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

Equal Sum Grid Partition I (LeetCode #3546) 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: Total sum must be even for equal partition.

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

The optimal solution for Equal Sum Grid Partition I runs in O(n) time and uses O(n) space. The complexity is linear due to the single pass needed for prefix sums.

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

The brute force approach for Equal Sum Grid Partition I is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible horizontal and vertical cut to see if the sums of the resulting sections are equal. This method is straightforward but inefficient for larger grids. The optimal Optimal Solution approach improves this to O(n).

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

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

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

Explain your thought process clearly. Discuss edge cases and constraints. Practice coding without an IDE.

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

Forgetting to check if total sum is even. Not handling edge cases with minimal grid sizes.

#3546

Equal Sum Grid Partition I

Medium

ArrayMatrixEnumerationPrefix 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)

Use prefix sums to efficiently calculate section sums. This allows us to determine if a cut can yield equal sums in linear time.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total sum of the grid and check if it's even.
2Step 2: Compute prefix sums for rows and check for horizontal cuts.
3Step 3: Compute prefix sums for columns and check for vertical cuts.

solution.py16 lines

1def canPartition(grid):
2    total_sum = sum(sum(row) for row in grid)
3    if total_sum % 2 != 0:
4        return False
5    target = total_sum // 2
6    prefix_sum = 0
7    for r in range(len(grid) - 1):
8        prefix_sum += sum(grid[r])
9        if prefix_sum == target:
10            return True
11    prefix_sum = 0
12    for c in range(len(grid[0]) - 1):
13        prefix_sum += sum(grid[i][c] for i in range(len(grid)))
14        if prefix_sum == target:
15            return True
16    return False

ℹ

Complexity note: The complexity is linear due to the single pass needed for prefix sums.

1Total sum must be even for equal partition.
2Prefix sums allow efficient sum calculations.

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