README_EN.md

comments	difficulty	edit_url	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/0700-0799/0719.Find%20K-th%20Smallest%20Pair%20Distance/README_EN.md	Array Two Pointers Binary Search Sorting

719. Find K-th Smallest Pair Distance

中文文档

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 <= 104
• 0 <= nums[i] <= 106
• 1 <= k <= n * (n - 1) / 2

Solutions

Solution 1

Python3

class Solution:
    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        def count(dist):
            cnt = 0
            for i, b in enumerate(nums):
                a = b - dist
                j = bisect_left(nums, a, 0, i)
                cnt += i - j
            return cnt

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

Java

class Solution {
    public int smallestDistancePair(int[] nums, int k) {
        Arrays.sort(nums);
        int left = 0, right = nums[nums.length - 1] - nums[0];
        while (left < right) {
            int mid = (left + right) >> 1;
            if (count(mid, nums) >= k) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private int count(int dist, int[] nums) {
        int cnt = 0;
        for (int i = 0; i < nums.length; ++i) {
            int left = 0, right = i;
            while (left < right) {
                int mid = (left + right) >> 1;
                int target = nums[i] - dist;
                if (nums[mid] >= target) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
            cnt += i - left;
        }
        return cnt;
    }
}

C++

class Solution {
public:
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int left = 0, right = nums.back() - nums.front();
        while (left < right) {
            int mid = (left + right) >> 1;
            if (count(mid, k, nums) >= k)
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }

    int count(int dist, int k, vector<int>& nums) {
        int cnt = 0;
        for (int i = 0; i < nums.size(); ++i) {
            int target = nums[i] - dist;
            int j = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
            cnt += i - j;
        }
        return cnt;
    }
};

Go

func smallestDistancePair(nums []int, k int) int {
	sort.Ints(nums)
	n := len(nums)
	left, right := 0, nums[n-1]-nums[0]
	count := func(dist int) int {
		cnt := 0
		for i, v := range nums {
			target := v - dist
			left, right := 0, i
			for left < right {
				mid := (left + right) >> 1
				if nums[mid] >= target {
					right = mid
				} else {
					left = mid + 1
				}
			}
			cnt += i - left
		}
		return cnt
	}
	for left < right {
		mid := (left + right) >> 1
		if count(mid) >= k {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

TypeScript

function smallestDistancePair(nums: number[], k: number): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    let left = 0,
        right = nums[n - 1] - nums[0];
    while (left < right) {
        let mid = (left + right) >> 1;
        let count = 0,
            i = 0;
        for (let j = 0; j < n; j++) {
            // 索引[i, j]距离nums[j]的距离<=mid
            while (nums[j] - nums[i] > mid) {
                i++;
            }
            count += j - i;
        }
        if (count >= k) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
function smallestDistancePair(nums, k) {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    let left = 0,
        right = nums[n - 1] - nums[0];

    while (left < right) {
        const mid = (left + right) >> 1;
        let count = 0,
            i = 0;

        for (let j = 0; j < n; j++) {
            while (nums[j] - nums[i] > mid) {
                i++;
            }
            count += j - i;
        }

        if (count >= k) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }

    return left;
}