_t* nums1, size_t nums1_size, int32_t* nums2, size_t nums2_size)
{
int32_t ret = 0;
UTILS_CHECK_PTR(nums1);
UTILS_CHECK_PTR(nums2);
int32_t min_1 = INT32_MAX;
int32_t min_2 = INT32_MAX;
int32_t common = INT32_MAX;
if ((0 == nums1_size) &&
(0 == nums2_size)) {
goto finish;
}
for (size_t i = 0; i < nums1_size; i ++) {
int32_t e = nums1[i];
min_1 = UTILS_MIN(e, min_1);
for (size_t j = 0; j < nums2_size; j ++) {
min_2 = UTILS_MIN(nums2[j], min_2);
if (e == nums2[j]) {
common = (common > e) ? e : common;
}
}
}
if (common == INT32_MAX) {
if (min_1 > min_2) {
ret = min_2 * 10 + min_1;
} else {
ret = min_1 * 10 + min_2;
}
} else {
ret = common;
}
finish:
return ret;
}
Description
Given two arrays of unique digits nums1 and nums2, return the smallest number that contains at least one digit from each array.
Example 1:
Input: nums1 = [4,1,3], nums2 = [5,7] Output: 15 Explanation: The number 15 contains the digit 1 from nums1 and the digit 5 from nums2. It can be proven that 15 is the smallest number we can have.
Example 2:
Input: nums1 = [3,5,2,6], nums2 = [3,1,7] Output: 3 Explanation: The number 3 contains the digit 3 which exists in both arrays.
Constraints:
1 <= nums1.length, nums2.length <= 9 1 <= nums1[i], nums2[i] <= 9 All digits in each array are unique.