grid47 Exploring patterns and algorithms
Sep 23, 2024
5 min read
Solution to LeetCode 447: Number of Boomerangs Problem
You are given n distinct points in the 2D plane. A boomerang is defined as a tuple of three points (i, j, k) where the distance between points i and j equals the distance between points i and k. Count the total number of boomerangs that can be formed from the given points.
Problem
Approach
Steps
Complexity
Input: A list of n distinct points in the 2D plane, each represented by a pair of integers [xi, yi].
Example: [[0,0], [1,0], [2,0]]
Constraints:
• 1 <= n <= 500
• -10^4 <= xi, yi <= 10^4
• All points are distinct.
Output: The output is a single integer representing the number of boomerangs formed from the given points.
Example: 2
Constraints:
• The number of boomerangs can be zero if no such triplets can be formed.
Goal: Count the total number of boomerangs by comparing distances between points.
Steps:
• 1. For each point, calculate the distances to all other points.
• 2. Store the frequency of each distance in a map (or hashmap).
• 3. For each distinct distance, calculate the number of ways to pick two points that share the same distance from the current point.
• 4. The number of boomerangs is the sum of all valid pairs for all distances.
Goal: The points are distinct, and the coordinates are within a specified range.
Steps:
• 1 <= n <= 500
• -10^4 <= xi, yi <= 10^4
• All points are distinct.
Assumptions:
• All points are distinct and represented as pairs of integers.
• Input: [[0,0], [1,0], [2,0]]
• Explanation: The two boomerangs formed are: [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]].
• Input: [[1,1], [2,2], [3,3]]
• Explanation: The two boomerangs formed are: [[1,1], [2,2], [3,3]] and [[2,2], [1,1], [3,3]].
Approach: The approach involves calculating distances between all pairs of points and counting valid triplets where the distance between two points is equal to the distance between another two points.
Observations:
• For each point, we can calculate distances to all other points, which gives us potential boomerangs.
• By storing the frequency of distances, we can efficiently calculate the number of valid boomerangs.
• We can optimize the process by using a hashmap to store the distances from each point.
Steps:
• 1. Iterate over each point in the list of points.
• 2. For each point, calculate the distance to every other point and store the results in a hashmap.
• 3. For each distance, calculate the number of valid pairs of points (i, j) such that the distance is the same.
• 4. Return the total count of boomerangs.
Empty Inputs:
• The input will always contain at least one point, so no need to handle empty inputs.
Large Inputs:
• For large inputs (up to 500 points), ensure that the solution efficiently handles the calculations of distances and the counting of pairs.
Special Values:
• Ensure that the solution handles cases with no possible boomerangs, such as when there are fewer than 3 points.
Constraints:
• The algorithm should efficiently handle the maximum constraint of 500 points.
intnumberOfBoomerangs(vector<vector<int>>& points) {
map<int, int> mp;
int res =0;
for(int i =0; i < points.size(); i++) {
for(int j =0; j < points.size(); j++) {
int d = getDist(points[i], points[j]);
mp[d]++;
}
for(auto [_, val]: mp)
res += val * (val -1);
mp.clear();
}
return res;
}
intgetDist(vector<int> a, vector<int> b) {
int x = a[0] - b[0];
int y = a[1] - b[1];
return x * x + y * y;
}
