solve: Maximal Square

Author: 0x01f7

README.md

@@ -77,7 +77,7 @@
 | 224 | [Basic Calculator][p224]                                            | [Ruby][s224]  | Medium      |
 | 223 | [Rectangle Area][p223]                                              | [Ruby][s223]  | Easy        |
 | 222 | [Count Complete Tree Nodes][p222]                                   | [Ruby][s222]  | Medium      |
-| 221 | [Maximal Square][p221]                                              |               | Medium      |
+| 221 | [Maximal Square][p221]                                              | [Ruby][s221]  | Medium      |
 | 220 | [Contains Duplicate III][p220]                                      | [Ruby][s220]  | Medium      |
 | 219 | [Contains Duplicate II][p219]                                       | [Ruby][s219]  | Easy        |
 | 218 | [The Skyline Problem][p218]                                         |               | Hard        |
@@ -605,6 +605,7 @@
 [s224]:./algorithms/basic_calculator.rb
 [s223]:./algorithms/rectangle_area.rb
 [s222]:./algorithms/count_complete_tree_nodes.rb
+[s221]:./algorithms/maximal_square.rb
 [s220]:./algorithms/contains_duplicate_iii.rb
 [s219]:./algorithms/contains_duplicate_ii.rb
 [s217]:./algorithms/contains_duplicate.rb

algorithms/maximal_square.rb

@@ -0,0 +1,38 @@
+# https://leetcode.com/problems/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.
+
+
+# @param {Character[][]} matrix
+# @return {Integer}
+def maximal_square(matrix)
+  return 0 if matrix.empty?
+
+  m, n = matrix.size, matrix[0].size
+    dp = Array.new(m) { Array.new(n, 0) }
+
+  maxwidth = 0
+
+  0.upto(m - 1) do |i|
+    0.upto(n - 1) do |j|
+      if i == 0 || j == 0 || matrix[i][j] == '0'
+        dp[i][j] = matrix[i][j].to_i
+      else
+        dp[i][j] = [dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]].min + 1
+      end
+      maxwidth = dp[i][j] if maxwidth < dp[i][j]
+    end
+  end
+
+  maxwidth * maxwidth
+end