Problem 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:

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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: 4

solution1:
time: O(mn)
space: O(mn)

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

solution2:
time: O(mn)
space: O(n)

dp[j] is to store the square that can be made in this

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class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        int m = matrix.size(), n = m>0 ? matrix[0].size(): 0;
        
        vector<int> dp(n+1, 0);
        int maxlen = 0, prev= 0;
        
        for(int i = 1; i<= m; ++i){
            for(int j= 1; j<= n; ++j){
                int temp = dp[j];
                //cout<< "dp["<<j<<"]:"<<dp[j]<<endl;
                //cout<<"temp:"<< temp<<endl;
                if(matrix[i-1][j-1] == '1'){
                    dp[j] = min(dp[j-1], min(prev, dp[j]))+1;
                    //+1 is because if this element is '1', it at least can form a square by itself even previous is 0
                    //cout<< "dp["<<j<<"-1]:"<<dp[j-1];
                    //cout<< ", prev:"<<prev;
                    //cout<< ", dp["<<j<<"]:"<<dp[j]<<endl;
                    //dp[i][j] = min(dp[i][j-1], min(dp[i-1][j], dp[i-1][j-1])) +1;
                    maxlen = max(maxlen, dp[j]);
                    //cout<<"max="<<maxlen<<endl;
                }
                else{
                    dp[j] = 0;
                    //cout<< "else dp["<<j<<"]:"<<dp[j]<<endl;
                }
                prev= temp;
                //cout<<"----"<<endl;
            }
        }
        for(auto n: dp)
            cout<<n<<" "<<endl;
        return maxlen*maxlen;
    }
};