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3164. Find the Number of Good Pairs II

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 <= 105
  • 1 <= nums1[i], nums2[j] <= 106
  • 1 <= k <= 103

Solutions

Solution 1: Hash Table + Enumerate Multiples

We use a hash table cnt1 to record the occurrence times of each number divided by $k$ in array nums1, and a hash table cnt2 to record the occurrence times of each number in array nums2.

Next, we enumerate each number $x$ in array nums2. For each number $x$, we enumerate its multiples $y$, where the range of $y$ is $[x, \textit{mx}]$, where mx is the maximum key value in cnt1. Then we count the sum of cnt1[y], denoted as $s$. Finally, we add $s \times v$ to the answer, where $v$ is cnt2[x].

The time complexity is $O(n + m + (M / k) \times \log m)$, and the space complexity is $O(n + m)$. Where $n$ and $m$ are the lengths of arrays nums1 and nums2 respectively, and $M$ is the maximum value in array nums1.

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class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:
        cnt1 = Counter(x // k for x in nums1 if x % k == 0)
        if not cnt1:
            return 0
        cnt2 = Counter(nums2)
        ans = 0
        mx = max(cnt1)
        for x, v in cnt2.items():
            s = sum(cnt1[y] for y in range(x, mx + 1, x))
            ans += s * v
        return ans

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class Solution {
    public long numberOfPairs(int[] nums1, int[] nums2, int k) {
        Map<Integer, Integer> cnt1 = new HashMap<>();
        for (int x : nums1) {
            if (x % k == 0) {
                cnt1.merge(x / k, 1, Integer::sum);
            }
        }
        if (cnt1.isEmpty()) {
            return 0;
        }
        Map<Integer, Integer> cnt2 = new HashMap<>();
        for (int x : nums2) {
            cnt2.merge(x, 1, Integer::sum);
        }
        long ans = 0;
        int mx = Collections.max(cnt1.keySet());
        for (var e : cnt2.entrySet()) {
            int x = e.getKey(), v = e.getValue();
            int s = 0;
            for (int y = x; y <= mx; y += x) {
                s += cnt1.getOrDefault(y, 0);
            }
            ans += 1L * s * v;
        }
        return ans;
    }
}

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class Solution {
public:
    long long numberOfPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        unordered_map<int, int> cnt1;
        for (int x : nums1) {
            if (x % k == 0) {
                cnt1[x / k]++;
            }
        }
        if (cnt1.empty()) {
            return 0;
        }
        unordered_map<int, int> cnt2;
        for (int x : nums2) {
            ++cnt2[x];
        }
        int mx = 0;
        for (auto& [x, _] : cnt1) {
            mx = max(mx, x);
        }
        long long ans = 0;
        for (auto& [x, v] : cnt2) {
            long long s = 0;
            for (int y = x; y <= mx; y += x) {
                s += cnt1[y];
            }
            ans += s * v;
        }
        return ans;
    }
};

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func numberOfPairs(nums1 []int, nums2 []int, k int