How do you solve Separate Squares I on LeetCode?

Separate Squares I (LeetCode #3453) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(maxY)) time complexity and O(1) space complexity. Key insight: Understanding how areas above and below a line can be calculated is crucial.

What is the brute force solution for Separate Squares I?

The brute force approach for Separate Squares I is Brute Force, with O(n²) time complexity and O(n) space complexity. We can check every possible horizontal line by calculating the area of squares above and below it. This is simple but inefficient, as it requires checking many lines. The optimal Optimal Solution approach improves this to O(n log(maxY)).

What algorithm or data structure is used to solve Separate Squares I?

Separate Squares I uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Area Calculation.

What are the interview tips for Separate Squares I?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, such as squares that are completely above or below the line. Practice explaining your thought process while coding to simulate interview conditions.

What are common mistakes when solving Separate Squares I?

Failing to account for overlapping areas correctly. Not using binary search effectively, leading to inefficient solutions.

#3453

Separate Squares I

Medium

ArrayBinary SearchBinary SearchArea Calculation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log(maxY))Space O(1)

We can use binary search to find the minimum y-coordinate efficiently by leveraging the cumulative area above and below a line. This reduces the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total area of all squares.
2Step 2: Use binary search on the range of y-coordinates to find the line where the area above equals the area below.
3Step 3: For each mid-point in the binary search, calculate the areas above and below that line.

solution.py19 lines

1# Full working Python code
2def min_y_coordinate(squares):
3    total_area = sum(l * l for _, _, l in squares)
4    low, high = 0, 10**9
5    
6    while low < high:
7        mid = (low + high) / 2
8        area_above = 0
9        area_below = 0
10        for x, y, l in squares:
11            if y + l > mid:
12                area_above += (min(y + l, mid) - mid) * l
13            if y < mid:
14                area_below += (min(mid, y + l) - y) * l
15        if area_above < area_below:
16            low = mid + 1
17        else:
18            high = mid
19    return low

ℹ

Complexity note: The time complexity is O(n log(maxY)) because we perform a binary search over the range of possible y-coordinates, checking the area for each square in O(n) time.

1Understanding how areas above and below a line can be calculated is crucial.
2Binary search can significantly reduce the number of checks needed to find the correct y-coordinate.

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