leetcode - 719. Find K-th Smallest Pair Distance

Description

The distance of a pair of integers a and b is defined as the absolute difference between a and b.

Given an integer array nums and an integer k, return the kth smallest distance among all the pairs nums[i] and nums[j] where 0 <= i < j < nums.length.

Example 1:

Input: nums = [1,3,1], k = 1
Output: 0
Explanation: Here are all the pairs:
(1,3) -> 2
(1,1) -> 0
(3,1) -> 2
Then the 1st smallest distance pair is (1,1), and its distance is 0.

Example 2:

Input: nums = [1,1,1], k = 2
Output: 0

Example 3:

Input: nums = [1,6,1], k = 3
Output: 5

Constraints:

n == nums.length
2 <= n <= 10^4
0 <= nums[i] <= 10^6
1 <= k <= n * (n - 1) / 2

Solution

Brute Force

Iterate all the possible pairs, then find the shortest distance. Use a max-root heap of size k to keep all the distance we have calculated.

Time complexity: $o(n^2)$
Space complexity: $o(k)$

Binary Search

The search range is [0, max(nums) - min(nums)], the difficult part is to find the monotonic part.

For binary search, we need to find a condition that, if condition(k) is True then condition(k + 1) is True.

Consider we have a given distance, there would be a point that, all the distance that is larger than the given distance have more k pairs, and all the distance that is smaller than given distance have less than k pairs.

	[d1, d2, d3, d4, d5, d6...]
					  ^

If d5 is our target distance, then d6 and larger distances would have more than k pairs.

So we need a function, with the distance as input, and the output is bool, where it’s True if input distance has k or more pairs, and False otherwise.

For this function, we could use 2 pointers. Image we have a fixed index i, and we have j where nums[j] - nums[i] <= distance, we keep moving j to right until nums[j] - nums[i] > distance, then we get j - i - 1 pairs.
Then we move i forward for 1 step, and check if nums[j] - nums[i] <= distance, if so, we keep moving j until it’s not.
Keep this process until i reaches the end of the list.

Time complexity: $o(n\log n)$, $o(n\log n)$ for sorting and binary search, $o(n)$ for 2 pointers.
Space complexity: $o(1)$

Code

Brute Force

class Solution:
    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        k_small_distance = []
        for i in range(len(nums)):
            for j in range(i + 1, len(nums)):
                distance = abs(nums[i] - nums[j])
                heapq.heappush(k_small_distance, -distance)
                if len(k_small_distance) > k:
                    heapq.heappop(k_small_distance)
        return -min(k_small_distance)

Binary Search

class Solution:
    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        def is_enough(distance: int) -> bool:
            """
            Returns:
                True: if there are k or more pairs with the distance or less
                False: if there are less than k pairs with the distance or less
            """
            count = 0
            i, j = 0, 0
            while i < len(nums) or j < len(nums):
                while j < len(nums) and nums[j] - nums[i] <= distance:
                    j += 1
                count += (j - i - 1)
                i += 1
            return count >= k

        nums.sort()
        left, right = 0, nums[-1] - nums[0]
        while left < right:
            mid = (left + right) >> 1
            if is_enough(mid):
                right = mid
            else:
                left = mid + 1
        return (left + right) >> 1