LeetCode #3548 — HARD

Equal Sum Grid Partition II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an m x n matrix grid of positive integers. Your task is to determine if it is possible to make either one horizontal or one vertical cut on the grid such that:

  • Each of the two resulting sections formed by the cut is non-empty.
  • The sum of elements in both sections is equal, or can be made equal by discounting at most one single cell in total (from either section).
  • If a cell is discounted, the rest of the section must remain connected.

Return true if such a partition exists; otherwise, return false.

Note: A section is connected if every cell in it can be reached from any other cell by moving up, down, left, or right through other cells in the section.

Example 1:

Input: grid = [[1,4],[2,3]]

Output: true

Explanation:

  • A horizontal cut after the first row gives sums 1 + 4 = 5 and 2 + 3 = 5, which are equal. Thus, the answer is true.

Example 2:

Input: grid = [[1,2],[3,4]]

Output: true

Explanation:

  • A vertical cut after the first column gives sums 1 + 3 = 4 and 2 + 4 = 6.
  • By discounting 2 from the right section (6 - 2 = 4), both sections have equal sums and remain connected. Thus, the answer is true.

Example 3:

Input: grid = [[1,2,4],[2,3,5]]

Output: false

Explanation:

  • A horizontal cut after the first row gives 1 + 2 + 4 = 7 and 2 + 3 + 5 = 10.
  • By discounting 3 from the bottom section (10 - 3 = 7), both sections have equal sums, but they do not remain connected as it splits the bottom section into two parts ([2] and [5]). Thus, the answer is false.

Example 4:

Input: grid = [[4,1,8],[3,2,6]]

Output: false

Explanation:

No valid cut exists, so the answer is false.

Constraints:

  • 1 <= m == grid.length <= 105
  • 1 <= n == grid[i].length <= 105
  • 2 <= m * n <= 105
  • 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 matrix grid of positive integers. Your task is to determine if it is possible to make either one horizontal or one vertical cut on the grid such that: Each of the two resulting sections formed by the cut is non-empty. The sum of elements in both sections is equal, or can be made equal by discounting at most one single cell in total (from either section). If a cell is discounted, the rest of the section must remain connected. Return true if such a partition exists; otherwise, return false. Note: A section is connected if every cell in it can be reached from any other cell by moving up, down, left, or right through other cells in the section.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[[1,4],[2,3]]

Example 2

[[1,2],[3,4]]

Example 3

[[1,2,4],[2,3,5]]
Step 02

Core Insight

What unlocks the optimal approach

  • In a grid (or any subgrid), when can a section be disconnected? Can disconnected components occur if the section spans more than one row and more than one column?
  • Handle single rows or single columns separately. For all other partitions, maintain the sums and value frequencies of each section to check whether removing at most one element from one section can make the two sums equal.
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
Largest constraint values
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 #3548: Equal Sum Grid Partition II
class Solution {
  public boolean canPartitionGrid(int[][] grid) {
    final long sum = Arrays.stream(grid).flatMapToInt(Arrays::stream).asLongStream().sum();
    return canPartition(grid, sum) || canPartition(reversed(grid), sum) ||
        canPartition(reversed(transposed(grid)), sum) || canPartition(transposed(grid), sum);
  }

  private boolean canPartition(int[][] grid, long sum) {
    long topSum = 0;
    Set<Integer> seen = new HashSet<>();
    for (int i = 0; i < grid.length; ++i) {
      topSum += Arrays.stream(grid[i]).asLongStream().sum();
      final long botSum = sum - topSum;
      Arrays.stream(grid[i]).forEach(seen::add);
      if (topSum - botSum == 0 || topSum - botSum == grid[0][0] ||
          topSum - botSum == grid[0][grid[0].length - 1] || topSum - botSum == grid[i][0])
        return true;
      if (grid[0].length > 1 && i > 0 && seen.contains((int) (topSum - botSum)))
        return true;
    }
    return false;
  }

  private int[][] transposed(int[][] grid) {
    int[][] res = new int[grid[0].length][grid.length];
    for (int i = 0; i < grid.length; ++i)
      for (int j = 0; j < grid[0].length; ++j)
        res[j][i] = grid[i][j];
    return res;
  }

  private int[][] reversed(int[][] grid) {
    return Arrays.stream(grid).collect(Collectors.collectingAndThen(Collectors.toList(), list -> {
      Collections.reverse(list);
      return list.toArray(new int[0][]);
    }));
  }
}
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(n)
Space
O(1)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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