LeetCode #1828 — MEDIUM

Queries on Number of Points Inside a Circle

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array points where points[i] = [x_i, y_i] is the coordinates of the i^th point on a 2D plane. Multiple points can have the same coordinates.

You are also given an array queries where queries[j] = [x_j, y_j, r_j] describes a circle centered at (x_j, y_j) with a radius of r_j.

For each query queries[j], compute the number of points inside the j^th circle. Points on the border of the circle are considered inside.

Return an array answer, where answer[j] is the answer to the j^th query.

Example 1:

Input: points = [[1,3],[3,3],[5,3],[2,2]], queries = [[2,3,1],[4,3,1],[1,1,2]]
Output: [3,2,2]
Explanation: The points and circles are shown above.
queries[0] is the green circle, queries[1] is the red circle, and queries[2] is the blue circle.

Example 2:

Input: points = [[1,1],[2,2],[3,3],[4,4],[5,5]], queries = [[1,2,2],[2,2,2],[4,3,2],[4,3,3]]
Output: [2,3,2,4]
Explanation: The points and circles are shown above.
queries[0] is green, queries[1] is red, queries[2] is blue, and queries[3] is purple.

Constraints:

1 <= points.length <= 500
points[i].length == 2
0 <= x_i, y_i <= 500
1 <= queries.length <= 500
queries[j].length == 3
0 <= x_j, y_j <= 500
1 <= r_j <= 500
All coordinates are integers.

Follow up: Could you find the answer for each query in better complexity than O(n)?

Step 01

Brute Force Baseline

Problem summary: You are given an array points where points[i] = [xi, yi] is the coordinates of the ith point on a 2D plane. Multiple points can have the same coordinates. You are also given an array queries where queries[j] = [xj, yj, rj] describes a circle centered at (xj, yj) with a radius of rj. For each query queries[j], compute the number of points inside the jth circle. Points on the border of the circle are considered inside. Return an array answer, where answer[j] is the answer to the jth query.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[[1,3],[3,3],[5,3],[2,2]]
[[2,3,1],[4,3,1],[1,1,2]]

Example 2

[[1,1],[2,2],[3,3],[4,4],[5,5]]
[[1,2,2],[2,2,2],[4,3,2],[4,3,3]]

Core Insight

What unlocks the optimal approach

For a point to be inside a circle, the euclidean distance between it and the circle's center needs to be less than or equal to the radius.
Brute force for each circle and iterate overall points and find those inside it.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1828: Queries on Number of Points Inside a Circle
class Solution {
    public int[] countPoints(int[][] points, int[][] queries) {
        int m = queries.length;
        int[] ans = new int[m];
        for (int k = 0; k < m; ++k) {
            int x = queries[k][0], y = queries[k][1], r = queries[k][2];
            for (var p : points) {
                int i = p[0], j = p[1];
                int dx = i - x, dy = j - y;
                if (dx * dx + dy * dy <= r * r) {
                    ++ans[k];
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1828: Queries on Number of Points Inside a Circle
func countPoints(points [][]int, queries [][]int) (ans []int) {
	for _, q := range queries {
		x, y, r := q[0], q[1], q[2]
		cnt := 0
		for _, p := range points {
			i, j := p[0], p[1]
			dx, dy := i-x, j-y
			if dx*dx+dy*dy <= r*r {
				cnt++
			}
		}
		ans = append(ans, cnt)
	}
	return
}

# Accepted solution for LeetCode #1828: Queries on Number of Points Inside a Circle
class Solution:
    def countPoints(
        self, points: List[List[int]], queries: List[List[int]]
    ) -> List[int]:
        ans = []
        for x, y, r in queries:
            cnt = 0
            for i, j in points:
                dx, dy = i - x, j - y
                cnt += dx * dx + dy * dy <= r * r
            ans.append(cnt)
        return ans

// Accepted solution for LeetCode #1828: Queries on Number of Points Inside a Circle
impl Solution {
    pub fn count_points(points: Vec<Vec<i32>>, queries: Vec<Vec<i32>>) -> Vec<i32> {
        queries
            .iter()
            .map(|v| {
                let cx = v[0];
                let cy = v[1];
                let r = v[2].pow(2);
                let mut count = 0;
                for p in points.iter() {
                    if (p[0] - cx).pow(2) + (p[1] - cy).pow(2) <= r {
                        count += 1;
                    }
                }
                count
            })
            .collect()
    }
}

// Accepted solution for LeetCode #1828: Queries on Number of Points Inside a Circle
function countPoints(points: number[][], queries: number[][]): number[] {
    return queries.map(([cx, cy, r]) => {
        let res = 0;
        for (const [px, py] of points) {
            if (Math.sqrt((cx - px) ** 2 + (cy - py) ** 2) <= r) {
                res++;
            }
        }
        return res;
    });
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.