How do you solve Get Biggest Three Rhombus Sums in a Grid on LeetCode?

Get Biggest Three Rhombus Sums in a Grid (LeetCode #1878) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * min(m, n)) time complexity and O(m * n) space complexity. Key insight: Understanding how to calculate rhombus sums is crucial for solving this problem.

What is the brute force solution for Get Biggest Three Rhombus Sums in a Grid?

The brute force approach for Get Biggest Three Rhombus Sums in a Grid is Brute Force, with O(m * n * min(m, n)) time complexity and O(1) space complexity. The brute force approach involves iterating through each cell in the grid and calculating the rhombus sums for all possible sizes centered at that cell. This method is straightforward but can be inefficient as it checks every possible rhombus shape. The optimal Optimal Solution approach improves this to O(m * n * min(m, n)).

What are the interview tips for Get Biggest Three Rhombus Sums in a Grid?

Always consider edge cases and how they might affect your calculations. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice calculating sums in different shapes to get comfortable with the logic.

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

Failing to account for edge cases where the rhombus may extend beyond the grid boundaries. Not using a set to store distinct sums, which can lead to incorrect results.

#1878

Get Biggest Three Rhombus Sums in a Grid

Medium

ArrayMathSortingHeap (Priority Queue)MatrixPrefix SumPrefix SumSet for distinct values

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * min(m, n))	O(m * n * min(m, n))
Space	O(1)	O(m * n)

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Intuition

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

The optimal solution leverages prefix sums to efficiently calculate rhombus sums without recalculating overlapping areas. This reduces redundant calculations and speeds up the process significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum array to store cumulative sums of the grid.
2Step 2: For each cell, calculate the rhombus sums using the prefix sums for all possible sizes.
3Step 3: Use a max-heap to keep track of the largest three distinct rhombus sums.

solution.py15 lines

1def getBiggestThree(grid):
2    m, n = len(grid), len(grid[0])
3    prefix_sum = [[0] * (n + 1) for _ in range(m + 1)]
4    for i in range(1, m + 1):
5        for j in range(1, n + 1):
6            prefix_sum[i][j] = grid[i - 1][j - 1] + prefix_sum[i - 1][j] + prefix_sum[i][j - 1] - prefix_sum[i - 1][j - 1]
7    rhombus_sums = set()
8    for i in range(m):
9        for j in range(n):
10            for size in range(min(i, j, m - i - 1, n - j - 1) + 1):
11                rhombus_sum = 0
12                for k in range(-size, size + 1):
13                    rhombus_sum += (prefix_sum[i + k + 1][j + size + 1] - prefix_sum[i + k + 1][j - (size - abs(k))] - prefix_sum[i + k][j + 1] + prefix_sum[i + k][j - (size - abs(k))])
14                rhombus_sums.add(rhombus_sum)
15    return sorted(rhombus_sums, reverse=True)[:3]

ℹ

Complexity note: The complexity is similar to the brute force approach, but we gain efficiency in calculating rhombus sums through the use of prefix sums, reducing redundant calculations.

1Understanding how to calculate rhombus sums is crucial for solving this problem.
2Using prefix sums can significantly reduce the number of calculations needed for overlapping areas.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.