## Description

Given a 2D binary matrix filled with 0’s and 1’s, find the largest square containing only 1’s and return its area.

Example:

Input:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4


## Solutions

### 1. Brute Force

# Time: O(mn*min(m,n))
# Space: O(n)
class Solution:
def maximalSquare(self, matrix: List[List[str]]) -> int:
if not matrix:
return 0
r, c = len(matrix), len(matrix[0])
max_area = 0
for i in range(r):
for j in range(c):
for l in range(min(r, c)):
if self.check_square(matrix, i, j, i + l, j + l):
max_area = max(max_area, (l+1)**2)
return max_area

def check_square(self, matrix, x, y , bottom_x, bottom_y):
r, c = len(matrix), len(matrix[0])
if bottom_x >= r or bottom_y >= c:
return False
for i in range(x, bottom_x+1):
for j in range(y, bottom_y+1):
if matrix[i][j] == '0':
return False
return True
# Time Limit Exceeded
# 69/69 cases passed (N/A)


### 2. todo


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// Author: Huahua
// Time complexity: O(n^3)
// Running time: 39 ms
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.empty()) return 0;
int m = matrix.size();
int n = matrix[0].size();

// sums[i][j] = sum(matrix[0][0] ~ matrix[i-1][j-1])
vector<vector<int>> sums(m + 1, vector<int>(n + 1, 0));
for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
sums[i][j] = matrix[i - 1][j - 1] - '0'
+ sums[i - 1][j]
+ sums[i][j - 1]
- sums[i - 1][j - 1];

int ans = 0;
for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
for (int k = min(m - i + 1, n - j + 1); k > 0; --k) {
int sum = sums[i + k - 1][j + k - 1]
- sums[i + k - 1][j - 1]
- sums[i - 1][j + k - 1]
+ sums[i - 1][j - 1];
// full of 1
if (sum == k*k) {
ans = max(ans, sum);
break;
}
}

return ans;
}
};


### 3. 2d-DP

# Time: O(mn)
# Space:O(mn)
class Solution:
def maximalSquare(self, matrix: List[List[str]]) -> int:
if not matrix: return 0
m , n = len(matrix), len(matrix[0])
dp = [[ 0 if matrix[i][j] == '0' else 1 for j in range(0, n)] for i in range(0, m)]

for i in range(1, m):
for j in range(1, n):
if matrix[i][j] == '1':
dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
else:
dp[i][j] = 0

res = max(max(row) for row in dp)
return res ** 2

# 69/69 cases passed (200 ms)
# Your runtime beats 73.57 % of python3 submissions
# Your memory usage beats 81.82 % of python3 submissions (13.9 MB)