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2447. Number of Subarrays With GCD Equal to K

Description

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

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

The greatest common divisor of an array is the largest integer that evenly divides all the array elements.

 

Example 1:

Input: nums = [9,3,1,2,6,3], k = 3
Output: 4
Explanation: The subarrays of nums where 3 is the greatest common divisor of all the subarray's elements are:
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]

Example 2:

Input: nums = [4], k = 7
Output: 0
Explanation: There are no subarrays of nums where 7 is the greatest common divisor of all the subarray's elements.

 

Constraints:

  • 1 <= nums.length <= 1000
  • 1 <= nums[i], k <= 109

Solutions

Solution 1: Direct Enumeration

We can enumerate $nums[i]$ as the left endpoint of the subarray, and then enumerate $nums[j]$ as the right endpoint of the subarray, where $i \le j$. During the enumeration of the right endpoint, we can use a variable $g$ to maintain the greatest common divisor of the current subarray. Each time we enumerate a new right endpoint, we update the greatest common divisor $g = \gcd(g, nums[j])$. If $g=k$, then the greatest common divisor of the current subarray equals $k$, and we increase the answer by $1$.

After the enumeration ends, return the answer.

The time complexity is $O(n \times (n + \log M))$, where $n$ and $M$ are the length of the array $nums$ and the maximum value in the array $nums$, respectively.

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class Solution:
    def subarrayGCD(self, nums: List[int], k: int) -> int:
        ans = 0
        for i in range(len(nums)):
            g = 0
            for x in nums[i:]:
                g = gcd(g, x)
                ans += g == k
        return ans

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class Solution {
    public int subarrayGCD(int[] nums, int k) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int g = 0;
            for (int j = i; j < n; ++j) {
                g = gcd(g, nums[j]);
                if (g == k) {
                    ++ans;
                }
            }
        }
        return ans;
    }

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

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class Solution {
public:
    int subarrayGCD(vector<int>& nums, int k) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int g = 0;
            for (int j = i; j < n; ++j) {
                g = gcd(g, nums[j]);
                ans += g == k;
            }
        }
        return ans;
    }
};

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func subarrayGCD(nums []int, k int) (ans int) {
    for i := range nums {
        g := 0
        for _, x := range nums[i:] {
            g = gcd(g, x)
            if g == k {
                ans++
            }
        }
    }
    return
}

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

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function subarrayGCD(nums: number[], k: number): number {
    let ans = 0;
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        let g = 0;
        for (let j = i; j < n; ++j) {
            g = gcd(g, nums[j]);
            if (g === k) {
                ++ans;
            }
        }
    }
    return ans;
}

function gcd(a: number, b: number): number {
    return b === 0 ? a : gcd(b, a % b);
}

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