carloscn
commented
1 year ago Analysis
static int32_t max_div_score(int32_t *nums, size_t nums_size, int32_t *divisors, size_t divisors_size)
{
int32_t ret = 0;
UTILS_CHECK_PTR(nums);
UTILS_CHECK_PTR(divisors);
UTILS_CHECK_LEN(nums_size);
UTILS_CHECK_LEN(divisors_size);
int32_t max_score = INT32_MIN;
int32_t min_div = INT32_MAX;
for (size_t i = 0; i < divisors_size; i ++) {
int32_t score = 0;
int32_t e = divisors[i];
for (size_t j = 0; j < nums_size; j ++) {
if (!(nums[j] % e)) {
score ++;
}
}
if (score > max_score) {
max_score = score;
min_div = e;
} else if (score == max_score) {
min_div = UTILS_MIN(min_div, e);
}
}
ret = min_div;
finish:
return ret;
}
Description
You are given two 0-indexed integer arrays nums and divisors.
The divisibility score of divisors[i] is the number of indices j such that nums[j] is divisible by divisors[i].
Return the integer divisors[i] with the maximum divisibility score. If there is more than one integer with the maximum score, return the minimum of them.
Example 1:
Input: nums = [4,7,9,3,9], divisors = [5,2,3] Output: 3 Explanation: The divisibility score for every element in divisors is: The divisibility score of divisors[0] is 0 since no number in nums is divisible by 5. The divisibility score of divisors[1] is 1 since nums[0] is divisible by 2. The divisibility score of divisors[2] is 3 since nums[2], nums[3], and nums[4] are divisible by 3. Since divisors[2] has the maximum divisibility score, we return it.
Example 2:
Input: nums = [20,14,21,10], divisors = [5,7,5] Output: 5 Explanation: The divisibility score for every element in divisors is: The divisibility score of divisors[0] is 2 since nums[0] and nums[3] are divisible by 5. The divisibility score of divisors[1] is 2 since nums[1] and nums[2] are divisible by 7. The divisibility score of divisors[2] is 2 since nums[0] and nums[3] are divisible by 5. Since divisors[0], divisors[1], and divisors[2] all have the maximum divisibility score, we return the minimum of them (i.e., divisors[2]).
Example 3:
Input: nums = [12], divisors = [10,16] Output: 10 Explanation: The divisibility score for every element in divisors is: The divisibility score of divisors[0] is 0 since no number in nums is divisible by 10. The divisibility score of divisors[1] is 0 since no number in nums is divisible by 16. Since divisors[0] and divisors[1] both have the maximum divisibility score, we return the minimum of them (i.e., divisors[0]).
Constraints:
1 <= nums.length, divisors.length <= 1000 1 <= nums[i], divisors[i] <= 109