LeetCode #3546 — MEDIUM

Equal Sum Grid Partition I

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

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 the elements in both sections is equal.

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

Example 1:

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

Output: true

Explanation:

A horizontal cut between row 0 and row 1 results in two non-empty sections, each with a sum of 5. Thus, the answer is true.

Example 2:

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

Output: false

Explanation:

No horizontal or vertical cut results in two non-empty sections with equal sums. Thus, 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

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 the elements in both sections is equal. Return true if such a partition exists; otherwise return false.

Baseline thinking

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

Pattern signal: Array

Example 1

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

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

There are two types of cuts: a <code>horizontal</code> cut or a <code>vertical</code> cut.
For a <code>horizontal</code> cut at row <code>r</code> (0 <= r <m - 1), split <code>grid</code> into rows 0...r vs. r+1...m-1 and compare their sums.
For a <code>vertical</code> cut at column <code>c</code> (0 <= c < n - 1), split <code>grid</code> into columns 0...c vs. c+1...n-1 and compare their sums.
Brute‑force all possible <code>r</code> and <code>c</code> cuts; if any yields equal section sums, return <code>true</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #3546: Equal Sum Grid Partition I
class Solution {
    public boolean canPartitionGrid(int[][] grid) {
        long s = 0;
        for (var row : grid) {
            for (int x : row) {
                s += x;
            }
        }
        if (s % 2 != 0) {
            return false;
        }
        int m = grid.length, n = grid[0].length;
        long pre = 0;
        for (int i = 0; i < m; ++i) {
            for (int x : grid[i]) {
                pre += x;
            }
            if (pre * 2 == s && i < m - 1) {
                return true;
            }
        }
        pre = 0;
        for (int j = 0; j < n; ++j) {
            for (int i = 0; i < m; ++i) {
                pre += grid[i][j];
            }
            if (pre * 2 == s && j < n - 1) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #3546: Equal Sum Grid Partition I
func canPartitionGrid(grid [][]int) bool {
	s := 0
	for _, row := range grid {
		for _, x := range row {
			s += x
		}
	}
	if s%2 != 0 {
		return false
	}
	m, n := len(grid), len(grid[0])
	pre := 0
	for i, row := range grid {
		for _, x := range row {
			pre += x
		}
		if pre*2 == s && i+1 < m {
			return true
		}
	}
	pre = 0
	for j := 0; j < n; j++ {
		for i := 0; i < m; i++ {
			pre += grid[i][j]
		}
		if pre*2 == s && j+1 < n {
			return true
		}
	}
	return false
}

# Accepted solution for LeetCode #3546: Equal Sum Grid Partition I
class Solution:
    def canPartitionGrid(self, grid: List[List[int]]) -> bool:
        s = sum(sum(row) for row in grid)
        if s % 2:
            return False
        pre = 0
        for i, row in enumerate(grid):
            pre += sum(row)
            if pre * 2 == s and i != len(grid) - 1:
                return True
        pre = 0
        for j, col in enumerate(zip(*grid)):
            pre += sum(col)
            if pre * 2 == s and j != len(grid[0]) - 1:
                return True
        return False

// Accepted solution for LeetCode #3546: Equal Sum Grid Partition I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3546: Equal Sum Grid Partition I
// class Solution {
//     public boolean canPartitionGrid(int[][] grid) {
//         long s = 0;
//         for (var row : grid) {
//             for (int x : row) {
//                 s += x;
//             }
//         }
//         if (s % 2 != 0) {
//             return false;
//         }
//         int m = grid.length, n = grid[0].length;
//         long pre = 0;
//         for (int i = 0; i < m; ++i) {
//             for (int x : grid[i]) {
//                 pre += x;
//             }
//             if (pre * 2 == s && i < m - 1) {
//                 return true;
//             }
//         }
//         pre = 0;
//         for (int j = 0; j < n; ++j) {
//             for (int i = 0; i < m; ++i) {
//                 pre += grid[i][j];
//             }
//             if (pre * 2 == s && j < n - 1) {
//                 return true;
//             }
//         }
//         return false;
//     }
// }

// Accepted solution for LeetCode #3546: Equal Sum Grid Partition I
function canPartitionGrid(grid: number[][]): boolean {
    let s = 0;
    for (const row of grid) {
        s += row.reduce((a, b) => a + b, 0);
    }
    if (s % 2 !== 0) {
        return false;
    }
    const [m, n] = [grid.length, grid[0].length];
    let pre = 0;
    for (let i = 0; i < m; ++i) {
        pre += grid[i].reduce((a, b) => a + b, 0);
        if (pre * 2 === s && i + 1 < m) {
            return true;
        }
    }
    pre = 0;
    for (let j = 0; j < n; ++j) {
        for (let i = 0; i < m; ++i) {
            pre += grid[i][j];
        }
        if (pre * 2 === s && j + 1 < n) {
            return true;
        }
    }
    return false;
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

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.