Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B.
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.
Note:
2 <= len(nums) <= 10000.
0 <= nums[i] < 1000000.
1 <= k <= len(nums) * (len(nums) - 1) / 2.
class Solution
public int smallestDistancePair(int[] nums, int k) {
Arrays.sort(nums);
int lo = 0;
int hi = nums[nums.length - 1] - nums[0];
while (lo < hi) {
int mi = (lo + hi) / 2;
int count = 0, left = 0;
for (int right = 0; right < nums.length; ++right) {
while (nums[right] - nums[left] > mi)
left++;
count += right - left;
}
//count = number of pairs with distance <= mi
if (count >= k)
hi = mi;
else
lo = mi + 1;
}
return
