Leetcode: Maximal Square

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

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.

Understand the problem:
This question can be solved by using DP. 
  -- Define dp[i][j] as the length of the maximal square of which the right bottom point ended with matrix[i][j]. 
  -- Initial value dp[0][j] = matrix[0][j]; dp[i][0] = matrix[i][0];
  -- Transit function: If matrix[i][j] == 1, dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1;
  -- Final state, max(dp[i][j] * dp[i][j])


Code (Java):

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public class Solution {     public int maximalSquare(char[][] matrix) {         if (matrix == null || matrix.length == 0) {             return 0;         }                   int rows = matrix.length;         int cols = matrix[0].length;                   int[][] dp = new int[rows][cols];                   // Initialization         for (int i = 0; i < cols; i++) {             dp[0][i] = matrix[0][i] - '0';         }                   for (int i = 0; i < rows; i++) {             dp[i][0] = matrix[i][0] - '0';         }                   for (int i = 1; i < rows; i++) {             for (int j = 1; j < cols; 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;                 }             }         }                   int maxArea = 0;         for (int i = 0; i < rows; i++) {             for (int j = 0; j < cols; j++) {                 maxArea = Math.max(maxArea, dp[i][j] * dp[i][j]);             }         }                   return maxArea;     } }

Update on 4/2/19: Rolling array

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public class Solution {     /**      * @param matrix: a matrix of 0 and 1      * @return: an integer      */     public int maxSquare(int[][] matrix) {         if (matrix == null || matrix.length == 0) {             return 0;         }                   int m = matrix.length;         int n = matrix[0].length;                   int maxLen = 0;                   int[][] dp = new int[2][n];         for (int i = 0; i < n; i++) {             if (matrix[0][i] == 1) {                 dp[0][i] = 1;                 maxLen = Math.max(maxLen, dp[0][i]);             }         }                   int cur = 0;         int old = 0;                   for (int i = 1; i < m; i++) {             old = cur;             cur = 1 - cur;             dp[cur][0] = matrix[i][0] == 1 ? 1 : 0;             maxLen = Math.max(maxLen, dp[cur][0]);                           for (int j = 1; j < n; j++) {                 if (matrix[i][j] == 1) {                     dp[cur][j] = Math.min(dp[old][j - 1], Math.min(dp[old][j], dp[cur][j - 1])) + 1;                 } else {                     dp[cur][j] = 0;                 }                                   maxLen = Math.max(maxLen, dp[cur][j]);             }         }                   return maxLen * maxLen;               } }