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Formatted question description: https://leetcode.ca/all/1727.html

1727. Largest Submatrix With Rearrangements

Level

Medium

Description

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:

Image text

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:

Image text

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.

Example 4:

Input: matrix = [[0,0],[0,0]]

Output: 0

Explanation: As there are no 1s, no submatrix of 1s can be formed and the area is 0.

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m * n <= 10^5
  • matrix[i][j] is 0 or 1.

Solution

For each element in matrix, calculate the number of consecutive 1’s above the element. Concretely, if matrix[i][j] == 1, then find the minimum k such that for all k <= p <= i, there is matrix[p][j] == 1. Create newMatrix with the same size as matrix, and set newMatrix[i][j] = i - k + 1 where k is defined above.

Then for each row curRow in newMatrix, sort curRow and loop over curRow backwards. For curRow[j], the current area is calculated as (n - j) * curRow[j]. Maintain the maximum area in the process.

Finally, return the maximum area.

  • class Solution {
    public int largestSubmatrix(int[][] matrix) {
        int maxArea = 0;
        int rows = matrix.length, columns = matrix[0].length;
        int[][] newMatrix = new int[rows][columns];
        for (int j = 0; j < columns; j++)
            newMatrix[0][j] = matrix[0][j];
        for (int i = 1; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                if (matrix[i][j] == 1)
                    newMatrix[i][j] = newMatrix[i - 1][j] + 1;
            }
        }
        for (int i = 0; i < rows; i++) {
            int[] curRow = new int[columns];
            System.arraycopy(newMatrix[i], 0, curRow, 0, columns);
            Arrays.sort(curRow);
            for (int j = columns - 1; j >= 0; j--) {
                if (curRow[j] == 0)
                    break;
                int area = (columns - j) * curRow[j];
                maxArea = Math.max(maxArea, area);
            }
        }
        return maxArea;
    }
}

############

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;
    }
}

  • // OJ: https://leetcode.com/problems/largest-submatrix-with-rearrangements/
// Time: O(MNlogN)
// Space: O(N)
class Solution {
public:
    int largestSubmatrix(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        vector<int> h(N);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                h[j] = A[i][j] == 0 ? 0 : (h[j] + 1);
            }
            map<int, int> m;
            for (int n : h) m[n]++;
            int w = 0;
            for (auto it = m.rbegin(); it != m.rend(); ++it) {
                w += it->second;
                ans = max(ans, w * it->first);
            }
        }
        return ans;
    }
};

  • 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

############

# 1727. Largest Submatrix With Rearrangements
# https://leetcode.com/problems/largest-submatrix-with-rearrangements/

class Solution:
    def largestSubmatrix(self, A: List[List[int]]):
        rows, cols = len(A), len(A[0])

        for i in range(1,rows):
            for j in range(cols):
                if A[i][j] == 1:
                    A[i][j] += A[i-1][j]
        
        res = 0
        for i in range(rows):
            row = sorted(A[i])
            for j in range(cols):
                res = max(res, row[j] * (cols - j))
        
        return res
                
            

  • 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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

  • 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;
}

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