花花酱 LeetCode 1727. Largest Submatrix With Rearrangements

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.

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 <= 105
• matrix[i][j] is 0 or 1.

Solution: DP + Sorting

Preprocess each column, for col j, matrix[i][j] := length consecutive ones of col j.

[0,0,1]    [0,0,1]
[1,1,1] => [1,1,2]
[1,0,1]    [2,0,3]

Then we enumerate ending row, for each ending row i, we sort row[i] in deceasing order

e.g. i = 2

[0,0,1]                  [-,-,-]
[1,1,2] sort by row 2 => [-,-,-]
[2,0,3]                  [3,2,0]

row[2][1] = 3, means there is a 3×1 all ones sub matrix, area = 3
row[2][2] = 2, means there is a 2×2 all ones sub matrix, area = 4.

Time complexity: O(m*n*log(n))
Space complexity: O(1)