Maximal Square

Dynamic Programming

Question

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

Example

For example, given the following matrix:

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

Return 4.

Tags

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Related Problems

Hard Maximal Rectangle

Analysis

考虑以f[i][j]为右下角顶点可以拓展的正方形边长,那么可以由f[i-1][j]f[i][j-1]f[i-1][j-1]三者的最小值决定f[i][j]的最长边长。

当我们判断以某个点为正方形右下角时最大的正方形时,那它的上方,左方和左上方三个点也一定是某个正方形的右下角,否则该点为右下角的正方形最大就是它自己了。这是定性的判断,那具体的最大正方形边长呢?我们知道,该点为右下角的正方形的最大边长,最多比它的上方,左方和左上方为右下角的正方形的边长多1,最好的情况是是它的上方,左方和左上方为右下角的正方形的大小都一样的,这样加上该点就可以构成一个更大的正方形。但如果它的上方,左方和左上方为右下角的正方形的大小不一样,合起来就会缺了某个角落,这时候只能取那三个正方形中最小的正方形的边长加1了。假设dp[i][j]表示以i,j为右下角的正方形的最大边长,则有

dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1

当然,如果这个点在原矩阵中本身就是0的话,那dp[i][j]肯定就是0了。

时间O(mn), 空间O(mn)

DP四要素

  • 状态 State

    • f[i][j] 表示以i和j作为正方形右下角可以拓展的最大边长
  • 方程 Function

    • if matrix[i][j] == 1
      • f[i][j] = min(f[i - 1][j], f[i][j-1], f[i-1][j-1]) + 1;
    • if matrix[i][j] == 0
      • f[i][j] = 0
  • 初始化 Intialization

    • f[i][0] = matrix[i][0];
    • f[0][j] = matrix[0][j];
  • 答案 Answer

    • max{f[i][j]}

可以使用二维滚动数组优化:

  1. 状态 State
    • f[i][j] 表示以i和j作为正方形右下角可以拓展的最大边长
  2. 方程 Function
    • if matrix[i][j] == 1
      • f[i%2][j] = min(f[(i - 1)%2][j], f[i%2][j-1], f[(i-1)%2][j-1]) + 1;
    • if matrix[i][j] == 0
    • f[i%2][j] = 0
  3. 初始化 Intialization
    • f[i%2][0] = matrix[i][0];
    • f[0][j] = matrix[0][j];
  4. 答案 Answer
    • max{f[i%2][j]}

Solution

Standard Dynamic Programming

public class Solution {
    /**
     * @param matrix: a matrix of 0 and 1
     * @return: an integer
     */
    public int maxSquare(int[][] matrix) {
        if (matrix.length == 0) return 0;

        int m = matrix.length, n = matrix[0].length;
        int max = 0;
        int[][] dp = new int[m][n];

        // 第一列赋值
        for (int i = 0; i < m; i++) {
            dp[i][0] = matrix[i][0];
            max = Math.max(max, dp[i][0]);
        }

        // 第一行赋值
        for (int i = 0; i < n; i++) {
            dp[0][i] = matrix[0][i];
            max = Math.max(max, dp[0][i]);
        }

        // 递推
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = matrix[i][j] == 1 ? Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1 : 0;

                max = Math.max(max, dp[i][j]);
            }
        }
        return max * max;
    }
}

LeetCode version (char[][] as input) - (12ms 30.09%)

class Solution {
    public int maximalSquare(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int m = matrix.length;
        int n = matrix[0].length;

        int[][] dp = new int[m][n];
        int maxSqLen = 0;
        for (int i = 0; i < m; i++) {
            dp[i][0] = matrix[i][0] == '1' ? 1 : 0;
            maxSqLen = Math.max(dp[i][0], maxSqLen);
        }
        for (int j = 0; j < n; j++) {
            dp[0][j] = matrix[0][j] == '1' ? 1 : 0;
            maxSqLen = Math.max(dp[0][j], maxSqLen);
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[i][j] == '1') {
                    dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
                    maxSqLen = Math.max(dp[i][j], maxSqLen);
                } 
            }
        }
        return maxSqLen * maxSqLen;
    }
}

A smart way to bypass initialization, use m + 1, n + 1 as dp[][] dimension:

(notice matrix[i-1][j-1] is used for determining dp[i][j]) - O(mn) time, O(mn) space - (13 ms, faster than 23.53%)

public class Solution {
    public int maximalSquare(char[][] matrix) {
        int rows = matrix.length, cols = rows > 0 ? matrix[0].length : 0;
        int[][] dp = new int[rows + 1][cols + 1];
        int maxsqlen = 0;
        for (int i = 1; i <= rows; i++) {
            for (int j = 1; j <= cols; j++) {
                if (matrix[i-1][j-1] == '1'){
                    dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
                    maxsqlen = Math.max(maxsqlen, dp[i][j]);
                }
            }
        }
        return maxsqlen * maxsqlen;
    }
}

Space optimized DP - O(mn) time, O(n) space

public class Solution {
    public int maximalSquare(char[][] matrix) {
        int rows = matrix.length, cols = rows > 0 ? matrix[0].length : 0;
        int[] dp = new int[cols + 1];
        int maxsqlen = 0, prev = 0;
        for (int i = 1; i <= rows; i++) {
            for (int j = 1; j <= cols; j++) {
                int temp = dp[j];
                if (matrix[i - 1][j - 1] == '1') {
                    dp[j] = Math.min(Math.min(dp[j - 1], prev), dp[j]) + 1;
                    maxsqlen = Math.max(maxsqlen, dp[j]);
                } else {
                    dp[j] = 0;
                }
                prev = temp;
            }
        }
        return maxsqlen * maxsqlen;
    }
}

Reference

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