#1878

Medium

LeetCode Problem Workspace

Get Biggest Three Rhombus Sums in a Grid

Find the three largest distinct rhombus sums from a given grid using array and math techniques.

array math sorting heap priority queue matrix

Problem Statement

Given a grid of integers with m rows and n columns, you need to identify the three largest rhombus sums. A rhombus sum is the sum of the elements forming the border of a rhombus, which is shaped like a square rotated by 45 degrees in the grid. The rhombus can have different sizes, and even an area of 0 is valid when no elements surround the center.

The problem involves finding the three largest distinct sums of rhombuses within the grid. Note that the grid can contain various sizes of rhombuses, ranging from a single element (0 area) to larger structures. The challenge is to compute and track the highest values, ensuring efficiency in your approach to handle up to 50x50 grids.

Examples

Example 1

Input: grid = [[3,4,5,1,3],[3,3,4,2,3],[20,30,200,40,10],[1,5,5,4,1],[4,3,2,2,5]]

Output: [228,216,211]

The rhombus shapes for the three biggest distinct rhombus sums are depicted above.

Blue: 20 + 3 + 200 + 5 = 228
Red: 200 + 2 + 10 + 4 = 216
Green: 5 + 200 + 4 + 2 = 211

Example 2

Input: grid = [[1,2,3],[4,5,6],[7,8,9]]

Output: [20,9,8]

The rhombus shapes for the three biggest distinct rhombus sums are depicted above.

Blue: 4 + 2 + 6 + 8 = 20
Red: 9 (area 0 rhombus in the bottom right corner)
Green: 8 (area 0 rhombus in the bottom middle)

Example 3

Input: grid = [[7,7,7]]

Output: [7]

All three possible rhombus sums are the same, so return [7].

Constraints

m == grid.length
n == grid[i].length
1 <= m, n <= 50
1 <= grid[i][j] <= 105

Solution Approach

Brute-force Sum Calculation

Start by calculating the sum of all possible rhombuses in the grid. Iterate through each potential rhombus center, calculate the sum for different sizes of rhombuses, and store the results. This approach will be inefficient for larger grids but helps illustrate the problem.

Optimized Search with Sorting

After computing the rhombus sums, sort them in descending order and select the top three distinct sums. This step ensures that you only retain the largest values, eliminating duplicates and unnecessary calculations. Sorting can be done using a heap or quicksort.

Efficient Sum Calculation with Prefix Sum

Use prefix sums to efficiently calculate the sum of any rhombus in constant time. By precomputing cumulative sums for all subgrids, you can quickly extract the sum of any rhombus, significantly reducing the time complexity.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach. The brute-force method has a time complexity of O(m * n * k), where k is the average size of the rhombus. Sorting the sums takes O(k log k) time. Using prefix sums can reduce the complexity to O(m * n), making it more efficient for larger grids.

What Interviewers Usually Probe

Can the candidate optimize the brute-force method for larger grids?
Does the candidate understand the importance of only keeping the top three distinct sums?
Can the candidate effectively apply sorting and prefix sums in their solution?

Common Pitfalls or Variants

Common pitfalls

Not efficiently calculating rhombus sums for larger grids, leading to timeouts.
Ignoring the requirement to maintain only the top three distinct sums, potentially causing unnecessary memory usage.
Misunderstanding the rhombus shape, resulting in incorrect sum calculations or missed valid rhombuses.

Follow-up variants

Finding the top 5 or 10 distinct rhombus sums instead of the top 3.
Applying the solution to grids with different dimensions, such as rectangular grids.
Adapting the approach to handle non-square rhombus shapes.

FAQ

What is a rhombus sum in the 'Get Biggest Three Rhombus Sums in a Grid' problem?

A rhombus sum is the sum of elements forming the border of a rhombus-shaped region in the grid. The rhombus is a square rotated by 45 degrees, with the center being one of the grid cells.

How do I ensure that I only keep the top three distinct rhombus sums?

After calculating all rhombus sums, sort them in descending order and select the first three distinct sums, ensuring no duplicates are retained.

Can I use prefix sums to optimize the 'Get Biggest Three Rhombus Sums in a Grid' problem?

Yes, prefix sums can be used to calculate rhombus sums efficiently by precomputing cumulative sums for all subgrids, reducing the need to repeatedly calculate sums for the same region.

What are common mistakes when solving the 'Get Biggest Three Rhombus Sums in a Grid' problem?

Common mistakes include inefficient sum calculations, failing to track distinct sums, or misunderstanding the rhombus shape, leading to incorrect results.

How can GhostInterview help with the 'Get Biggest Three Rhombus Sums in a Grid' problem?

GhostInterview provides optimized solutions, guides through sorting and prefix sum applications, and helps ensure efficient handling of large grids while maintaining the top distinct sums.

terminal

Solution

Solution 1: Enumerate Diamond Center + Prefix Sum + Ordered Set

We can preprocess to get two prefix sum arrays $s_1$ and $s_2$, where $s_1[i][j]$ represents the sum of the elements on the upper left diagonal ending at $(i, j)$, and $s_2[i][j]$ represents the sum of the elements on the upper right diagonal ending at $(i, j)$.

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class Solution:
    def getBiggestThree(self, grid: List[List[int]]) -> List[int]:
        m, n = len(grid), len(grid[0])
        s1 = [[0] * (n + 2) for _ in range(m + 1)]
        s2 = [[0] * (n + 2) for _ in range(m + 1)]
        for i, row in enumerate(grid, 1):
            for j, x in enumerate(row, 1):
                s1[i][j] = s1[i - 1][j - 1] + x
                s2[i][j] = s2[i - 1][j + 1] + x
        ss = SortedSet()
        for i, row in enumerate(grid, 1):
            for j, x in enumerate(row, 1):
                l = min(i - 1, m - i, j - 1, n - j)
                ss.add(x)
                for k in range(1, l + 1):
                    a = s1[i + k][j] - s1[i][j - k]
                    b = s1[i][j + k] - s1[i - k][j]
                    c = s2[i][j - k] - s2[i - k][j]
                    d = s2[i + k][j] - s2[i][j + k]
                    ss.add(
                        a + b + c + d - grid[i + k - 1][j - 1] + grid[i - k - 1][j - 1]
                    )
                while len(ss) > 3:
                    ss.remove(ss[0])
        return list(ss)[::-1]

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