add: LargestSquareInAMatrix

Author: Jer3myYu

java/Dynamic Programming/LargestSquareInAMatrix.java

@@ -0,0 +1,40 @@
+public class LargestSquareInAMatrix {
+    public int largestSquareInAMatrix(int[][] matrix) {
+        if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) {
+            return 0;
+        }
+        int m = matrix.length;
+        int n = matrix[0].length;
+        int[][] dp = new int[m][n];
+        int maxLen = 0;
+        // Base case: If a cell in row 0 is 1, the largest square ending there has a
+        // length of 1.
+        for (int j = 0; j < n; j++) {
+            if (matrix[0][j] == 1) {
+                dp[0][j] = 1;
+                maxLen = 1;
+            }
+        }
+        // Base case: If a cell in column 0 is 1, the largest square ending there has
+        // a length of 1.
+        for (int i = 0; i < m; i++) {
+            if (matrix[i][0] == 1) {
+                dp[i][0] = 1;
+                maxLen = 1;
+            }
+        }
+        // Populate the rest of the DP table.
+        for (int i = 1; i < m; i++) {
+            for (int j = 1; j < n; j++) {
+                if (matrix[i][j] == 1) {
+                    // The length of the largest square ending here is determined by
+                    // the smallest square ending at the neighboring cells (left,
+                    // top-left, top), plus 1 to include this cell.
+                    dp[i][j] = 1 + Math.min(dp[i - 1][j], Math.min(dp[i - 1][j - 1], dp[i][j - 1]));
+                }
+                maxLen = Math.max(maxLen, dp[i][j]);
+            }
+        }
+        return maxLen * maxLen;
+    }
+}