#2857

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Count Pairs of Points With Distance k

Solve the problem of counting pairs of points with distance k using array scanning and hash table techniques.

Problem Statement

You are given a 2D integer array coordinates and an integer k, where each entry coordinates[i] represents the coordinates of a point in a 2D plane. The distance between any two points (x1, y1) and (x2, y2) is defined as (x1 XOR x2) + (y1 XOR y2), where XOR is the bitwise XOR operation.

You are tasked with finding the number of pairs (i, j) such that i < j and the distance between the points at coordinates[i] and coordinates[j] equals k. The challenge is to solve this efficiently with large input sizes.

Examples

Example 1

Input: coordinates = [[1,2],[4,2],[1,3],[5,2]], k = 5

Output: 2

We can choose the following pairs:

(0,1): Because we have (1 XOR 4) + (2 XOR 2) = 5.
(2,3): Because we have (1 XOR 5) + (3 XOR 2) = 5.

Example 2

Input: coordinates = [[1,3],[1,3],[1,3],[1,3],[1,3]], k = 0

Output: 10

Any two chosen pairs will have a distance of 0. There are 10 ways to choose two pairs.

Constraints

2 <= coordinates.length <= 50000
0 <= xi, yi <= 106
0 <= k <= 100

Solution Approach

Array Scanning and Hash Table Lookup

To solve the problem efficiently, use array scanning combined with a hash table to track previously seen points. This avoids the need for a brute force O(n^2) solution, allowing for faster lookup of previously processed coordinates. The XOR operation can be exploited to calculate the distance quickly.

Utilize XOR Properties

Using the XOR properties, such as x2 = x XOR x1 and y2 = y XOR y1, enables you to quickly find candidate points that could form valid pairs. By leveraging these properties, we reduce redundant calculations and focus on potential point pairs directly.

Optimizing with a Hash Map

Store the coordinates in a hash map and increment the count when the current point can form a valid pair with any previous point. This approach reduces the problem to O(n) time complexity by leveraging hash lookups for constant time checks.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the chosen approach. Using a hash table for constant time lookups results in an overall time complexity of O(n). Space complexity is also O(n) due to the need to store the coordinates in the hash table.

What Interviewers Usually Probe

Look for understanding of XOR operations and how they can be applied to solve the problem.
Assess familiarity with efficient array scanning combined with hash lookups to optimize performance.
Check if the candidate can explain the reduction of problem complexity through hash map usage.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the problem with brute force solutions that do not scale to large input sizes.
Forgetting to utilize the XOR properties, which leads to unnecessary computations when checking point pairs.
Not handling edge cases properly, such as coordinates with the same values or zero distance.

Follow-up variants

Modify the problem to handle a non-zero distance threshold, testing for pairs within a specific distance range.
Extend the problem by adding constraints where points may have negative coordinates.
Adjust the approach for real-time updates by adding or removing points and recalculating the count of valid pairs.

FAQ

What is the core pattern in the "Count Pairs of Points With Distance k" problem?

The core pattern in this problem is using array scanning combined with hash lookups to efficiently find pairs of points that meet the distance condition defined by XOR operations.

How does the XOR operation help in solving the problem?

The XOR operation simplifies the calculation of distances between points, allowing for faster lookups of potential pairs without directly comparing all point pairs.

What is the time complexity of the solution?

Using a hash table for constant-time lookups, the time complexity of the solution is O(n), where n is the number of points.

Can this problem be solved without using a hash table?

While possible, solving the problem without a hash table would result in a slower, brute force solution with O(n^2) time complexity.

What are some edge cases to consider for this problem?

Edge cases include handling points with identical coordinates, ensuring that the XOR operation correctly handles these scenarios, and managing cases where the distance k is zero.

terminal

Solution

Solution 1: Hash Table + Enumeration

We can use a hash table $cnt$ to count the occurrence of each point in the array $coordinates$.

1

2

3

4

5

6

7

8

9

10

11

class Solution:
    def countPairs(self, coordinates: List[List[int]], k: int) -> int:
        cnt = Counter()
        ans = 0
        for x2, y2 in coordinates:
            for a in range(k + 1):
                b = k - a
                x1, y1 = a ^ x2, b ^ y2
                ans += cnt[(x1, y1)]
            cnt[(x2, y2)] += 1
        return ans

Continue Practicing

#2869 minimum operations to collect elements #2897 apply operations on array to maximize sum of squares #2932 maximum strong pair xor i