LeetCode #1878 — MEDIUM

Get Biggest Three Rhombus Sums in a Grid

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

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 corner.

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.

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]
Explanation: 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]
Explanation: 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]
Explanation: 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

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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 corner. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[[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]]

Example 2

[[1,2,3],[4,5,6],[7,8,9]]

Example 3

[[7,7,7]]

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Step 02

Core Insight

What unlocks the optimal approach

  • You need to maintain only the biggest 3 distinct sums
  • The limits are small enough for you to iterate over all rhombus sizes then iterate over all possible borders to get the sums
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1878: Get Biggest Three Rhombus Sums in a Grid
class Solution {
    public int[] getBiggestThree(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] s1 = new int[m + 1][n + 2];
        int[][] s2 = new int[m + 1][n + 2];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                s1[i][j] = s1[i - 1][j - 1] + grid[i - 1][j - 1];
                s2[i][j] = s2[i - 1][j + 1] + grid[i - 1][j - 1];
            }
        }
        TreeSet<Integer> ss = new TreeSet<>();
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int l = Math.min(Math.min(i - 1, m - i), Math.min(j - 1, n - j));
                ss.add(grid[i - 1][j - 1]);
                for (int k = 1; k <= l; ++k) {
                    int a = s1[i + k][j] - s1[i][j - k];
                    int b = s1[i][j + k] - s1[i - k][j];
                    int c = s2[i][j - k] - s2[i - k][j];
                    int 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 (ss.size() > 3) {
                    ss.pollFirst();
                }
            }
        }
        int[] ans = new int[ss.size()];
        for (int i = 0; i < ans.length; ++i) {
            ans[i] = ss.pollLast();
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(m × n × \min(m, n)
Space
O(m × n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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