# Accepted dp solution in JAVA with detailed explaination.

• ``````public class Solution {
public int maximalSquare(char[][] matrix) {
if(matrix == null||matrix.length == 0||matrix[0].length == 0)
return 0;
int height = matrix.length;
int width = matrix[0].length;

int[][] arr = new int[height][width];//create a same size int[][] to translate char into int for later use, and this makes this method run faster than that comparing characters.
for(int row = 0 ; row < height; row++)
{
for(int col = 0 ; col < width; col++)
{
arr[row][col] = matrix[row][col] - '0';
}
}
int[][] dp = new int[height][width];//create same size dp[][]
boolean containsOne = false;//copy the first row and col into dp[][] and detect if there is a 1 in it;
for(int row = 0 ; row < height; row++)
{
if(arr[row][0] == 1) containsOne = true;
dp[row][0] = arr[row][0];
}
for(int col = 0 ; col < width; col++)
{
if(arr[0][col] == 1) containsOne = true;
dp[0][col] = arr[0][col];
}
if(height == 1 || width == 1)
return containsOne? 1 : 0;
int res = containsOne?1:0;

for(int row = 1 ; row < height; row++)
{
for(int col = 1 ; col < width; col++)
{
if(arr[row][col] == 1)
{
//dp[row][col] represents the size of square whose lower right corner is arr[row][col]
//every position's value is 0 or square numbers like 1 4 9...
// If the current position is 1, then the minimum value of the three adjacent positions dp[row-1][col-1],dp[row ][col-1] and dp[row-1][col] determines the square value of current position
//sqrt(num) means the length of side of the square. if the new corner if going to be the lower right corner of a bigger square, the new area will be the 1 + 2*length + area of little square. You can understand this easily by drawing a matrix contains all 1s and map it into dp[][];