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Posts tagged as “divisor”

花花酱 LeetCode 2644. Find the Maximum Divisibility Score

By zxi on April 28, 2023

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

Solution: Brute Force

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

C++

花花酱 LeetCode 1952. Three Divisors

By zxi on December 31, 2021

Given an integer n, return true if n has exactly three positive divisors. Otherwise, return false.

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)

C++

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)

C++

花花酱 LeetCode 1390. Four Divisors

By zxi on March 22, 2020

Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors.

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)

C++