LeetCode #1727 — MEDIUM

Largest Submatrix With Rearrangements

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

The Problem

Problem Statement

You are given a binary matrix matrix of size m x n, and you are allowed to rearrange the columns of the matrix in any order.

Return the area of the largest submatrix within matrix where every element of the submatrix is 1 after reordering the columns optimally.

Example 1:

Input: matrix = [[0,0,1],[1,1,1],[1,0,1]]
Output: 4
Explanation: You can rearrange the columns as shown above.
The largest submatrix of 1s, in bold, has an area of 4.

Example 2:

Input: matrix = [[1,0,1,0,1]]
Output: 3
Explanation: You can rearrange the columns as shown above.
The largest submatrix of 1s, in bold, has an area of 3.

Example 3:

Input: matrix = [[1,1,0],[1,0,1]]
Output: 2
Explanation: Notice that you must rearrange entire columns, and there is no way to make a submatrix of 1s larger than an area of 2.

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m * n <= 10⁵
matrix[i][j] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: You are given a binary matrix matrix of size m x n, and you are allowed to rearrange the columns of the matrix in any order. Return the area of the largest submatrix within matrix where every element of the submatrix is 1 after reordering the columns optimally.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[[0,0,1],[1,1,1],[1,0,1]]

Example 2

[[1,0,1,0,1]]

Example 3

[[1,1,0],[1,0,1]]

Core Insight

What unlocks the optimal approach

For each column, find the number of consecutive ones ending at each position.
For each row, sort the cumulative ones in non-increasing order and "fit" the largest submatrix.

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 #1727: Largest Submatrix With Rearrangements
class Solution {
    public int largestSubmatrix(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        for (int i = 1; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 1) {
                    matrix[i][j] = matrix[i - 1][j] + 1;
                }
            }
        }
        int ans = 0;
        for (var row : matrix) {
            Arrays.sort(row);
            for (int j = n - 1, k = 1; j >= 0 && row[j] > 0; --j, ++k) {
                int s = row[j] * k;
                ans = Math.max(ans, s);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1727: Largest Submatrix With Rearrangements
func largestSubmatrix(matrix [][]int) int {
	m, n := len(matrix), len(matrix[0])
	for i := 1; i < m; i++ {
		for j := 0; j < n; j++ {
			if matrix[i][j] == 1 {
				matrix[i][j] = matrix[i-1][j] + 1
			}
		}
	}
	ans := 0
	for _, row := range matrix {
		sort.Ints(row)
		for j, k := n-1, 1; j >= 0 && row[j] > 0; j, k = j-1, k+1 {
			ans = max(ans, row[j]*k)
		}
	}
	return ans
}

# Accepted solution for LeetCode #1727: Largest Submatrix With Rearrangements
class Solution:
    def largestSubmatrix(self, matrix: List[List[int]]) -> int:
        for i in range(1, len(matrix)):
            for j in range(len(matrix[0])):
                if matrix[i][j]:
                    matrix[i][j] = matrix[i - 1][j] + 1
        ans = 0
        for row in matrix:
            row.sort(reverse=True)
            for j, v in enumerate(row, 1):
                ans = max(ans, j * v)
        return ans

// Accepted solution for LeetCode #1727: Largest Submatrix With Rearrangements
struct Solution;

impl Solution {
    fn largest_submatrix(matrix: Vec<Vec<i32>>) -> i32 {
        let n = matrix.len();
        let m = matrix[0].len();
        let mut rows = vec![vec![0; m]; n];
        for i in 0..n {
            for j in 0..m {
                if matrix[i][j] == 1 {
                    rows[i][j] = if i > 0 { rows[i - 1][j] } else { 0 } + 1;
                }
            }
        }
        let mut res = 0;
        for i in 0..n {
            res = res.max(Self::rearrange(&mut rows[i]));
        }
        res
    }

    fn rearrange(row: &mut Vec<i32>) -> i32 {
        row.sort_unstable();
        let m = row.len();
        let mut max = 0;
        for i in 0..m {
            max = max.max(row[i] * (m - i) as i32);
        }
        max
    }
}

#[test]
fn test() {
    let matrix = vec_vec_i32![[0, 0, 1], [1, 1, 1], [1, 0, 1]];
    let res = 4;
    assert_eq!(Solution::largest_submatrix(matrix), res);
    let matrix = vec_vec_i32![[1, 0, 1, 0, 1]];
    let res = 3;
    assert_eq!(Solution::largest_submatrix(matrix), res);
    let matrix = vec_vec_i32![[1, 1, 0], [1, 0, 1]];
    let res = 2;
    assert_eq!(Solution::largest_submatrix(matrix), res);
    let matrix = vec_vec_i32![[0, 0], [0, 0]];
    let res = 0;
    assert_eq!(Solution::largest_submatrix(matrix), res);
}

// Accepted solution for LeetCode #1727: Largest Submatrix With Rearrangements
function largestSubmatrix(matrix: number[][]): number {
    for (let column = 0; column < matrix[0].length; column++) {
        for (let row = 0; row < matrix.length; row++) {
            let tempRow = row;
            let count = 0;

            while (tempRow < matrix.length && matrix[tempRow][column] === 1) {
                count++;
                tempRow++;
            }

            while (count !== 0) {
                matrix[row][column] = count;
                count--;
                row++;
            }
        }
    }

    for (let row = 0; row < matrix.length; row++) {
        matrix[row].sort((a, b) => a - b);
    }

    let maxSubmatrixArea = 0;

    for (let row = 0; row < matrix.length; row++) {
        for (let col = matrix[row].length - 1; col >= 0; col--) {
            maxSubmatrixArea = Math.max(
                maxSubmatrixArea,
                matrix[row][col] * (matrix[row].length - col),
            );
        }
    }

    return maxSubmatrixArea;
}

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.