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

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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)
  
  

• 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
  }
  

• 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;
  }
  
  

• /**
   * @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;
  }
  
  

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