Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.
For example, given the following matrix:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Return 4.
Solution 1. For each entry of 1, use it as a possible bottom right point of a square of all 1s, then find the largest square of all 1s for this entry. The runtime is O(n^4): O(n^2) possible entries and O(n^2) time for each check.
The problem of this solution is that for each entry of 1, we start from scratch to check. We basically throw all the previous check results for previous entries of 1. Take the following example, when checking the entry of red color, we've already checked entries of yellow, green and purple that they all form a length 3 square. However solution 1 discard these infos and checked all the entries in the 4 by 4 grid to determine it actually forms a length 4 square. Instead, its neighboring results(left, top and top left) can be used to get a O(1) check as shown in solution 2.
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
Solution 2. Dynamic Programming, O(n^2) runtime, O(n^2) space
State: dp[i][j] : the max length of the square of all 1s whose bottom right point is matrix[i][j].
State Transition: dp[i][j] == 0, if matrix[i][j] == 0; dp[i][j] = 1 + min{dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]}, if matrix[i][j] == 1.
Initialization: dp[i][0] = matrix[i][0]; dp[0][j] = matrix[0][j].
class Solution {
public int maximalSquare(char[][] matrix) {
if(matrix.length == 0) {
return 0;
}
int n = matrix.length, m = matrix[0].length;
int[][] dp = new int[n][m]; //dp[i][j]: the max side length of square whose bottom right corner is (i, j).
int ans = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(matrix[i][j] == '1') {
dp[i][j] = 1;
if(i > 0 && j > 0) {
dp[i][j] = Math.max(dp[i][j], Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1);
}
ans = Math.max(ans, dp[i][j] * dp[i][j]);
}
}
}
return ans;
}
}
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