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2354. Number of Excellent Pairs 👍

  • Time: $O(30n) = O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  long long countExcellentPairs(vector<int>& nums, int k) {
    constexpr int kMaxBit = 30;
    // bits(num1 | num2) + bits(num1 & num2) = bits(num1) + bits(num2)
    long ans = 0;
    vector<long> count(kMaxBit);

    for (const unsigned num : unordered_set<int>(nums.begin(), nums.end()))
      ++count[popcount(num)];

    for (int i = 0; i < kMaxBit; ++i)
      for (int j = 0; j < kMaxBit; ++j)
        if (i + j >= k)
          ans += count[i] * count[j];

    return ans;
  }
};

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class Solution {
  public long countExcellentPairs(int[] nums, int k) {
    final int kMaxBit = 30;
    // bits(num1 | num2) + bits(num1 & num2) = bits(num1) + bits(num2)
    long ans = 0;
    long[] count = new long[kMaxBit];

    for (final int num : Arrays.stream(nums).boxed().collect(Collectors.toSet()))
      ++count[Integer.bitCount(num)];

    for (int i = 0; i < kMaxBit; ++i)
      for (int j = 0; j < kMaxBit; ++j)
        if (i + j >= k)
          ans += count[i] * count[j];

    return ans;
  }
}

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class Solution:
  def countExcellentPairs(self, nums: List[int], k: int) -> int:
    count = collections.Counter(map(int.bit_count, set(nums)))
    return sum(count[i] * count[j]
               for i in count
               for j in count
               if i + j >= k)