The Maximum Square Area by Removing Fences From a Field coding problem involves a large rectangular field with horizontal and vertical fences at specific coordinates. You can remove any number of fences to create a larger "cell." Your goal is to find the maximum possible area of a square field that can be formed by the remaining fences.
Atlassian uses this problem to test a candidate's ability to use Hash Tables for efficient lookups. The problem requires you to find a common "distance" that exists between both horizontal fences and vertical fences. It evaluates your skill in enumeration and your ability to handle large coordinates by focusing on the differences between them.
This follows the Array, Hash Table, and Enumeration interview pattern.
d that exists in both sets defines a d x d square. The answer is d² (modulo 10^9 + 7).Field: 4x4. Horiz fences: [2]. Vert fences: [3]. Boundaries: H: [0, 2, 4], V: [0, 3, 4]. Horizontal distances:
The most common mistake is a brute-force approach that checks every combination of four fences, which is O(H² * V²) and too slow. Another error is forgetting to include the field's outer boundaries (0 and m/n) as "fences." Some candidates also fail to return -1 if no square (other than potentially the original fences) can be formed.
For "Hash Table, Enumeration interview pattern" problems, think about what property needs to be "shared." Here, the square property requires width == height. By storing all possible heights and then checking all possible widths against them, you efficiently find the largest shared value.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Find the Maximum Number of Elements in Subset | Medium | Solve | |
| Number of Black Blocks | Medium | Solve | |
| Count Special Quadruplets | Easy | Solve | |
| Form Smallest Number From Two Digit Arrays | Easy | Solve | |
| Sum of Imbalance Numbers of All Subarrays | Hard | Solve |