K Closest Points to Origin
Problem Description
Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0).
The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2).
You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in).
Example 1:
Input: points = [[1,3],[-2,2]], k = 1
Output: [[-2,2]]
Explanation:
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest k = 1 points from the origin, so the answer is just [[-2,2]].
Example 2:
Input: points = [[3,3],[5,-1],[-2,4]], k = 2
Output: [[3,3],[-2,4]]
Explanation: The answer [[-2,4],[3,3]] would also be accepted.
Constraints:
- 1 <= k <= points.length <= 104
- -104 <= xi, yi <= 104
Solve it on LeetCode
Approaches
1. Naive Approach: Sorting
Intuition:
The most straightforward way to solve this problem is to calculate the distance of each point from the origin and then sort the points based on these distances.
Steps:
- Calculate the Euclidean distance for each point from the origin.
- Store the distances along with their respective points.
- Sort the points based on the calculated distances.
- Return the first
Kpoints from the sorted list.
Code:
- Time Complexity: O(N log N), where N is the number of points. This is due to the sorting step.
- Space Complexity: O(1), as we are sorting the array in place.
2. Optimized Approach: Heap/Priority Queue
Intuition:
Instead of sorting the entire array, we can maintain a max heap of size K to keep track of the K closest points. This allows for more efficient insertion and removal operations as we process each point.
Steps:
- Create a max heap (priority queue) that will store the points based on their distance from the origin.
- Iterate through each point and calculate its distance from the origin.
- Maintain the heap to only contain the
Kclosest points by removing the farthest point when the heap size exceedsK. - Convert the result from the heap to an array and return it.
Code:
- Time Complexity: O(N log K), where N is the number of points and K is the number of closest points required. This is due to the priority queue operations.
- Space Complexity: O(K), for storing the
Kclosest points in the heap.
3. Optimal Approach: Quickselect
Intuition:
Quickselect is an optimization over quicksort that allows us to partition the array and only focus on the K closest elements. It can achieve an average time complexity better than sorting or using a heap.
Steps:
- Implement a partitioning method to organize points relative to a pivot such that all points closer than the pivot appear before all points farther than the pivot.
- Recursively adjust the partitioning based on the position of K relative to the pivot's final index until the partition yielding the closest K points is achieved.
Code:
- Time Complexity: O(N) on average, but O(N^2) in the worst case due to the recursive partitioning process.
- Space Complexity: O(1), as it is an in-place sorting algorithm.