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

Solution: Brute Force

Time complexity: O(m*n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int maxDivScore(vector<int>& nums, vector<int>& divisors) {
    int maxScore = 0;
    int ans = INT_MAX;
    for (int d : divisors) {
      int score = 0;
      for (int x : nums) {
        score += x % d == 0;
      }
      if (score > maxScore) {
        maxScore = score;
        ans = d;
      }
      if (score == maxScore)
        ans = min(ans, d);
    }
    return ans;
  }
};

The post 花花酱 LeetCode 2644. Find the Maximum Divisibility Score appeared first on Huahua's Tech Road.

An integer m is a divisor of n if there exists an integer k such that n = k * m.

Example 1:

Input: n = 2
Output: false
Explantion: 2 has only two divisors: 1 and 2.

Example 2:

Input: n = 4
Output: true
Explantion: 4 has three divisors: 1, 2, and 4.

Constraints:

• 1 <= n <= 104

Solution: Enumerate divisors.

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  bool isThree(int n) {
    int c = 0;
    for (int i = 1; i <= n; ++i)
      if (n % i == 0) ++c;        
    return c == 3;
  }
};

Optimization

Only need to enumerate divisors up to sqrt(n). Special handle for the d * d == n case.

Time complexity: O(sqrt(n))
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  bool isThree(int n) {
    int c = 0;
    for (int i = 1; i <= sqrt(n); ++i)
      if (n % i == 0)
        c += 1 + (i * i != n);
    return c == 3;
  }
};

The post 花花酱 LeetCode 1952. Three Divisors appeared first on Huahua's Tech Road.

If there is no such integer in the array, return 0.

Example 1:

Input: nums = [21,4,7]
Output: 32
Explanation:
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.

Constraints:

• 1 <= nums.length <= 10^4
• 1 <= nums[i] <= 10^5

Solution: Math

If a number is a perfect square (e.g. 9 = 3 * 3), it will have odd number of divisors. (9: 1, 3, 9).

Time complexity: O(sum(sqrt(num_i))
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int sumFourDivisors(vector<int>& nums) {
    int ans = 0;
    for (int n : nums) {
      int r = sqrt(n);
      if (n <= 4 || r * r == n) continue;
      int count = 2;
      int sum = 1 + n;
      for (int d = 2; d <= r; ++d)
        if (n % d == 0) {
          count += 2;
          sum += n / d + d;
        }      
      if (count == 4) ans += sum;
    }
    return ans;
  }
};

The post 花花酱 LeetCode 1390. Four Divisors appeared first on Huahua's Tech Road.