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2470. Number of Subarrays With LCM Equal to K

Description

Given an integer array nums and an integer k, return the number of subarrays of nums where the least common multiple of the subarray's elements is k.

A subarray is a contiguous non-empty sequence of elements within an array.

The least common multiple of an array is the smallest positive integer that is divisible by all the array elements.

 

Example 1:

Input: nums = [3,6,2,7,1], k = 6
Output: 4
Explanation: The subarrays of nums where 6 is the least common multiple of all the subarray's elements are:
- [3,6,2,7,1]
- [3,6,2,7,1]
- [3,6,2,7,1]
- [3,6,2,7,1]

Example 2:

Input: nums = [3], k = 2
Output: 0
Explanation: There are no subarrays of nums where 2 is the least common multiple of all the subarray's elements.

 

Constraints:

  • 1 <= nums.length <= 1000
  • 1 <= nums[i], k <= 1000

Solutions

Solution 1: Enumeration

Enumerate each number as the first number of the subarray, and then enumerate each number as the last number of the subarray. Calculate the least common multiple of this subarray. If the least common multiple equals $k$, then increment the answer by one.

The time complexity is $O(n^2)$. Here, $n$ is the length of the array.

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class Solution:
    def subarrayLCM(self, nums: List[int], k: int) -> int:
        n = len(nums)
        ans = 0
        for i in range(n):
            a = nums[i]
            for b in nums[i:]:
                x = lcm(a, b)
                ans += x == k
                a = x
        return ans

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class Solution {
    public int subarrayLCM(int[] nums, int k) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int a = nums[i];
            for (int j = i; j < n; ++j) {
                int b = nums[j];
                int x = lcm(a, b);
                if (x == k) {
                    ++ans;
                }
                a = x;
            }
        }
        return ans;
    }

    private int lcm(int a, int b) {
        return a * b / gcd(a, b);
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

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class Solution {
public:
    int subarrayLCM(vector<int>& nums, int k) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int a = nums[i];
            for (int j = i; j < n; ++j) {
                int b = nums[j];
                int x = lcm(a, b);
                ans += x == k;
                a = x;
            }
        }
        return ans;
    }
};

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func subarrayLCM(nums []int, k int) (ans int) {
    for i, a := range nums {
        for _, b := range nums[i:] {
            x := lcm(a, b)
            if x == k {
                ans++
            }
            a = x
        }
    }
    return
}

func gcd(a, b int) int {
    if b == 0 {
        return a
    }
    return gcd(b, a%b)
}

func lcm(a, b int) int {
    return a * b / gcd(a, b)
}

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