LeetCode #3143 — MEDIUM

Maximum Points Inside the Square

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

The Problem

Problem Statement

You are given a 2D array points and a string s where, points[i] represents the coordinates of point i, and s[i] represents the tag of point i.

A valid square is a square centered at the origin (0, 0), has edges parallel to the axes, and does not contain two points with the same tag.

Return the maximum number of points contained in a valid square.

Note:

A point is considered to be inside the square if it lies on or within the square's boundaries.
The side length of the square can be zero.

Example 1:

Input: points = [[2,2],[-1,-2],[-4,4],[-3,1],[3,-3]], s = "abdca"

Output: 2

Explanation:

The square of side length 4 covers two points points[0] and points[1].

Example 2:

Input: points = [[1,1],[-2,-2],[-2,2]], s = "abb"

Output: 1

Explanation:

The square of side length 2 covers one point, which is points[0].

Example 3:

Input: points = [[1,1],[-1,-1],[2,-2]], s = "ccd"

Output: 0

Explanation:

It's impossible to make any valid squares centered at the origin such that it covers only one point among points[0] and points[1].

Constraints:

1 <= s.length, points.length <= 10⁵
points[i].length == 2
-10⁹ <= points[i][0], points[i][1] <= 10⁹
s.length == points.length
points consists of distinct coordinates.
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 2D array points and a string s where, points[i] represents the coordinates of point i, and s[i] represents the tag of point i. A valid square is a square centered at the origin (0, 0), has edges parallel to the axes, and does not contain two points with the same tag. Return the maximum number of points contained in a valid square. Note: A point is considered to be inside the square if it lies on or within the square's boundaries. The side length of the square can be zero.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search

Example 1

[[2,2],[-1,-2],[-4,4],[-3,1],[3,-3]]
"abdca"

Example 2

[[1,1],[-2,-2],[-2,2]]
"abb"

Example 3

[[1,1],[-1,-1],[2,-2]]
"ccd"

Core Insight

What unlocks the optimal approach

The smallest edge length of a square to include point <code>(x, y)</code> is <code>max(abs(x), abs(y)) * 2</code>.
Sort the points by <code>max(abs(x), abs(y))</code> and try each edge length, check the included point tags.

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 #3143: Maximum Points Inside the Square
class Solution {
    public int maxPointsInsideSquare(int[][] points, String s) {
        TreeMap<Integer, List<Integer>> g = new TreeMap<>();
        for (int i = 0; i < points.length; ++i) {
            int x = points[i][0], y = points[i][1];
            int key = Math.max(Math.abs(x), Math.abs(y));
            g.computeIfAbsent(key, k -> new ArrayList<>()).add(i);
        }
        boolean[] vis = new boolean[26];
        int ans = 0;
        for (var idx : g.values()) {
            for (int i : idx) {
                int j = s.charAt(i) - 'a';
                if (vis[j]) {
                    return ans;
                }
                vis[j] = true;
            }
            ans += idx.size();
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3143: Maximum Points Inside the Square
func maxPointsInsideSquare(points [][]int, s string) (ans int) {
	g := map[int][]int{}
	for i, p := range points {
		key := max(p[0], -p[0], p[1], -p[1])
		g[key] = append(g[key], i)
	}
	vis := [26]bool{}
	keys := []int{}
	for k := range g {
		keys = append(keys, k)
	}
	sort.Ints(keys)
	for _, k := range keys {
		idx := g[k]
		for _, i := range idx {
			j := s[i] - 'a'
			if vis[j] {
				return
			}
			vis[j] = true
		}
		ans += len(idx)
	}
	return
}

# Accepted solution for LeetCode #3143: Maximum Points Inside the Square
class Solution:
    def maxPointsInsideSquare(self, points: List[List[int]], s: str) -> int:
        g = defaultdict(list)
        for i, (x, y) in enumerate(points):
            g[max(abs(x), abs(y))].append(i)
        vis = set()
        ans = 0
        for d in sorted(g):
            idx = g[d]
            for i in idx:
                if s[i] in vis:
                    return ans
                vis.add(s[i])
            ans += len(idx)
        return ans

// Accepted solution for LeetCode #3143: Maximum Points Inside the Square
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3143: Maximum Points Inside the Square
// class Solution {
//     public int maxPointsInsideSquare(int[][] points, String s) {
//         TreeMap<Integer, List<Integer>> g = new TreeMap<>();
//         for (int i = 0; i < points.length; ++i) {
//             int x = points[i][0], y = points[i][1];
//             int key = Math.max(Math.abs(x), Math.abs(y));
//             g.computeIfAbsent(key, k -> new ArrayList<>()).add(i);
//         }
//         boolean[] vis = new boolean[26];
//         int ans = 0;
//         for (var idx : g.values()) {
//             for (int i : idx) {
//                 int j = s.charAt(i) - 'a';
//                 if (vis[j]) {
//                     return ans;
//                 }
//                 vis[j] = true;
//             }
//             ans += idx.size();
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3143: Maximum Points Inside the Square
function maxPointsInsideSquare(points: number[][], s: string): number {
    const n = points.length;
    const g: Map<number, number[]> = new Map();
    for (let i = 0; i < n; ++i) {
        const [x, y] = points[i];
        const key = Math.max(Math.abs(x), Math.abs(y));
        if (!g.has(key)) {
            g.set(key, []);
        }
        g.get(key)!.push(i);
    }
    const keys = Array.from(g.keys()).sort((a, b) => a - b);
    const vis: boolean[] = Array(26).fill(false);
    let ans = 0;
    for (const key of keys) {
        const idx = g.get(key)!;
        for (const i of idx) {
            const j = s.charCodeAt(i) - 'a'.charCodeAt(0);
            if (vis[j]) {
                return ans;
            }
            vis[j] = true;
        }
        ans += idx.length;
    }
    return ans;
}

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 × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Maximum Points Inside the Square

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Mutating counts without cleanup

Boundary update without `+1` / `-1`