Welcome to Subscribe On Youtube

3162. Find the Number of Good Pairs I

Description

You are given 2 integer arrays nums1 and nums2 of lengths n and m respectively. You are also given a positive integer k.

A pair (i, j) is called good if nums1[i] is divisible by nums2[j] * k (0 <= i <= n - 1, 0 <= j <= m - 1).

Return the total number of good pairs.

 

Example 1:

Input: nums1 = [1,3,4], nums2 = [1,3,4], k = 1

Output: 5

Explanation:

The 5 good pairs are (0, 0), (1, 0), (1, 1), (2, 0), and (2, 2).

Example 2:

Input: nums1 = [1,2,4,12], nums2 = [2,4], k = 3

Output: 2

Explanation:

The 2 good pairs are (3, 0) and (3, 1).

 

Constraints:

  • 1 <= n, m <= 50
  • 1 <= nums1[i], nums2[j] <= 50
  • 1 <= k <= 50

Solutions

Solution 1: Brute Force Enumeration

We directly enumerate all digits $(x, y)$, and check whether $x \mod (y \times k) = 0$. If it is, we increment the answer.

After the enumeration ends, we return the answer.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of arrays nums1 and nums2 respectively. The space complexity is $O(1)$.

  • class Solution {
    public int numberOfPairs(int[] nums1, int[] nums2, int k) {
        int ans = 0;
        for (int x : nums1) {
            for (int y : nums2) {
                if (x % (y * k) == 0) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int numberOfPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        int ans = 0;
        for (int x : nums1) {
            for (int y : nums2) {
                if (x % (y * k) == 0) {
                    ++ans;
                }
            }
        }
        return ans;
    }
};

  • class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:
        return sum(x % (y * k) == 0 for x in nums1 for y in nums2)


  • func numberOfPairs(nums1 []int, nums2 []int, k int) (ans int) {
	for _, x := range nums1 {
		for _, y := range nums2 {
			if x%(y*k) == 0 {
				ans++
			}
		}
	}
	return
}

  • function numberOfPairs(nums1: number[], nums2: number[], k: number): number {
    let ans = 0;
    for (const x of nums1) {
        for (const y of nums2) {
            if (x % (y * k) === 0) {
                ++ans;
            }
        }
    }
    return ans;
}

All Problems

All Solutions