Problem Description

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.

Intuition

To solve this problem, consider the properties of the squares centered at the origin with edges parallel to the axes. For any point (x, y), determine the edge length of the square such that the point lies on the boundary of or within the square. This can be calculated as max(abs(x), abs(y)), giving the distance to the furthest axis-aligned component from the origin.

The core idea is to group points based on this distance and consider consecutive increasing distances, starting from zero, adding layers of points into the square as long as a valid square configuration is maintained. A valid configuration requires that no two points within the current square share the same tag.

By storing these distances and corresponding points in a hash table and keeping track of the tags using a set while iterating through sorted distances, we can count the maximum number of valid points. If any duplicate tags are encountered while processing a group of points, the traversal stops for that square size, and the current count is returned as the solution.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The solution uses a combination of a hash table and sorting to efficiently determine the maximum number of points that can fit inside a valid square while adhering to the tag constraint. Here’s the detailed breakdown:

1. Hash Table Construction:

   • Iterate over each point (x, y) in the points array.
   • Calculate the distance d from the origin to the boundary of the prospective square. This distance is max(abs(x), abs(y)).
   • Use this distance d as a key in a hash table g, where each key stores a list of indices of points that have the same distance d from the origin. This grouping helps in processing points layer by layer starting from the smallest squares.

2. Distance Sorting:

   • Sort the distances stored in the hash table. This ensures that we process the points starting from the smallest possible square (which is crucial for maximizing the number of points).

3. Maintaining Validity with Tags:

   • Initialize a set vis to keep track of the tags of the points included in the current valid square.
   • Initialize a counter ans to store the maximum number of valid points inside a square.
   • Iterate over each distance d in the sorted order:
     • Retrieve all point indices corresponding to this distance.
     • For each point index:
       • Check the tag s[i] of the point. If this tag is already in vis, it implies adding this point would violate the unique tag constraint, so break the loop and return ans.
       • Otherwise, add the tag to vis.
     • Add the number of points processed at this distance to ans.

4. Result:

   • After processing all possible layers without violating the constraints, ans contains the maximum number of points that can fit in a valid square.

The use of a hash table helps in efficiently grouping points, while sorting ensures the correctness of layering points into the square.

Example Walkthrough

Let's illustrate the solution approach with a small example.

Suppose you have the following points represented by a 2D array points and their corresponding tags in a string s:

points = [[1, 2], [-1, -2], [3, 1], [-3, 1]]
s = "abcd"

Step-by-Step Breakdown:

1. Hash Table Construction:

   • Calculate the distance d from the origin for each point, using the formula max(abs(x), abs(y)).
     • For point (1, 2), d = max(abs(1), abs(2)) = 2.
     • For point (-1, -2), d = max(abs(-1), abs(-2)) = 2.
     • For point (3, 1), d = max(abs(3), abs(1)) = 3.
     • For point (-3, 1), d = max(abs(-3), abs(1)) = 3.
   • Build a hash table g:
     • g = {2: [0, 1], 3: [2, 3]}.

2. Distance Sorting:

   • Sort the keys (distances) in g to get [2, 3].

3. Maintaining Validity with Tags:

   • Initialize vis = set() to track unique tags.

   • Initialize ans = 0 to store the maximum valid points.

   • Iterate over sorted distances:

     For distance 2:

     • Points with indices [0, 1] correspond to tags 'a' and 'b'.
     • Neither 'a' nor 'b' are in vis.
     • Add 'a' and 'b' to vis.
     • Increment ans by 2 (ans = 2).

     For distance 3:

     • Points with indices [2, 3] correspond to tags 'c' and 'd'.
     • Neither 'c' nor 'd' are in vis.
     • Add 'c' and 'd' to vis.
     • Increment ans by 2 (ans = 4).

4. Result:

   • After processing, ans is 4. This is the maximum number of points that can fit in a valid square at the origin with no duplicate tags.

This approach ensures we count the maximum number of points following the rules of tags and distance from the origin. The key lies in grouping points by their distance from the origin and maintaining a unique set of tags.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n × log n). This is because the code sorts a list containing the keys of the defaultdict g, which has at most n elements, where each element corresponds to a unique maximum distance from the origin for a point. Sorting this list requires O(n × log n) operations. Then, iterating through each list in g and checking membership in a set vis contributes O(n) operations.

The space complexity of the code is O(n). This is due to the storage of the indices of points in the defaultdict g, which holds indices for each potential value of maximum distance, and also due to the storage of characters in the set vis, both of which are proportional to n.

Learn more about how to find time and space complexity quickly.