LeetCode - Maximal Rectangle
Problem description
Given a 2D binary matrix filled with 0’s and 1’s, find the largest rectangle containing only 1’s and return its area.
Example:
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Input:
[
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
]
Output: 6
Analysis
The first idea is to save continuous 1’s into a 2-d dp array.
So the sample array would be:
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[
[1,0,1,0,0],
[1,0,3,2,1],
[5,4,3,2,1],
[1,0,0,1,0]
]
And then we can check every column, and use the same algorithm in largest-rectangle to solve the maximum area in each column and get a O(MN) solution.
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class Solution {
public int maximalRectangle(char[][] matrix) {
if (matrix == null){
return 0;
}
int row = matrix.length;
if (row == 0){
return 0;
}
int col = matrix[0].length;
if (col == 0){
return 0;
}
int[][] dp = new int[row+1][col+1];
for (int i = 1; i<row+1; i++){
for (int j = 1; j< col+1; j++){
if (matrix[i-1][j-1] == '1' && dp[i][j - 1] != 0){
dp[i][j] = dp[i][j-1] - 1;
}
else if (matrix[i-1][j-1] == '1') {
int count = 0;
while (j+count-1 < col && matrix[i-1][j+count-1] == '1'){
count++;
}
dp[i][j] = count;
}
}
}
int max = 0;
for (int j = 1; j < col+1; j++){
int[] lessFromLeft = new int[row+1];
int[] lessFromRight = new int[row+1];
lessFromRight[row] = row+1;
lessFromLeft[1] = 0;
for (int i = 2; i < row+1; i++){
int p = i - 1;
while (p >= 1 && dp[p][j] >= dp[i][j])
p = lessFromLeft[p];
lessFromLeft[i] = p;
}
for (int i = row; i >= 1; i--){
int p = i + 1;
while (p < row + 1 && dp[p][j] >= dp[i][j])
p = lessFromRight[p];
lessFromRight[i] = p;
}
for (int i = 1; i < row+1; i++){
max = Math.max(max, (lessFromRight[i] - lessFromLeft[i] - 1) * dp[i][j]);
}
}
return max;
}
}
What to improve
- be sensitive about learned problem.