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leetcode_java/src/main/java/LeetCodeJava/DynamicProgramming/MaximalSquare.java
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| package LeetCodeJava.DynamicProgramming; | ||
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| // https://leetcode.com/problems/maximal-square/description/ | ||
| /** | ||
| * 221. Maximal Square | ||
| * | ||
| * Medium | ||
| * Topics | ||
| * Companies | ||
| * Given an m x n binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area. | ||
| * | ||
| * | ||
| * | ||
| * Example 1: | ||
| * | ||
| * | ||
| * Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]] | ||
| * Output: 4 | ||
| * Example 2: | ||
| * | ||
| * | ||
| * Input: matrix = [["0","1"],["1","0"]] | ||
| * Output: 1 | ||
| * Example 3: | ||
| * | ||
| * Input: matrix = [["0"]] | ||
| * Output: 0 | ||
| * | ||
| * | ||
| * Constraints: | ||
| * | ||
| * m == matrix.length | ||
| * n == matrix[i].length | ||
| * 1 <= m, n <= 300 | ||
| * matrix[i][j] is '0' or '1'. | ||
| * | ||
| */ | ||
| public class MaximalSquare { | ||
| // V0 | ||
| // TODO : implement | ||
| // public int maximalSquare(char[][] matrix) { | ||
| // | ||
| // } | ||
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| // V1 | ||
| // IDEA : DP | ||
| // https://leetcode.com/problems/maximal-square/solutions/61876/accepted-clean-java-dp-solution/ | ||
| public int maximalSquare_1(char[][] a) { | ||
| if (a == null || a.length == 0 || a[0].length == 0) | ||
| return 0; | ||
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| int max = 0, n = a.length, m = a[0].length; | ||
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| // dp(i, j) represents the length of the square | ||
| // whose lower-right corner is located at (i, j) | ||
| // dp(i, j) = min{ dp(i-1, j-1), dp(i-1, j), dp(i, j-1) } | ||
| int[][] dp = new int[n + 1][m + 1]; | ||
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| for (int i = 1; i <= n; i++) { | ||
| for (int j = 1; j <= m; j++) { | ||
| if (a[i - 1][j - 1] == '1') { | ||
| dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1; | ||
| max = Math.max(max, dp[i][j]); | ||
| } | ||
| } | ||
| } | ||
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| // return the area | ||
| return max * max; | ||
| } | ||
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| // V2 | ||
| // IDEA : DP | ||
| // https://leetcode.com/problems/maximal-square/solutions/61805/evolve-from-brute-force-to-dp/ | ||
| public int maximalSquare_2(char[][] matrix) { | ||
| int r=matrix.length; | ||
| if(r==0) return 0; | ||
| int c=matrix[0].length,edge=0; | ||
| int[][] dp=new int[r+1][c+1]; | ||
| for(int i=1;i<=r;i++) | ||
| for(int j=1;j<=c;j++) { | ||
| if(matrix[i-1][j-1]=='0') continue; | ||
| dp[i][j]=1+Math.min(dp[i-1][j],Math.min(dp[i-1][j-1],dp[i][j-1])); | ||
| edge=Math.max(edge,dp[i][j]); | ||
| } | ||
| return edge*edge; | ||
| } | ||
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| // V3_1 | ||
| // IDEA : DP | ||
| // https://leetcode.com/problems/maximal-square/solutions/61828/my-java-dp-ac-solution-simple-and-easy-to-understand-with-explanation/ | ||
| public int maximalSquare_3_1(char[][] matrix) { | ||
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| //illegal check - no square can be formed | ||
| if(matrix == null || matrix.length == 0) return 0; | ||
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| int result = 0; | ||
| int[][] count = new int[matrix.length][matrix[0].length]; | ||
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| //initialize first row and first column | ||
| for(int i = 0; i < matrix.length; i ++) { | ||
| count[i][0] = matrix[i][0] == '0' ? 0 : 1; | ||
| result = Math.max(result, count[i][0]); | ||
| } | ||
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| for(int i = 0; i < matrix[0].length; i ++) { | ||
| count[0][i] = matrix[0][i] == '0' ? 0 : 1; | ||
| result = Math.max(result, count[0][i]); | ||
| } | ||
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| //start to transfer status to iterate each cell from (1, 1) to (m, n) | ||
| //if i am a 0, the square stops, reset | ||
| for(int i = 1; i < matrix.length; i++) { | ||
| for(int j = 1; j < matrix[0].length; j++) { | ||
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| //I break the square reset myself to zero | ||
| if(matrix[i][j] == '0') { | ||
| count[i][j] = 0; | ||
| continue; | ||
| } | ||
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| //if I am 1, it depends if I can grow the size of the square, if I have a 0 guy around me, | ||
| //I can only be a top left guy | ||
| if(count[i - 1][j - 1] == 0 || count[i - 1][j] == 0 || count[i][j - 1] == 0) { | ||
| count[i][j] = 1; | ||
| } | ||
| //if guys around are the same size, I can be the right-bottom guy of a bigger square | ||
| else if(count[i - 1][j - 1] == count[i - 1][j] && count[i - 1][j] == count[i][j - 1]) { | ||
| count[i][j] = count[i - 1][j - 1] + 1; | ||
| } | ||
| //guys around me not the same, I can only be the right-bottom guy of a least square | ||
| else { | ||
| count[i][j] = Math.min(Math.min(count[i - 1][j - 1], count[i - 1][j]), | ||
| count[i][j - 1]) + 1; | ||
| } | ||
| result = Math.max(result, count[i][j]); | ||
| } | ||
| } | ||
| return result * result; | ||
| } | ||
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| // V3_2 | ||
| // https://leetcode.com/problems/maximal-square/solutions/61828/my-java-dp-ac-solution-simple-and-easy-to-understand-with-explanation/ | ||
| public int maximalSquare_3_2(char[][] matrix) { | ||
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| if(matrix == null || matrix.length == 0) return 0; | ||
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| int result = 0; | ||
| int[][] count = new int[matrix.length][matrix[0].length]; | ||
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| for(int i = 0; i < matrix.length; i ++) { | ||
| count[i][0] = matrix[i][0] == '0' ? 0 : 1; | ||
| result = Math.max(result, count[i][0]); | ||
| } | ||
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| for(int i = 0; i < matrix[0].length; i ++) { | ||
| count[0][i] = matrix[0][i] == '0' ? 0 : 1; | ||
| result = Math.max(result, count[0][i]); | ||
| } | ||
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| for(int i = 1; i < matrix.length; i++) { | ||
| for(int j = 1; j < matrix[0].length; j++) { | ||
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| if(matrix[i][j] == '0') { | ||
| count[i][j] = 0; | ||
| continue; | ||
| } | ||
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| count[i][j] = Math.min(Math.min(count[i - 1][j - 1], count[i - 1][j]), | ||
| count[i][j - 1]) + 1; | ||
| result = Math.max(result, count[i][j]); | ||
| } | ||
| } | ||
| return result * result; | ||
| } | ||
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| } |
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