Find K-th Smallest Pair Distance

We want to find the kth smallest distance among all pairs in the array nums. One efficient way to do this is by using binary search along with the two pointers technique. Here’s the plan:

  1. Sort the array nums.
  2. Perform binary search over the possible distances, ranging from 0 to the maximum possible distance.
  3. Use two pointers to count the number of pairs with a distance less than or equal to the mid-value of the binary search.
  4. Adjust the binary search bounds according to the count in comparison to k.

Here’s the code:

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class Solution:
    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        nums.sort()
        low, high = 0, nums[-1] - nums[0]  # Possible range of distances

        while low < high:
            mid = (low + high) // 2
            count = 0
            j = 0
            # Use two pointers technique to count pairs with distance <= mid
            for i in range(len(nums)):
                while j < len(nums) and nums[j] - nums[i] <= mid:
                    j += 1
                count += j - i - 1

            # If there are fewer than k pairs with distance <= mid, then increase low
            if count < k:
                low = mid + 1
            else: # Otherwise, reduce high
                high = mid

        return low

The binary search helps in identifying the kth smallest distance efficiently, and sorting the array allows for a quick count of pairs with a specific distance. This code leverages both sorting and binary search, resulting in a more efficient algorithm that meets the problem’s constraints.

Identifying Problem Isomorphism

“Find K-th Smallest Pair Distance” involves searching for the Kth smallest distance among all pairs in a sorted array. This problem can be solved using a binary search approach on the range of possible distances.

An isomorphic problem is “378. Kth Smallest Element in a Sorted Matrix”. In this problem, you’re tasked with finding the Kth smallest element in a row and column-wise sorted matrix. This problem also applies a binary search technique, but the search space is the range of numbers present in the matrix.

The common thread between these problems is the binary search methodology. In both, the search is not performed directly on the data but on a potential solution space. The difference lies in the data structure (array vs matrix) and what the ‘Kth’ item refers to (a distance vs an element). Hence, it is an approximate mapping.

By studying and understanding the solutions to these problems, you’ll gain insights into using binary search in a non-traditional manner, which can be valuable for tackling similar problems.

“Find K-th Smallest Pair Distance” can be mapped to “Kth Smallest Element in a Sorted Matrix”.

The “Find K-th Smallest Pair Distance” problem requires you to find the K-th smallest distance among all possible pairs in an array. This involves sorting pairs according to their distance and returning the K-th element.

The “Kth Smallest Element in a Sorted Matrix” problem requires you to find the K-th smallest element in a matrix that is sorted both row-wise and column-wise. This involves traversing a sorted structure (the matrix) and returning the K-th smallest element.

These problems share the key concept of finding the K-th smallest element in a sorted structure, which is the reason for this mapping. However, the types of structures and the definitions of ‘distance’ or ’element’ differ, making this an approximate mapping. “Kth Smallest Element in a Sorted Matrix” is simpler as it involves a straightforward traversal of a sorted matrix, while the other requires calculating distances among pairs.

10 Prerequisite LeetCode Problems

“719. Find K-th Smallest Pair Distance” requires understanding of binary search and sorting. Here are 10 problems as preparation:

  1. “278. First Bad Version”: This is a simple problem to get started with binary search.

  2. “704. Binary Search”: This is a direct problem on binary search and will help you understand its basics.

  3. “35. Search Insert Position”: In this problem, you apply binary search in a slightly different context, where you have to find the insert position of a target.

  4. “349. Intersection of Two Arrays”: This problem requires sorting and binary search and is a bit more complex than the previous ones.

  5. “167. Two Sum II - Input array is sorted”: This problem deals with finding pair with a certain sum in a sorted array.

  6. “34. Find First and Last Position of Element in Sorted Array”: In this problem, you will use binary search to find the first and last position of an element.

  7. “658. Find K Closest Elements”: This problem requires finding the k closest elements to a given value, which involves sorting and binary search.

  8. “392. Is Subsequence”: This problem uses a binary search concept to check if a sequence is a subsequence of another.

  9. “540. Single Element in a Sorted Array”: This problem requires understanding of binary search in depth to find a single element in a sorted array.

  10. “275. H-Index II”: In this problem, binary search is applied on a sorted citations array to find the h-index.

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def smallestDistancePair(nums, k):
    nums.sort()

    n = len(nums)
    l = 0
    r = nums[n - 1] - nums[0]

    while l < r:
        cnt = 0
        m = l + (r - l) // 2

        i = 0
        j = 0
        while i < n:
            while j < n and nums[j] <= nums[i] + m: 
                j += 1
            cnt += j - i - 1
            i += 1

        if cnt < k:
            l = m + 1
        else:
            r = m

    return l

Problem Classification

This problem falls under the domain of “Combinatorics” and “Array manipulation”. It combines elements of counting (enumerating all pairs), sorting (finding the kth smallest pair), and array manipulation (computing the difference between pairs).

‘What’ Components:

  1. Integer Array (“nums”): This is the primary data structure that we’re dealing with in the problem. Each of its elements are used to form pairs.

  2. Pairs of integers: We’re asked to generate all possible pairs of integers from the given array. This is a combinatorial task.

  3. Difference calculation: For each pair, we’re asked to calculate the absolute difference between the two integers. This is a simple arithmetic operation.

  4. Kth smallest difference: We’re asked to find the kth smallest difference among all pairs. This is a sorting/ranking task.

This problem can be classified as a “Combinatorial Search” problem. We’re tasked with generating all possible pairs of elements, performing a calculation on each pair, and then sorting these results to find the kth smallest.

The key challenges in this problem include generating all pairs efficiently, calculating differences, and finding the kth smallest difference. While the first two tasks are relatively straightforward, the last task can be quite challenging, especially given the problem’s constraints.

One naive solution might be to generate all pairs, calculate their differences, sort the differences, and return the kth smallest. However, this approach is likely to be too slow given the size of the input array. So, we’ll need to consider more efficient search/sorting techniques, such as binary search or heap-based methods.

Language Agnostic Coding Drills

  1. Dissecting the code:

The code can be broken down into several discrete programming concepts:

a. Variable and Array Initialization: The basic concept of initializing variables and arrays. Here we initialize variables like ’l’, ‘r’, ’n’, ’m’, and ‘cnt’ as well as sort an input array.

b. Loops: The concept of using loops (specifically nested while-loops in this case) to iterate over elements.

c. Conditions: Using if-else conditions to perform different operations based on whether the condition is met.

d. Binary Search: The problem uses a binary search to efficiently find the kth smallest pair difference. This requires understanding of how binary search works and how to implement it in code.

e. Two Pointers: The algorithm uses a two-pointer technique in the inner while-loop to keep track of the elements being compared. This technique often requires a good understanding of the problem and careful management of the pointers.

  1. Coding Drills by Increasing Difficulty:

a. Variable and Array Initialization: This is usually the first concept introduced to beginners. It is about how to create and initialize variables and arrays.

b. Loops: This concept is slightly more advanced than simple variable initialization, as it involves understanding control flow and how to iterate over arrays.

c. Conditions: Using if-else conditions is slightly more complex, as it requires understanding of Boolean logic.

d. Binary Search: Implementing a binary search requires a combination of the previously mentioned concepts, along with a good understanding of how binary search works.

e. Two Pointers: Using two pointers efficiently typically requires good understanding of the problem at hand and careful planning.

  1. Problem-Solving Approach:

The solution starts by sorting the array, which makes the subsequent steps easier. The algorithm then initializes the search space to [0, max(nums) - min(nums)]. It enters a loop, in which it calculates the midpoint of the current search space. Using two pointers ‘i’ and ‘j’, it counts the number of pairs with difference less than or equal to ’m’. This count is compared to ‘k’. If the count is less than ‘k’, we know our desired pair difference lies in the upper half of the search space. Else, it lies in the lower half. This process is repeated until our search space is narrowed down to a single element, which is our answer.

Targeted Drills in Python

  1. Python-based Coding Drills for Identified Concepts:

a. Variable and Array Initialization:

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n = 5  # Initialize a variable
arr = [1, 2, 3, 4, 5]  # Initialize an array
arr.sort()  # Sort an array

b. Loops:

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for i in range(n):  # Loop through numbers from 0 to n-1
    print(i)

i = 0
while i < n:  # Loop while a certain condition is met
    print(i)
    i += 1

c. Conditions:

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if n > 2:  # Use if-else condition
    print("n is greater than 2")
else:
    print("n is not greater than 2")

d. Binary Search:

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def binary_search(arr, target):
    l, r = 0, len(arr) - 1
    while l <= r:
        mid = l + (r - l) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            l = mid + 1
        else:
            r = mid - 1
    return -1

e. Two Pointers:

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def two_pointers(arr, target):
    i, j = 0, len(arr) - 1
    while i < j:
        if arr[i] + arr[j] == target:
            return [i, j]
        elif arr[i] + arr[j] < target:
            i += 1
        else:
            j -= 1
    return -1
  1. Problem-specific Concepts:

The two problem-specific concepts in this problem are the usage of binary search over pair differences (instead of array elements) and the two-pointer technique to count the number of pairs with difference less than or equal to mid. These drills would look similar to the binary search and two pointer drills mentioned above, but with a problem-specific twist.

  1. Integrating the Drills:

The drills can be integrated together in the following manner:

  • Start with array initialization and sorting.
  • Initialize the variables used for binary search (l and r).
  • Enter a loop, which will continue until the binary search condition is met.
  • Inside the loop, initialize the other variables (m and cnt).
  • Use the two-pointer technique inside the loop to count the number of pairs with difference less than or equal to m.
  • Use a condition to check if cnt is less than k. If it is, update l to m + 1. Else, update r to m.
  • Finally, return l, which will be the kth smallest pair difference.