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3164. Find the Number of Good Pairs II 👍

  • Time: $O(m\sqrt{n})$
  • Space: $O(m)$
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class Solution {
 public:
  long long numberOfPairs(vector<int>& nums1, vector<int>& nums2, int k) {
    unordered_map<int, int> count;
    long ans = 0;

    for (const int num : nums2)
      ++count[num * k];

    for (const int num : nums1)
      for (int divisor = 1; divisor <= sqrt(num); ++divisor)
        if (num % divisor == 0) {
          ans += count.contains(divisor) ? count[divisor] : 0;
          if (num / divisor != divisor)
            ans += count.contains(num / divisor) ? count[num / divisor] : 0;
        }

    return ans;
  }
};

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class Solution {
  public long numberOfPairs(int[] nums1, int[] nums2, int k) {
    HashMap<Integer, Integer> count = new HashMap<>();
    long ans = 0;

    for (final int num : nums2)
      count.merge(num * k, 1, Integer::sum);

    for (final int num : nums1)
      for (int divisor = 1; divisor <= (int) Math.sqrt(num); ++divisor)
        if (num % divisor == 0) {
          ans += count.getOrDefault(divisor, 0);
          if (num / divisor != divisor)
            ans += count.getOrDefault(num / divisor, 0);
        }

    return ans;
  }
}

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class Solution:
  def numberOfPairs(self, nums1: list[int], nums2: list[int], k: int) -> int:
    count = collections.Counter(num * k for num in nums2)
    ans = 0

    for num in nums1:
      for divisor in range(1, int(num ** 0.5) + 1):
        if num % divisor == 0:
          ans += count[divisor]
          if num // divisor != divisor:
            ans += count[num // divisor]

    return ans