How do you solve Maximum Points Inside the Square on LeetCode?

Maximum Points Inside the Square (LeetCode #3143) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: The maximum distance to include a point is determined by the maximum of its x and y coordinates.

What is the time complexity of Maximum Points Inside the Square?

The optimal solution for Maximum Points Inside the Square runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n) time, and we use a hash map to track unique tags, which requires O(n) space.

What is the brute force solution for Maximum Points Inside the Square?

The brute force approach for Maximum Points Inside the Square is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible square size and counts the points inside it. This method is straightforward but inefficient, as it involves checking all pairs of points. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Points Inside the Square?

Maximum Points Inside the Square uses the following concepts: Array, Hash Table, String, Binary Search, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Points Inside the Square?

Always clarify the constraints and edge cases with the interviewer. Discuss your thought process and reasoning behind each step. Consider both time and space complexity in your solution.

What are common mistakes when solving Maximum Points Inside the Square?

Not considering the uniqueness of tags when counting points. Failing to sort points based on their distance before applying the sliding window.

#3143

Maximum Points Inside the Square

Medium

ArrayHash TableStringBinary SearchSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach sorts points based on their distance from the origin and uses a sliding window technique to efficiently count unique tags within a square of varying side lengths.

⚙️

Algorithm

3 steps

1Step 1: Calculate the maximum distance for each point using max(abs(x), abs(y)) and store it along with the corresponding tag.
2Step 2: Sort the points based on this maximum distance.
3Step 3: Use a sliding window to check how many unique tags fit within the current square size defined by the maximum distance of the point at the right end of the window.

solution.py24 lines

1# Full working Python code
2from collections import defaultdict
3
4def maxPoints(points, s):
5    n = len(points)
6    point_info = [(max(abs(x), abs(y)), s[i]) for i, (x, y) in enumerate(points)]
7    point_info.sort()
8    max_count = 0
9    left = 0
10    tags = defaultdict(int)
11
12    for right in range(n):
13        tags[point_info[right][1]] += 1
14        while tags[point_info[right][1]] > 1:
15            tags[point_info[left][1]] -= 1
16            if tags[point_info[left][1]] == 0:
17                del tags[point_info[left][1]]
18            left += 1
19        max_count = max(max_count, len(tags))
20    return max_count
21
22points = [[2,2],[-1,-2],[-4,4],[-3,1],[3,-3]]
23s = 'abdca'
24print(maxPoints(points, s))

ℹ

Complexity note: The sorting step takes O(n log n) time, and we use a hash map to track unique tags, which requires O(n) space.

1The maximum distance to include a point is determined by the maximum of its x and y coordinates.
2Using a sliding window helps efficiently count unique tags without rechecking all points.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.