How do you solve Equal Sum Grid Partition II on LeetCode?

Equal Sum Grid Partition II (LeetCode #3548) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Prefix sums help in reducing redundant calculations.

What is the time complexity of Equal Sum Grid Partition II?

The optimal solution for Equal Sum Grid Partition II runs in O(n) time and uses O(n) space. We compute prefix sums in linear time, allowing us to check conditions efficiently, leading to O(n) complexity.

What is the brute force solution for Equal Sum Grid Partition II?

The brute force approach for Equal Sum Grid Partition II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible horizontal and vertical cut to see if the sums of the two sections are equal or can be made equal by removing one cell. This involves calculating sums for each cut and checking conditions. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Equal Sum Grid Partition II?

Equal Sum Grid Partition II uses the following concepts: Array, Hash Table, Matrix, Enumeration, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Equal Sum Grid Partition II?

Explain your thought process clearly. Discuss edge cases like single row/column grids. Practice similar problems to build confidence.

What are common mistakes when solving Equal Sum Grid Partition II?

Forgetting to check connectivity after discounting a cell. Not considering all possible cuts.

#3548

Equal Sum Grid Partition II

Hard

ArrayHash TableMatrixEnumerationPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking all cuts, we can calculate prefix sums for rows and columns, allowing us to efficiently determine the sums of sections and check for possible discounts.

⚙️

Algorithm

3 steps

1Step 1: Calculate total sum of the grid and prefix sums for rows and columns.
2Step 2: For each possible cut, check if the sums are equal or can be made equal by removing one cell.
3Step 3: Use a frequency map to track cell values for quick lookup.

solution.py16 lines

1def canPartition(grid):
2    total = sum(sum(row) for row in grid)
3    m, n = len(grid), len(grid[0])
4    row_sums = [sum(row) for row in grid]
5    col_sums = [sum(grid[i][j] for i in range(m)) for j in range(n)]
6    for i in range(m-1):
7        top_sum = sum(row_sums[:i+1])
8        bottom_sum = total - top_sum
9        if top_sum == bottom_sum or abs(top_sum - bottom_sum) <= max(max(row) for row in grid):
10            return True
11    for j in range(n-1):
12        left_sum = sum(col_sums[:j+1])
13        right_sum = total - left_sum
14        if left_sum == right_sum or abs(left_sum - right_sum) <= max(max(row) for row in grid):
15            return True
16    return False

ℹ

Complexity note: We compute prefix sums in linear time, allowing us to check conditions efficiently, leading to O(n) complexity.

1Prefix sums help in reducing redundant calculations.
2Checking for possible discounts can be done efficiently with frequency maps.

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