# Solving Leetcode Interviews in Seconds with AI: Find K-th Smallest Pair Distance


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "719" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's an efficient solution to find the kth smallest distance among all pairs in an integer array:

*   **Binary Search:** The core idea is to use binary search to find the kth smallest distance. The search space is between 0 (minimum possible distance) and the maximum possible distance (max(nums) - min(nums)).
*   **Counting Pairs:** For each mid-value during the binary search, we count the number of pairs with a distance less than or equal to mid. This count is used to determine if the mid-value is too high or too low.
*   **Optimization with Sorting:** Sorting the input array `nums` allows for an efficient way to count the number of pairs with distances less than or equal to a given mid value. We can use a two-pointer approach after sorting.

*   **Runtime Complexity:** O(n log n + n log W), where n is the length of `nums` and W is the range of possible distances (max(nums) - min(nums)). This is dominated by the sorting and binary search process.
*   **Storage Complexity:** O(1) - constant extra space. Sorting can be done in place, or if not in place, it uses O(n) auxiliary space, but we assume O(1) extra space otherwise.

	
	# Code
	```python
	def smallest_distance_pair(nums, k):
    """
    Finds the kth smallest distance among all pairs in an integer array.

    Args:
        nums: A list of integers.
        k: An integer representing the desired rank of the distance.

    Returns:
        The kth smallest distance among all pairs.
    """
    nums.sort()
    n = len(nums)
    low = 0
    high = nums[-1] - nums[0]

    while low <= high:
        mid = (low + high) // 2
        count = 0
        left = 0
        for right in range(n):
            while nums[right] - nums[left] > mid:
                left += 1
            count += right - left

        if count >= k:
            high = mid - 1
        else:
            low = mid + 1

    return low
	```