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

Updated
3 min read

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, 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

    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

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