How do you solve Number of Boomerangs on LeetCode?

Number of Boomerangs (LeetCode #447) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using a hashmap to count occurrences of distances can significantly reduce the number of checks needed.

What is the brute force solution for Number of Boomerangs?

The brute force approach for Number of Boomerangs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible triplets of points to see if they form a boomerang. This is straightforward but inefficient, as it involves checking every combination of three points. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Number of Boomerangs?

Number of Boomerangs uses the following concepts: Array, Hash Table, Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Boomerangs?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as when there are fewer than three points. Practice explaining your thought process as you code to simulate the interview environment.

What are common mistakes when solving Number of Boomerangs?

Not considering the order of points when counting boomerangs. Forgetting to check if points are distinct before calculating distances.

#447

Number of Boomerangs

Medium

ArrayHash TableMathHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

The optimal solution uses a hashmap to count distances from each anchor point to all other points. This allows us to quickly find how many points are at the same distance, significantly reducing the number of checks needed.

⚙️

Algorithm

4 steps

1Step 1: Initialize a count variable to zero.
2Step 2: For each point, use a hashmap to store the frequency of distances to all other points.
3Step 3: For each distance in the hashmap, if there are 'k' points at that distance, add k * (k - 1) to the count (as each pair can form a boomerang).
4Step 4: Return the total count.

solution.py11 lines

1def numberOfBoomerangs(points):
2    count = 0
3    for p in points:
4        distance_count = {}
5        for q in points:
6            if p != q:
7                dist = (p[0] - q[0]) ** 2 + (p[1] - q[1]) ** 2
8                distance_count[dist] = distance_count.get(dist, 0) + 1
9        for k in distance_count.values():
10            count += k * (k - 1)
11    return count

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we optimize the distance checks using a hashmap. The space complexity is O(n) for storing distances in the hashmap.

1Using a hashmap to count occurrences of distances can significantly reduce the number of checks needed.
2The order of points matters in boomerangs, so we must account for permutations.

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