What is the time complexity of Right Triangles?

The optimal solution has a time complexity of O(rows * cols).

What is the space complexity of Right Triangles?

The space complexity is O(rows + cols).

What data structures are used to solve Right Triangles?

This problem uses Array, Hash Table, Math, Combinatorics, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Right Triangles?

Insight 1: Realizing that directly iterating through all possible combinations of three '1' cells would be inefficient. Instead, focus on iterating through each '1' cell and using it as the potential 'right angle' of the triangle. Insight 2: Recognize that for a cell with a value of '1' to be part of a right triangle, there must be at least one other '1' in the same row and at least one other '1' in the same column.

Which companies ask LeetCode 3128. Right Triangles in interviews?

This problem is frequently asked at Mitsogo.

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Right Triangles - LeetCode 3128 Solution

Right Triangles - Complete Solution Guide

Right Triangles is LeetCode problem 3128, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 2D boolean matrix grid . A collection of 3 elements of grid is a right triangle if one of its elements is in the same row with another element and in the same column with the third element. The 3 elements may not be next to each other. Return an integer that is the number of right triangles that can be made with 3 elements of grid such that all of them have a value of 1. Example 1: 0 1 0 0 1 1 0 1 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1 1 0 1 0 Input: grid = [[0,1,0],[0,1,1],[0,1,0]] Outpu

Detailed Explanation

The problem asks us to count the number of 'right triangles' that can be formed from a given 2D boolean matrix (grid). A 'right triangle' is defined as a set of three cells in the grid, all having a value of 1, such that one cell shares the same row as another cell and the same column as the third cell. The cells are not necessarily adjacent.

Solution Approach

The solution efficiently counts the number of right triangles by first pre-computing the number of '1's in each row and column. Then, for each cell with a value of '1', it calculates the number of other '1's in its row and the number of other '1's in its column. The product of these two numbers gives the number of right triangles that can be formed with that cell as the right-angle vertex. By summing these values for all '1' cells, we obtain the total number of right triangles.

Step-by-Step Algorithm

Step 1: Initialize two arrays, `row_ones` and `col_ones`, to store the number of '1's in each row and column, respectively. Initialize them to 0.
Step 2: Iterate through the grid. For each cell `grid[r][c]`, if its value is '1', increment `row_ones[r]` and `col_ones[c]`.
Step 3: Initialize a variable `total_triangles` to 0. This variable will store the final count of right triangles.
Step 4: Iterate through the grid again. For each cell `grid[r][c]`, if its value is '1', check if there are at least two '1's in that row (i.e., `row_ones[r] > 1`) and at least two '1's in that column (i.e., `col_ones[c] > 1`).
Step 5: If both conditions in Step 4 are true, it means we can form right triangles using `grid[r][c]` as the right angle. The number of such triangles is `(row_ones[r] - 1) * (col_ones[c] - 1)`. Add this value to `total_triangles`.
Step 6: After iterating through all cells, return the value of `total_triangles`.

Key Insights

Insight 1: Realizing that directly iterating through all possible combinations of three '1' cells would be inefficient. Instead, focus on iterating through each '1' cell and using it as the potential 'right angle' of the triangle.
Insight 2: Recognize that for a cell with a value of '1' to be part of a right triangle, there must be at least one other '1' in the same row and at least one other '1' in the same column.
Insight 3: Efficiently calculating the number of '1's in each row and column beforehand using auxiliary arrays enables us to avoid repeated iterations and significantly optimize the solution's runtime.

Complexity Analysis

Time Complexity: O(rows * cols)

Space Complexity: O(rows + cols)

Topics

This problem involves: Array, Hash Table, Math, Combinatorics, Counting.

Companies

Asked at: Mitsogo.

Right Triangles | LeetCode 3128

Medium

Array Hash Table Math Combinatorics Counting

Time: O(rows * cols)Space: O(rows + cols)

Solution Code

Available in:

class Solution {
    public long numberOfRightTriangles(int[][] grid) {
        int rows = grid.length;
        int cols = grid[0].length;

        int[] row_ones = new int[rows];
        int[] col_ones = new int[cols];

        for (int r = 0; r < rows; r++) {
            for (int c = 0; c < cols; c++) {
                if (grid[r][c] == 1) {
                    row_ones[r] += 1;
                    col_ones[c] += 1;
                }
            }
        }
        
        long total_triangles = 0;

        for (int r = 0; r < rows; r++) {
            for (int c = 0; c < cols; c++) {
                if (grid[r][c] == 1) {
                    if (row_ones[r] > 1 && col_ones[c] > 1) {
                        total_triangles += (long)(row_ones[r] - 1) * (col_ones[c] - 1);
                    }
                }
            }
        }

        return total_triangles;
    }
}

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Problem Info

Description

Problem 3128: Right Triangles - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Combinatorics Counting

Companies

Mitsogo

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3128

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize two arrays, `row_ones` and `col_ones`, to store the number of '1's in each row and column, respectively. Initialize them to 0.
2
Step 2: Iterate through the grid. For each cell `grid[r][c]`, if its value is '1', increment `row_ones[r]` and `col_ones[c]`.
3
Step 3: Initialize a variable `total_triangles` to 0. This variable will store the final count of right triangles.
4
Step 4: Iterate through the grid again. For each cell `grid[r][c]`, if its value is '1', check if there are at least two '1's in that row (i.e., `row_ones[r] > 1`) and at least two '1's in that column (i.e., `col_ones[c] > 1`).
5
Step 5: If both conditions in Step 4 are true, it means we can form right triangles using `grid[r][c]` as the right angle. The number of such triangles is `(row_ones[r] - 1) * (col_ones[c] - 1)`. Add this value to `total_triangles`.
6
Step 6: After iterating through all cells, return the value of `total_triangles`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(rows * cols)

💾

Space Complexity

O(rows + cols)

📝

Why?

The time complexity is derived from the nested loops that iterate through the grid. The space complexity is derived from the space used by the `row_ones` and `col_ones` arrays, which store counts for each row and column.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-Force Approach

Iterate through all possible combinations of three cells in the grid. For each combination, check if all three cells have a value of '1' and if they form a right triangle according to the problem's definition.

Time: O( (rows * cols)^3 ), as we need to iterate through all possible combinations of three cells.Space: O(1), since no extra space is needed other than a few variables.

Trade-offs: The brute-force approach is much simpler to implement but has a significantly higher time complexity, making it impractical for large grids. The provided solution is much more efficient due to its pre-computation step.

✨

Key Insights

1Insight 1: Realizing that directly iterating through all possible combinations of three '1' cells would be inefficient. Instead, focus on iterating through each '1' cell and using it as the potential 'right angle' of the triangle.
2Insight 2: Recognize that for a cell with a value of '1' to be part of a right triangle, there must be at least one other '1' in the same row and at least one other '1' in the same column.
3Insight 3: Efficiently calculating the number of '1's in each row and column beforehand using auxiliary arrays enables us to avoid repeated iterations and significantly optimize the solution's runtime.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Empty grid (rows or cols = 0). The code handles this implicitly because the loops won't execute, and the function will return 0.
Edge case 2: Grid with no '1's. The `row_ones` and `col_ones` arrays will remain all zeros, and the `total_triangles` will remain 0, so the function will return the correct result (0).
Edge case 3: Grid with only one '1'. No right triangles can be formed and algorithm correctly return zero.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of right triangles by not subtracting '1' from `row_ones[r]` and `col_ones[c]` when counting other '1's in the same row and column.

✗Common mistake 2: Ignoring the constraint that all three vertices of the right triangle must have a value of '1'.

✗Common mistake 3: Not handling edge cases like empty grids or grids with no '1's.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: In social network analysis, you can represent relationships as a grid. Finding right triangles can help identify patterns of connections (e.g., 'friend of a friend').

🏦Real-world scenario 2: In image processing, consider a binary image. This algorithm helps analyze pixel arrangements and potentially identify geometric shapes or patterns that involve right angles.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Right Triangles"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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