Get Biggest Three Rhombus Sums in a Grid

Get Biggest Three Rhombus Sums in a Grid - Problem

Array Math Sorting Heap (Priority Queue) Matrix Prefix Sum Medium

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.

A rhombus can have an area of 0 (single cell), which means it consists of just one cell. The rhombus border includes all cells that form the diamond shape outline.

Return the biggest three distinct rhombus sums in the grid in descending order. If there are less than three distinct values, return all of them.

Input & Output

Example 1 — Basic Grid with Multiple Rhombuses

$ 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]

💡 Note: The largest rhombus sum is 228 (border of large diamond), followed by 216 and 211. We return the three biggest distinct values in descending order.

Example 2 — Small Grid

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

› Output: [20,16,13]

💡 Note: In a 3x3 grid, we can form various rhombuses. The largest sum is 20 from the border 2+6+8+4, then 16 and 13 from other diamond patterns.

Example 3 — Single Row

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

› Output: [7]

💡 Note: With only single cells possible, all rhombuses have the same sum of 7. We return only one distinct value.

Constraints

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

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8 f Facebook 6

The key insight is to systematically generate all possible rhombus shapes by treating each cell as a potential center and trying all valid radii. A rhombus of radius r has four edges that can be traversed systematically. The optimized approach uses a set for automatic duplicate removal and clean border calculation, achieving O(m×n×min(m,n)²) time complexity while maintaining code clarity.

Common Approaches

✓ Brute Force - Check All Rhombuses

⏱️ Time: O(m×n×min(m,n)×min(m,n)) Space: O(m×n)

For each cell as center, try all possible rhombus sizes and calculate the sum of border elements. Collect all sums, remove duplicates, sort descending, and return top 3.

Frequency Count Approach

⏱️ Time: O(m×n×min(m,n)²) Space: O(m×n)

Use a hash map to count frequencies of rhombus sums, which provides better insight into the distribution of values and can be more memory efficient for sparse results.

Optimized with Set and Early Termination

⏱️ Time: O(m×n×min(m,n)²) Space: O(m×n)

Optimize the brute force approach by using a set to automatically handle duplicates and improve the rhombus border traversal to avoid redundant calculations.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

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Ln 1, Col 1

Smart Actions

💡 Explanation

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