[LeetCode 85] Maximal Rectangle

Given a rows x cols binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.

Example 1:

Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 6
Explanation: The maximal rectangle is shown in the above picture.

Example 2:

Input: matrix = []
Output: 0

Example 3:

Input: matrix = [["0"]]
Output: 0

Example 4:

Input: matrix = [["1"]]
Output: 1

Example 5:

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

Constraints:

rows == matrix.length
cols == matrix.length
0 <= row, cols <= 200
matrix[i][j] is '0' or '1'.

This problem's optimal solution builds on its related problem: Largest Rectangle in Histogram. If we treat each row as the new base line of a histogram, we can get the largest rectangle in O(Col) time. We need to do this O(Row) time by keeping a prefix sum array h[], where h[i] is the height of the ith column in the current histogram.

The runtime is O(Row * Col) and is BCR.

class Solution {
    public int maximalRectangle(char[][] matrix) {
        if(matrix.length == 0) {
            return 0;
        }
        int m = matrix.length, n = matrix[0].length, ans = 0;
        int[] h = new int[n];
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                h[j] = (matrix[i][j] == '0' ? 0 : h[j] + 1);
            }
            ans = Math.max(ans, largestRectangeleHistogram(h));
        }
        return ans;
    }
    private int largestRectangeleHistogram(int[] h) {
        ArrayDeque<Integer> dq = new ArrayDeque<>();
        dq.addLast(-1);
        int best = 0;
        for(int i = 0; i < h.length; i++) {
            while(dq.peekLast() >= 0 && h[i] < h[dq.peekLast()]) {
                int j = dq.pollLast();
                best = Math.max(best, h[j] * (i - 1 - dq.peekLast()));
            }
            dq.addLast(i);
        }
        while(dq.peekLast() >= 0) {
            int j = dq.pollLast(); 
            best = Math.max(best, h[j] * (h.length - 1 - dq.peekLast()));
        }
        return best;
    }
}