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719. Find K-th Smallest Pair Distance 👍

  • Time: $O(\texttt{sort}) + O(n\log (\max(\texttt{nums}) - \min(\texttt{nums})))$
  • Space: $O(\texttt{sort})$
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class Solution {
 public:
  int smallestDistancePair(vector<int>& nums, int k) {
    ranges::sort(nums);

    int l = 0;
    int r = nums.back() - nums.front();

    while (l < r) {
      const int m = (l + r) / 2;
      if (numPairDistancesNoGreaterThan(nums, m) >= k)
        r = m;
      else
        l = m + 1;
    }

    return l;
  }

 private:
  int numPairDistancesNoGreaterThan(const vector<int>& nums, int m) {
    int count = 0;
    int j = 1;
    // For each index i, find the first index j s.t. nums[j] > nums[i] + m,
    // so numPairDistancesNoGreaterThan for the index i will be j - i - 1.
    for (int i = 0; i < nums.size(); ++i) {
      while (j < nums.size() && nums[j] <= nums[i] + m)
        ++j;
      count += j - i - 1;
    }
    return count;
  }
};

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class Solution {
  public int smallestDistancePair(int[] nums, int k) {
    Arrays.sort(nums);

    int l = 0;
    int r = nums[nums.length - 1] - nums[0];

    while (l < r) {
      final int m = (l + r) / 2;
      if (numPairDistancesNoGreaterThan(nums, m) >= k)
        r = m;
      else
        l = m + 1;
    }

    return l;
  }

  private int numPairDistancesNoGreaterThan(int[] nums, int m) {
    int count = 0;
    int j = 1;
    // For each index i, find the first index j s.t. nums[j] > nums[i] + m,
    // so numPairDistancesNoGreaterThan for the index i will be j - i - 1.
    for (int i = 0; i < nums.length; ++i) {
      while (j < nums.length && nums[j] <= nums[i] + m)
        ++j;
      count += j - i - 1;
    }
    return count;
  }
}

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class Solution:
  def smallestDistancePair(self, nums: list[int], k: int) -> int:
    nums.sort()

    def numPairDistancesNoGreaterThan(m: int) -> int:
      count = 0
      j = 1
      # For each index i, find the first index j s.t. nums[j] > nums[i] + m,
      # so numPairDistancesNoGreaterThan for the index i will be j - i - 1.
      for i, num in enumerate(nums):
        while j < len(nums) and nums[j] <= num + m:
          j += 1
        count += j - i - 1
      return count

    return bisect.bisect_left(range(nums[-1] - nums[0]), k,
                              key=numPairDistancesNoGreaterThan)