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Merge pull request #177 from TheAlgorithms/ones-and-zeroes
Ones and Zeros: Dynamic Programming approach
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* [Count Sorted Vowel Strings](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/count_sorted_vowel_strings.rb)
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* [Fibonacci](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/fibonacci.rb)
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* [House Robber](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/house_robber.rb)
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* [Ones And Zeros](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/ones_and_zeros.rb)
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* [Pascal Triangle Ii](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/pascal_triangle_ii.rb)
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## Maths
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56
dynamic_programming/ones_and_zeros.rb
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56
dynamic_programming/ones_and_zeros.rb
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# You are given an array of binary strings strs and two integers m and n.
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#
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# 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.
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#
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# A set x is a subset of a set y if all elements of x are also elements of y.
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#
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# Example 1:
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#
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# Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3
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# Output: 4
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# Explanation: The largest subset with at most 5 0's and 3 1's is {"10", "0001", "1", "0"}, so the answer is 4.
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# Other valid but smaller subsets include {"0001", "1"} and {"10", "1", "0"}.
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# {"111001"} is an invalid subset because it contains 4 1's, greater than the maximum of 3.
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#
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# Example 2:
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#
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# Input: strs = ["10","0","1"], m = 1, n = 1
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# Output: 2
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# Explanation: The largest subset is {"0", "1"}, so the answer is 2.
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#
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# Approach #1 Dynamic Programming
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#
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# @param {String[]} strs
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# @param {Integer} m
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# @param {Integer} n
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# @return {Integer}
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def find_max_form(strs, m, n)
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dp = (m + 1).times.map { [0] * (n + 1) }
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strs.each do |str|
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zeros = str.count('0')
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ones = str.count('1')
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m.downto(zeros) do |i|
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n.downto(ones) do |j|
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dp[i][j] = [dp[i][j], dp[i - zeros][j - ones] + 1].max
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end
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end
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end
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dp[m][n]
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end
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strs = %w[10 0001 111001 1 0]
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m = 5
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n = 3
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puts find_max_form(strs, m, n)
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# Output: 4
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strs = %w[10 0 1]
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m = 1
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n = 1
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puts find_max_form(strs, m, n)
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# Output: 2
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