LeetCode | DP专题详解

221 medium

 

 

221. Maximal Square

Medium

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

 

 

To appy DP, we define the state as the maximal size (square = size * size) of the square that can be formed till point (i, j), denoted as dp[i][j].

For the topmost row (i = 0) and the leftmost column (j = 0), we have dp[i][j] = matrix[i][j] - '0', meaning that it can at most form a square of size 1 when the matrix has a '1'in that cell.

When i > 0 and j > 0, if matrix[i][j] = '0', then dp[i][j] = 0 since no square will be able to contain the '0' at that cell. If matrix[i][j] = '1', we will have dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1, which means that the square will be limited by its left, upper and upper-left neighbors.

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

In the above code, it uses O(mn) space. Actually each time when we update dp[i][j], we only need dp[i-1][j-1], dp[i-1][j] (the previous row) and dp[i][j-1] (the current row). So we may just keep two rows.

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        if (matrix.empty()) {
            return 0;
        }
        int m = matrix.size(), n = matrix[0].size(), sz = 0;
        vector<int> pre(n, 0), cur(n, 0);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (!i || !j || matrix[i][j] == '0') {
                    cur[j] = matrix[i][j] - '0';
                } else {
                    cur[j] = min(pre[j - 1], min(pre[j], cur[j - 1])) + 1;
                }
                sz = max(cur[j], sz);
            }
            fill(pre.begin(), pre.end(), 0);
            swap(pre, cur);
        }
        return sz * sz;
    }
};

Furthermore, we may only use just one vector (thanks to @stellari for sharing the idea).

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        if (matrix.empty()) {
            return 0;
        }
        int m = matrix.size(), n = matrix[0].size(), sz = 0, pre;
        vector<int> cur(n, 0);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int temp = cur[j];
                if (!i || !j || matrix[i][j] == '0') {
                    cur[j] = matrix[i][j] - '0';
                } else {
                    cur[j] = min(pre, min(cur[j], cur[j - 1])) + 1;
                }
                sz = max(cur[j], sz);
                pre = temp;
            }
        }
        return sz * sz;
    }
};