How do you solve Count Pairs of Points With Distance k on LeetCode?

Count Pairs of Points With Distance k (LeetCode #2857) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using XOR allows for efficient distance calculations based on bit manipulation.

What is the brute force solution for Count Pairs of Points With Distance k?

The brute force approach for Count Pairs of Points With Distance k is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of points to see if their calculated distance equals k. This is straightforward but inefficient for large datasets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Pairs of Points With Distance k?

Count Pairs of Points With Distance k uses the following concepts: Array, Hash Table, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Pairs of Points With Distance k?

Always clarify the problem statement and constraints before jumping to solutions. Think about edge cases, such as duplicate points or maximum values for k. Discuss your thought process and reasoning for each step with the interviewer.

What are common mistakes when solving Count Pairs of Points With Distance k?

Not considering the order of pairs (i < j) when counting. Forgetting to initialize the HashMap correctly or mishandling the key format.

#2857

Count Pairs of Points With Distance k

Medium

ArrayHash TableBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a HashMap to store the frequency of possible distances, allowing us to find pairs more efficiently. This reduces the need for nested loops and speeds up the process significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a HashMap to keep track of the frequency of points based on their coordinates.
2Step 2: Iterate through each point and calculate the required coordinates for the other point that would yield the distance k.
3Step 3: For each point, determine how many times the required coordinates have been seen before using the HashMap.
4Step 4: Update the HashMap with the current point's coordinates after checking.

solution.py15 lines

1# Full working Python code
2
3def countPairs(coordinates, k):
4    from collections import defaultdict
5    count = 0
6    freq_map = defaultdict(int)
7    for x1, y1 in coordinates:
8        for x2 in range(0, 2**20):  # 20 bits for 10^6
9            y2 = (k - (x1 ^ x2))
10            count += freq_map[(x2, y2)]
11        freq_map[(x1, y1)] += 1
12    return count
13
14# Example usage
15print(countPairs([[1,2],[4,2],[1,3],[5,2]], 5))

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the list of coordinates once, and the space complexity is O(n) due to the storage of point frequencies in the HashMap.

1Using XOR allows for efficient distance calculations based on bit manipulation.
2HashMaps can significantly reduce the time complexity by storing previously seen coordinates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.