How do you solve Right Triangles on LeetCode?

Right Triangles (LeetCode #3128) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The right triangle can be formed by ensuring one element is in the same row as another and in the same column as the third.

What is the brute force solution for Right Triangles?

The brute force approach for Right Triangles is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check every possible combination of three elements in the grid to see if they form a right triangle. This is straightforward but inefficient, as it involves checking many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Right Triangles?

Right Triangles uses the following concepts: Array, Hash Table, Math, Combinatorics, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Right Triangles?

Always start with a brute force approach to understand the problem better. Look for patterns in the problem that can simplify your solution. Be ready to explain your thought process clearly, especially when transitioning from brute force to an optimal solution.

What are common mistakes when solving Right Triangles?

Not considering all combinations of 1s in the brute force approach, leading to missed triangles. Confusing the conditions for forming a right triangle.

#3128

Right Triangles

Medium

ArrayHash TableMathCombinatoricsCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Instead of checking all combinations, we can count the number of 1s in each row and column. For each 1 found, the number of right triangles it can form is determined by the counts of 1s in its row and column.

⚙️

Algorithm

3 steps

1Step 1: Count the number of 1s in each row and each column.
2Step 2: For each cell that contains a 1, calculate the potential triangles it can form using the formula: (rowCount - 1) * (colCount - 1).
3Step 3: Sum these values to get the total number of right triangles.

solution.py20 lines

1# Full working Python code
2
3def count_right_triangles(grid):
4    rows = [0] * len(grid)
5    cols = [0] * len(grid[0])
6    count = 0
7
8    for i in range(len(grid)):
9        for j in range(len(grid[0])):
10            if grid[i][j] == 1:
11                rows[i] += 1
12                cols[j] += 1
13
14    for i in range(len(grid)):
15        for j in range(len(grid[0])):
16            if grid[i][j] == 1:
17                count += (rows[i] - 1) * (cols[j] - 1)
18
19    return count
20

ℹ

Complexity note: The time complexity is O(n) because we are iterating through the grid a couple of times, and the space complexity is O(n) due to the additional arrays for row and column counts.

1The right triangle can be formed by ensuring one element is in the same row as another and in the same column as the third.
2Counting the number of 1s in rows and columns allows us to efficiently calculate potential triangles.

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