Ones and Zeros: Dynamic Programming approach

dynamic_programming/ones_and_zeros.rb

@@ -0,0 +1,56 @@
+# You are given an array of binary strings strs and two integers m and n.
+#
+# Return the size of the largest subset of strs such that there are at most m 0's and n 1's in the subset.
+#
+# A set x is a subset of a set y if all elements of x are also elements of y.
+#
+# Example 1:
+#
+# Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3
+# Output: 4
+# Explanation: The largest subset with at most 5 0's and 3 1's is {"10", "0001", "1", "0"}, so the answer is 4.
+# Other valid but smaller subsets include {"0001", "1"} and {"10", "1", "0"}.
+# {"111001"} is an invalid subset because it contains 4 1's, greater than the maximum of 3.
+#
+# Example 2:
+#
+# Input: strs = ["10","0","1"], m = 1, n = 1
+# Output: 2
+# Explanation: The largest subset is {"0", "1"}, so the answer is 2.
+
+#
+# Approach #1 Dynamic Programming
+#
+
+# @param {String[]} strs
+# @param {Integer} m
+# @param {Integer} n
+# @return {Integer}
+def find_max_form(strs, m, n)
+  dp = (m + 1).times.map { [0] * (n + 1) }
+
+  strs.each do |str|
+    zeros = str.count('0')
+    ones = str.count('1')
+
+    m.downto(zeros) do |i|
+      n.downto(ones) do |j|
+        dp[i][j] = [dp[i][j], dp[i - zeros][j - ones] + 1].max
+      end
+    end
+  end
+
+  dp[m][n]
+end
+
+strs = %w[10 0001 111001 1 0]
+m = 5
+n = 3
+puts find_max_form(strs, m, n)
+# Output: 4
+
+strs = %w[10 0 1]
+m = 1
+n = 1
+puts find_max_form(strs, m, n)
+# Output: 2