🚀Day 2. K Closest Points to Origin 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given an array of points where each point is represented as points[i] = [xi, yi] 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 defined as: $\text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ You may return the k closest points in any order. The driver code will sort them before printing.

🔍 Example Walkthrough:

Example 1:

Input:

k = 2, points[] = [[1, 3], [-2, 2], [5, 8], [0, 1]]

Output:

[[-2, 2], [0, 1]]

Explanation:

The Euclidean distances from the origin (0,0) are:

• Point (1,3) = √10

• Point (-2,2) = √8

• Point (5,8) = √89

• Point (0,1) = √1

The two closest points to the origin are [-2,2] and [0,1].

Example 2:

Input:

k = 1, points = [[2, 4], [-1, -1], [0, 0]]

Output:

[[0, 0]]

Explanation:

The Euclidean distances from the origin (0,0) are:

• Point (2,4) = √20

• Point (-1,-1) = √2

• Point (0,0) = √0

The closest point to the origin is (0,0).

Constraints:

• $( 1 \leq k \leq \text{points.size()} \leq 10^5 )$

• $( -10^4 \leq x_i, y_i \leq 10^4 )$

🎯 My Approach:

Partial Sort using nth_element (O(N + K log K) Time, O(1) Space)

1. We need to find the k closest points to the origin based on their Euclidean distance.

2. Instead of sorting the entire array (which takes O(N log N)), we can use nth_element which partially sorts the array in O(N + K log K) time.

3. The first k elements in the sorted range will be the k closest points.

4. Since we don’t need exact sorting, nth_element is much more efficient than a full sort.

Algorithm Steps:

1. Use nth_element to place the k closest points at the beginning of the array.

2. Extract the first k points and return them as the result.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N + K log K), since nth_element partially sorts the array.

• Expected Auxiliary Space Complexity: O(1), as sorting is done in place without extra memory.

📝 Solution Code

Code (C++)

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& p, int k) {
        nth_element(p.begin(), p.begin() + k, p.end(), [](auto&a, auto&b){
            return a[0]*a[0] + a[1]*a[1] < b[0]*b[0] + b[1]*b[1];
        });
        return {p.begin(), p.begin()+k};
    }
};

⚡ Alternative Approaches (C++)

2️⃣ Min-Heap Approach (O(N log K) Time, O(K) Space)

1. Use a max-heap of size k to store the k closest points.

2. Iterate through all points, pushing them into the heap.

3. If the heap size exceeds k, remove the farthest point.

4. At the end, the heap contains the k closest points.

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& p, int k) {
        priority_queue<pair<int, vector<int>>> pq;
        for (auto& a : p) {
            int d = a[0] * a[0] + a[1] * a[1];
            pq.push({d, a});
            if (pq.size() > k) pq.pop();
        }
        vector<vector<int>> res;
        while (!pq.empty()) res.push_back(pq.top().second), pq.pop();
        return res;
    }
};

🔹 Pros: Efficient for streaming data. 🔹 Cons: Uses extra space (O(K)).

3️⃣ Sorting Approach (O(N log N) Time, O(1) Space)

1. Sort all points based on their Euclidean distance.

2. Pick the first k points.

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& p, int k) {
        sort(p.begin(), p.end(), [](auto& a, auto& b) {
            return a[0] * a[0] + a[1] * a[1] < b[0] * b[0] + b[1] * b[1];
        });
        return vector<vector<int>>(p.begin(), p.begin() + k);
    }
};

🔹 Pros: Simple to implement. 🔹 Cons: Inefficient for large N.

📊 Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	⚡ Method	✅ Pros	⚠️ Cons
Optimized Partial Sort	🟢 O(N + K log K)	🟢 O(1)	nth_element	Best runtime & space efficiency	None
Min-Heap (Priority Queue)	🟡 O(N log K)	🟡 O(K)	Heap-based	Good for streaming data	Extra space usage
Sorting Approach	🔴 O(N log N)	🟢 O(1)	Sorting	Simple & easy to implement	Slow for large N

💡 Best Choice?

• ✅ For best efficiency: Partial Sort (O(N + K log K), O(1)).

• ✅ For real-time data processing: Min-Heap (O(N log K), O(K)).

• ✅ For simplicity: Sorting Approach (O(N log N), O(1)).

Code (Java)

class Solution {
    public int[][] kClosest(int[][] p, int k) {
        Arrays.sort(p, Comparator.comparingInt(a -> a[0] * a[0] + a[1] * a[1]));
        return Arrays.copyOfRange(p, 0, k);
    }
}

Code (Python)

class Solution:
    def kClosest(self, p: list[list[int]], k: int) -> list[list[int]]:
        return sorted(p, key=lambda a: a[0]**2 + a[1]**2)[:k]

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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