Problem Description

You are provided with a collection of n distinct points in a two-dimensional plane, represented by the coordinates points[i] = [xi, yi]. A boomerang is a set of three points (i, j, k) with the condition that the distance between point i and point j is the same as the distance between point i and point k. It's important to note that the order of the points in this tuple matters — which means (i, j, k) and (i, k, j) are considered different boomerangs if j and k are different points.

Your task is to find and return the total number of boomerangs among the given points.

Intuition

To solve this problem, we iterate through each point and consider it as the center point i of a potential boomerang. Then, for each of these center points, we compute the distance to every other point j and k in the list. If several points are at the same distance from the center point, these points can form multiple boomerangs, because they can be reordered while keeping the center point fixed.

One intuitive approach is to use a hash map (dictionary in Python) to store the number of points at a certain distance from the center point. For every distance, we can calculate the number of possible permutations for boomerangs with two identical legs (the segments i-j and i-k). Specifically, for n points at the same distance from i, there are n * (n - 1) possible boomerangs, because you can pick the first point in n ways and the second point in (n - 1) ways.

To implement this, we can use the Counter class from the Python Collections module, which essentially creates a frequency table of the distances. Then, we iterate through the distance frequency table, calculate the number of possible boomerangs for each distance, and add them up to get the result for the current center point. We do this for each point in the list and sum the results to get the final answer.

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Solution Approach

The solution uses a combination of a brute-force approach and a hash map to efficiently count the number of boomerangs. Here's a step-by-step explanation of the implementation:

1. Initialize a variable ans to store the total number of boomerangs.

2. Loop through each point in the points list. This point will act as the point i in the potential boomerang (i, j, k).

3. Inside the loop, create a Counter object called counter. This will be used to track the frequency of distances from the current point i to all other points.

4. Now, loop through each point again within the outer loop to check the distance from point i to every other point (let's call this point q).

5. Calculate the squared distance between points i (the current point in the outer loop) and q (the current point in the inner loop) by using the formula: distance = (p[0] - q[0]) * (p[0] - q[0]) + (p[1] - q[1]) * (p[1] - q[1]). Squared distances are used instead of actual distances to avoid floating-point comparison issues and the need for computation-intensive square root operations.

6. Increment the count for this distance in the counter. The key is the distance, and the value is the number of points at that distance from i.

7. After filling the counter with all distances for the current point i, iterate through the counter values, and for each distance value val, calculate the number of possible boomerangs using the formula val * (val - 1). This represents selecting any two points (order matters) out of the val points that are equidistant from i.

8. Add the calculated potential boomerangs to the ans variable.

9. Continue the process until all points have been used as point i.

10. Finally, return the value stored in ans, which is the total number of boomerangs.

The algorithm essentially boils down to:

• For every point, count the number of points at each distinct distance.
• For each distance that has more than one point, calculate the possible arrangements of boomerang pairs and sum them up.

This method is O(n^2) in time complexity because it involves a nested loop over the points, with each loop being O(n). The space complexity is O(n) in the worst case, as the counter may contain up to n-1 distinct distance entries if all points are equidistant from point i.

Example Walkthrough

Consider an example where points = [[0,0], [1,0], [2,0]]. We will use the abovementioned solution approach to calculate the total number of boomerangs.

1. Initialize ans = 0 as there are no boomerangs counted yet.

2. Start with the first point [0,0] and create an empty Counter object: counter = Counter().

3. Loop through all points to compute distances from [0,0] to each point:

   • The distance from [0,0] to itself is 0. counter[0] becomes 1.
   • The distance from [0,0] to [1,0] is 1^2 + 0^2 = 1. counter[1] becomes 1.
   • The distance from [0,0] to [2,0] is 2^2 + 0^2 = 4. counter[4] becomes 1.

4. Since there are no two points with the same distance from [0,0], the total for this center point is 0. No updates to ans.

5. Repeat step 2 with the second point [1,0]:

   • The distance from [1,0] to [0,0] is 1. counter[1] becomes 1.
   • The distance from [1,0] to itself is 0. counter[0] becomes 1.
   • The distance from [1,0] to [2,0] is 1. counter[1] becomes 2 (since we already had one point at distance 1).

6. In this case, counter has {1: 2, 0: 1}. There are counter[1] = 2 points at distance 1, so we use the formula 2 * (2 - 1) to find the number of boomerangs with [1,0] as the center which is 2.

7. Update ans by adding the number of boomerangs calculated: ans += 2.

8. Repeat steps 2-7 for the third point [2,0]. This will result in the same count as with point [1,0] because it is the reflection of the situation with respect to the y-axis.

9. After processing all points, the ans is 2 + 2 = 4, which means there are 4 boomerangs in this set of points.

Thus, the function should return 4 based on the example input.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code snippet involves two nested loops. The outer loop iterates over each point p in the list of points. For each point p, the inner loop compares it to every other point q in the list to calculate the distances and store them in a hash table (counter).

The outer loop runs n times, where n is the number of points. For each iteration of the outer loop, the inner loop also runs n times to calculate the distances from point p to every other point q. Therefore, the two nested loops result in an O(n^2) time complexity for the distance calculations.

After calculating distances, the code iterates over the values in the hash table, which in the worst case contains n entries (this is the case when all distances from point p to every other point q are unique). For each unique distance, an O(1) operation is performed to compute the combination of boomerangs. The sum of combinations for each distance is also O(n) since it depends on the number of entries in the counter.

Therefore, the total time complexity of the code is O(n^2) for the loops, plus O(n) for the sum of combinations. As the O(n^2) term dominates, the overall time complexity is O(n^2).

Space Complexity

The space complexity of the code is primarily determined by the storage required for the hash table counter. In the worst case, if every distance calculated is unique, the counter will hold n entries, where n is the number of points. Therefore, the space complexity is O(n).

In addition to the counter, a fixed amount of space is used for variables such as ans and distance, which does not depend on the number of points and thus contributes an O(1) term.

As a result, the total space complexity of the code is O(n) due to the hash table.

Learn more about how to find time and space complexity quickly using problem constraints.