# 221. Maximal Square

## Description

Given an m x n binary matrix filled with 0's and 1's, find the largest square 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: 4


Example 2:

Input: matrix = [["0","1"],["1","0"]]
Output: 1


Example 3:

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


Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 300
• matrix[i][j] is '0' or '1'.

## Solutions

Solution 1: Dynamic Programming

We define $dp[i + 1][j + 1]$ as the maximum square side length with the lower right corner at index $(i, j)$. The answer is the maximum value among all $dp[i + 1][j + 1]$.

The state transition equation is:

$dp[i + 1][j + 1] = \begin{cases} 0 & \text{if } matrix[i][j] = '0' \\ \min(dp[i][j], dp[i][j + 1], dp[i + 1][j]) + 1 & \text{if } matrix[i][j] = '1' \end{cases}$

The time complexity is $O(m\times n)$, and the space complexity is $O(m\times n)$. Where $m$ and $n$ are the number of rows and columns of the matrix, respectively.

• class Solution {
public int maximalSquare(char[][] matrix) {
int m = matrix.length, n = matrix[0].length;
int[][] dp = new int[m + 1][n + 1];
int mx = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (matrix[i][j] == '1') {
dp[i + 1][j + 1] = Math.min(Math.min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
mx = Math.max(mx, dp[i + 1][j + 1]);
}
}
}
return mx * mx;
}
}

• class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
int mx = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (matrix[i][j] == '1') {
dp[i + 1][j + 1] = min(min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
mx = max(mx, dp[i + 1][j + 1]);
}
}
}
return mx * mx;
}
};

• '''
eg. a 10*10 sqaure with full of 1s
this square move 1 line down
this square move 1 line right

so there needs an extra single 1 at bottom right, to make it a larger full square
'''
class Solution:
def maximalSquare(self, matrix: List[List[str]]) -> int:
m, n = len(matrix), len(matrix[0])
dp = [[0] * (n + 1) for _ in range(m + 1)]
mx = 0
for i in range(m):
for j in range(n):
if matrix[i][j] == '1':
dp[i + 1][j + 1] = 1 + min(dp[i][j + 1], dp[i + 1][j], dp[i][j])
mx = max(mx, dp[i + 1][j + 1])
return mx * mx


• func maximalSquare(matrix [][]byte) int {
m, n := len(matrix), len(matrix[0])
dp := make([][]int, m+1)
for i := 0; i <= m; i++ {
dp[i] = make([]int, n+1)
}
mx := 0
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if matrix[i][j] == '1' {
dp[i+1][j+1] = min(min(dp[i][j+1], dp[i+1][j]), dp[i][j]) + 1
mx = max(mx, dp[i+1][j+1])
}
}
}
return mx * mx
}

• public class Solution {
public int MaximalSquare(char[][] matrix) {
int m = matrix.Length, n = matrix[0].Length;
var dp = new int[m + 1, n + 1];
int mx = 0;
for (int i = 0; i < m; ++i)
{
for (int j = 0; j < n; ++j)
{
if (matrix[i][j] == '1')
{
dp[i + 1, j + 1] = Math.Min(Math.Min(dp[i, j + 1], dp[i + 1, j]), dp[i, j]) + 1;
mx = Math.Max(mx, dp[i + 1, j + 1]);
}
}
}
return mx * mx;
}
}