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. 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. 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. 3.
    初始化 Intialization
    • f[i%2][0] = matrix[i][0];
    • f[0][j] = matrix[0][j];
  4. 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