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, 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
numsallows 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
numsand 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