What is the time complexity of Get Biggest Three Rhombus Sums in a Grid?

The optimal solution has a time complexity of O(m*n*min(m,n)).

What is the space complexity of Get Biggest Three Rhombus Sums in a Grid?

The space complexity is O(m*n*min(m,n)).

What data structures are used to solve Get Biggest Three Rhombus Sums in a Grid?

This problem uses Array, Math, Sorting, Heap (Priority Queue), Matrix, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Get Biggest Three Rhombus Sums in a Grid?

Insight 1: Realizing that a rhombus can be defined by its center and a 'radius' (k) representing the distance from the center to a corner. Insight 2: Efficiently calculating the sum of a rhombus by iterating through its sides and adding the values.

Which companies ask LeetCode 1878. Get Biggest Three Rhombus Sums in a Grid in interviews?

This problem is frequently asked at Capital One, Quora.

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Get Biggest Three Rhombus Sums in a Grid - LeetCode 1878 Solution

Get Biggest Three Rhombus Sums in a Grid - Complete Solution Guide

Get Biggest Three Rhombus Sums in a Grid is LeetCode problem 1878, 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 an m x n integer matrix grid . A rhombus sum is the sum of the elements that form the border of a regular rhombus shape in grid . The rhombus must have the shape of a square rotated 45 degrees with each of the corners centered in a grid cell. Below is an image of four valid rhombus shapes with the corresponding colored cells that should be included in each rhombus sum : Note that the rhombus can have an area of 0, which is depicted by the purple rhombus in the bottom right co

Detailed Explanation

The problem asks us to find the three largest distinct rhombus sums within a given m x n integer matrix (grid). A rhombus sum is calculated by summing the values of the elements that form the border of a rhombus shape. These rhombuses are essentially squares rotated 45 degrees. The rhombus corners are centered at grid cells. We need to return the three largest *distinct* sums in descending order. If there are fewer than three distinct sums, return all distinct sums.

Solution Approach

The solution iterates through each cell of the grid and considers it as the center of a potential rhombus. For each cell, it calculates rhombus sums for different sizes of rhombuses (different values of k). It uses a set to store the distinct rhombus sums. Finally, it sorts the sums in descending order and returns the top three (or fewer if there are fewer than three distinct sums).

Step-by-Step Algorithm

Step 1: Iterate through each cell (r, c) in the grid, treating it as the center of a potential rhombus.
Step 2: For each cell (r, c), calculate the maximum possible radius 'k' of a rhombus centered at that cell. 'k' is limited by the distance to the grid boundaries in all four directions.
Step 3: Calculate the rhombus sum for k = 0 (which is just the value of the cell itself) and add it to the set of rhombus sums.
Step 4: Iterate from k = 1 up to max_k. For each 'k', calculate the rhombus sum by summing the elements along the four sides of the rhombus.
Step 5: Add the calculated rhombus sum to the set of distinct rhombus sums.
Step 6: After iterating through all cells and all possible 'k' values, convert the set of rhombus sums to a list.
Step 7: Sort the list of rhombus sums in descending order.
Step 8: Return the first three elements (or all elements if there are fewer than three) of the sorted list.

Key Insights

Insight 1: Realizing that a rhombus can be defined by its center and a 'radius' (k) representing the distance from the center to a corner.
Insight 2: Efficiently calculating the sum of a rhombus by iterating through its sides and adding the values.
Insight 3: Using a set to store rhombus sums to ensure only distinct sums are considered, thus avoiding duplicates.
Insight 4: The maximum 'radius' (k) is limited by the distance to the edges of the grid from the center cell. It is crucial to correctly bound k.

Complexity Analysis

Time Complexity: O(m*n*min(m,n))

Space Complexity: O(m*n*min(m,n))

Topics

This problem involves: Array, Math, Sorting, Heap (Priority Queue), Matrix, Prefix Sum.

Companies

Asked at: Capital One, Quora.

Get Biggest Three Rhombus Sums in a Grid | LeetCode 1878

Medium

Array Math Sorting Heap (Priority Queue)Matrix Prefix Sum

Time: O(m*n*min(m,n))Space: O(m*n*min(m,n))

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public int[] getBiggestThree(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        Set<Integer> rhombusSums = new HashSet<>();

        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                // Rhombus of area 0 (k=0)
                rhombusSums.add(grid[r][c]);

                // Rhombuses with k > 0
                int maxK = Math.min(Math.min(r, m - 1 - r), Math.min(c, n - 1 - c));
                for (int k = 1; k <= maxK; k++) {
                    int currentSum = 0;

                    // top point: (r-k, c)
                    // right point: (r, c+k)
                    // bottom point: (r+k, c)
                    // left point: (r, c-k)

                    // top-left side (from top point to right point, exclusive of right)
                    for (int i = 0; i < k; i++) {
                        currentSum += grid[r - k + i][c + i];
                    }

                    // top-right side (from right point to bottom point, exclusive of bottom)
                    for (int i = 0; i < k; i++) {
                        currentSum += grid[r + i][c + k - i];
                    }

                    // bottom-right side (from bottom point to left point, exclusive of left)
                    for (int i = 0; i < k; i++) {
                        currentSum += grid[r + k - i][c - i];
                    }

                    // bottom-left side (from left point to top point, exclusive of top)
                    for (int i = 0; i < k; i++) {
                        currentSum += grid[r - i][c - k + i];
                    }

                    rhombusSums.add(currentSum);
                }
            }
        }

        List<Integer> sortedSumsList = new ArrayList<>(rhombusSums);
        sortedSumsList.sort((a, b) -> b - a);

        int[] result = new int[Math.min(3, sortedSumsList.size())];
        for (int i = 0; i < Math.min(3, sortedSumsList.size()); i++) {
            result[i] = sortedSumsList.get(i);
        }

        return result;
    }
}

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Problem Info

Description

Problem 1878: Get Biggest Three Rhombus Sums in a Grid - Detailed explanation and solution approaches.

Topics

Array Math Sorting Heap (Priority Queue)Matrix Prefix Sum

Companies

Capital OneQuora

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1878

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

8 steps

1
Step 1: Iterate through each cell (r, c) in the grid, treating it as the center of a potential rhombus.
2
Step 2: For each cell (r, c), calculate the maximum possible radius 'k' of a rhombus centered at that cell. 'k' is limited by the distance to the grid boundaries in all four directions.
3
Step 3: Calculate the rhombus sum for k = 0 (which is just the value of the cell itself) and add it to the set of rhombus sums.
4
Step 4: Iterate from k = 1 up to max_k. For each 'k', calculate the rhombus sum by summing the elements along the four sides of the rhombus.
5
Step 5: Add the calculated rhombus sum to the set of distinct rhombus sums.
6
Step 6: After iterating through all cells and all possible 'k' values, convert the set of rhombus sums to a list.
7
Step 7: Sort the list of rhombus sums in descending order.
8
Step 8: Return the first three elements (or all elements if there are fewer than three) of the sorted list.

💡

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(m*n*min(m,n))

💾

Space Complexity

O(m*n*min(m,n))

📝

Why?

The algorithm iterates through each cell in the grid (O(m*n)). For each cell, it iterates up to min(m,n) times to calculate rhombus sums (based on k). The rhombus sum calculation itself is O(k) but is bounded by min(m,n). So the time complexity of calculating all rhombus sums is O(m*n*min(m,n)). The set can potentially store up to O(m*n*min(m, n)) distinct sums in the worst case (where many different rhombus sums exist). Sorting the sums takes O(N log N) where N is O(m * n * min(m, n)), however since we extract the top 3 the sorting is dominated by the earlier terms. Therefore the time complexity is O(m * n * min(m, n)). The space complexity is dominated by the set which can store up to O(m * n * min(m, n)) distinct sums in the worst case.

📖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

Using a Priority Queue (Heap)

Instead of storing all rhombus sums in a set and then sorting, a min-heap of size 3 could be used to maintain the largest three distinct sums seen so far. This could offer a slightly different performance profile.

Time: O(m * n * min(m, n) * log(3)) = O(m * n * min(m, n))Space: O(3) = O(1)

Trade-offs: The heap approach might be slightly faster in practice if the number of distinct rhombus sums is very large because it avoids sorting a potentially large list. The space complexity is better (O(1) vs O(m*n*min(m,n))). However, the log(3) factor is small, so the difference might not be significant. The implementation would be more complex.

✨

Key Insights

1Insight 1: Realizing that a rhombus can be defined by its center and a 'radius' (k) representing the distance from the center to a corner.
2Insight 2: Efficiently calculating the sum of a rhombus by iterating through its sides and adding the values.
3Insight 3: Using a set to store rhombus sums to ensure only distinct sums are considered, thus avoiding duplicates.
4Insight 4: The maximum 'radius' (k) is limited by the distance to the edges of the grid from the center cell. It is crucial to correctly bound k.

📚

Key Terms Explained

For Students

Array

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Small grids (e.g., 1x1, 1xn, mx1). The solution correctly handles these cases because it computes max_k correctly, potentially resulting in only the cell values being added to the set. The code also correctly limits the returned array size to a maximum of 3.
Edge case 2: Grids where all elements are the same. The solution will correctly identify this and return a list containing only the distinct element (which will be only one value).
Edge case 3: Grid dimensions are small such that some rhombuses are invalid at the edges. The `max_k` calculation correctly restricts 'k' so that only valid rhombuses are considered.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the rhombus sum, especially when summing the elements along the sides. Off-by-one errors and incorrect indexing are common.

✗Common mistake 2: Not handling the edge cases correctly, such as small grids or when all elements are the same.

✗Common mistake 3: Failing to account for the 'distinct' requirement, leading to duplicate sums in the result. Not using a set will lead to this.

✗Common mistake 4: Incorrectly determining the maximum possible 'radius' (k) for a rhombus, leading to out-of-bounds array accesses or incorrect rhombus sums.

✅

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: Image processing - Identifying dominant patterns or shapes in an image. Rhombus shapes could be used to identify diagonal features.

🏦Real-world scenario 2: Data analysis - Finding prominent trends or clusters in a dataset represented as a grid. The 'rhombus sum' could represent the sum of related data points.

🎮Real-world scenario 3: Game development - Calculating influence maps or scores in grid-based games, where the rhombus shape represents an area of effect or influence.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 18, 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 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 "Get Biggest Three Rhombus Sums in a Grid"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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