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    LeetCode link: 474. Ones and Zeroes, difficulty: Medium.

    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.

    Constraints:

    • 1 <= strs.length <= 600
    • 1 <= strs[i].length <= 100
    • 'strs[i]' consists only of digits '0' and '1'
    • 1 <= m, n <= 100

    Intuition
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    This question is difficult. It is recommended to complete a simple question of the same type first 416. Partition Equal Subset Sum.

    • After completing 416, you will find that this question requires solving the 0/1 Knapsack Problem in two dimensions.
    • The solution is to first solve the problem in one dimension and then expand it to two dimensions.
    • It is no need to draw a grid that considers both dimensions together, that's too complicated. Let's first only consider the quantity limit of 0.

    Pattern of "Dynamic Programming"

    "Dynamic programming" is divided into five steps

    1. Determine the meaning of each value of the array dp.
    2. Initialize the value of the array dp.
    3. Fill in the dp grid data "in order" according to an example.
    4. Based on the dp grid data, derive the "recursive formula".
    5. Write a program and print the dp array. If it is not as expected, adjust it.

    Detailed description of these five steps

    1. Determine the meaning of each value of the array dp.
      • First determine whether dp is a one-dimensional array or a two-dimensional array. A one-dimensional rolling array means that the values ​​of the array are overwritten at each iteration. Most of the time, using one-dimensional rolling array instead of two-dimensional array can simplify the code; but for some problems, such as operating "two swappable arrays", for the sake of ease of understanding, it is better to use two-dimensional array.
      • Try to use the meaning of the return value required by the problem as the meaning of dp[i] (one-dimensional) or dp[i][j] (two-dimensional). It works about 60% of the time. If it doesn't work, try other meanings.
      • Try to save more information in the design. Repeated information only needs to be saved once in a dp[i].
      • Use simplified meanings. If the problem can be solved with boolean value, don't use numeric value.
    2. Initialize the value of the array dp. The value of dp involves two levels:
      1. The length of dp. Usually: condition array length plus 1 or condition array length.
      2. The value of dp[i] or dp[i][j]. dp[0] or dp[0][0] sometimes requires special treatment.
    3. Fill in the dp grid data "in order" according to an example.
      • The "recursive formula" is the core of the "dynamic programming" algorithm. But the "recursive formula" is obscure. If you want to get it, you need to make a table and use data to inspire yourself.
      • If the original example is not good enough, you need to redesign one yourself.
      • According to the example, fill in the dp grid data "in order", which is very important because it determines the traversal order of the code.
      • Most of the time, from left to right, from top to bottom. But sometimes it is necessary to traverse from right to left, from bottom to top, from the middle to the right (or left), such as the "palindrome" problems. Sometimes, it is necessary to traverse a line twice, first forward and then backward.
      • When the order is determined correctly, the starting point is determined. Starting from the starting point, fill in the dp grid data "in order". This order is also the order in which the program processes.
      • In this process, you will get inspiration to write a "recursive formula". If you can already derive the formula, you do not need to complete the grid.
    4. Based on the dp grid data, derive the "recursive formula".
      • There are three special positions to pay attention to: dp[i - 1][j - 1], dp[i - 1][j] and dp[i][j - 1], the current dp[i][j] often depends on them.
      • When operating "two swappable arrays", due to symmetry, we may need to use dp[i - 1][j] and dp[i][j - 1] at the same time.
    5. Write a program and print the dp array. If it is not as expected, adjust it.
      • Focus on analyzing those values that are not as expected.

    After reading the above, do you feel that "dynamic programming" is not that difficult? Try to solve this problem. 🤗

    Steps

    Common steps in '0/1 Knapsack Problem'

    These five steps are a pattern for solving Dynamic Programming problems.

    1. Determine the meaning of the dp[j]

      • Since we are only considering the zero count constraint for now, we can use a one-dimensional dp array.
      • dp[j] represents the maximum number of strings that can be selected with at most j zeros.
      • dp[j] is an integer.

    2. Determine the dp array's initial value

      • Use an example, example 1: Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3.
      • After initialization: python max_zero_count = m dp = [0] * (max_zero_count + 1)
      • The dp array size is one greater than the zero count constraint. This way, the index value equals the constraint value, making it easier to understand.
      • dp[0] = 0, indicating that with no zeros, we can select 0 strings.
      • dp[j] = 0 as the initial value because we will use max to increase them later.

    3. Determine the dp array's recurrence formula

      • Let's analyze the example step by step:

        # Initial state
#    0 1 2 3 4 5
#    0 0 0 0 0 0

# After processing "10" (1 zero)
#    0 1 2 3 4 5
#    0 1 1 1 1 1

# After processing "0001" (3 zeros)
#    0 1 2 3 4 5
#    0 1 1 1 2 2

# After processing "111001" (2 zeros)
#    0 1 2 3 4 5
#    0 1 1 2 2 2

# After processing "1" (0 zeros)
#    0 1 2 3 4 5
#    0 2 2 3 3 3

# After processing "0" (1 zero)
#    0 1 2 3 4 5
#    0 2 3 3 4 4

      • After analyzing the sample dp grid, we can derive the Recurrence Formula:

        dp[j] = max(dp[j], dp[j - zero_count] + 1)

      • This formula means: for each string, we can either:

        1. Not select it (keep the current value dp[j])
        2. Select it (add 1 to the value at dp[j - zero_count])

    4. Determine the dp array's traversal order

      • First iterate through the strings, then iterate through the zero count (in reverse order).
      • When iterating through the zero count, since dp[j] depends on dp[j] and dp[j - zero_count], we should traverse from right to left.
      • This ensures that we don't use the same string multiple times.

    5. Check the dp array's value

      • Print the dp to see if it is as expected.
      • The final answer will be at dp[max_zero_count].

    6. The code that only considers the quantity limit of 0 is:

      class Solution:
    def findMaxForm(self, strs: List[str], max_zero_count: int, n: int) -> int:
        dp = [0] * (max_zero_count + 1)

        for string in strs:
            zero_count = count_zero(string)

            for j in range(len(dp) - 1, zero_count - 1, -1): # must iterate in reverse order!
                dp[j] = max(dp[j], dp[j - zero_count] + 1)

        return dp[-1]

def count_zero(string):
    zero_count = 0

    for bit in string:
        if bit == '0':
            zero_count += 1

    return zero_count

    Now, you can consider another dimension: the quantity limit of 1.

    It should be handled similarly to 0 but in another dimension. Please see the complete code below.

    Complexity

    Time complexity

    O(N∗M∗Len)

    Space complexity

    O(N∗M)

    Solution in languages: Python C++ Java C# JavaScript Go Ruby

    Python # 

    class Solution:
    def findMaxForm(self, strs: List[str], max_zero_count: int, max_one_count: int) -> int:
        dp = [[0] * (max_one_count + 1) for _ in range(max_zero_count + 1)]

        for string in strs:
            zero_count, one_count = count_zero_one(string)

            for i in range(len(dp) - 1, zero_count - 1, -1):
                for j in range(len(dp[0]) - 1, one_count - 1, -1):
                    dp[i][j] = max(dp[i][j], dp[i - zero_count][j - one_count] + 1)

        return dp[-1][-1]


def count_zero_one(string):
    zero_count = 0
    one_count = 0

    for bit in string:
        if bit == '0':
            zero_count += 1
        else:
            one_count += 1

    return zero_count, one_count

    C++ # 

    class Solution {
public:
    int findMaxForm(vector<string>& strs, int max_zero_count, int max_one_count) {
        vector<vector<int>> dp(max_zero_count + 1, vector<int>(max_one_count + 1, 0));

        for (auto& str : strs) {
            auto zero_count = 0;
            auto one_count = 0;

            for (auto bit : str) {
                if (bit == '0') {
                    zero_count++;
                } else {
                    one_count++;
                }
            }

            for (auto i = max_zero_count; i >= zero_count; i--) {
                for (auto j = max_one_count; j >= one_count; j--) {
                    dp[i][j] = max(dp[i][j], dp[i - zero_count][j - one_count] + 1);
                }
            }
        }

        return dp[max_zero_count][max_one_count];
    }
};

    Java # 

    class Solution {
    public int findMaxForm(String[] strs, int maxZeroCount, int maxOneCount) {
        var dp = new int[maxZeroCount + 1][maxOneCount + 1];

        for (var str : strs) {
            var zeroCount = 0;
            var oneCount = 0;

            for (var bit : str.toCharArray()) {
                if (bit == '0') {
                    zeroCount++;
                } else {
                    oneCount++;
                }
            }

            for (var i = maxZeroCount; i >= zeroCount; i--) {
                for (var j = maxOneCount; j >= oneCount; j--) {
                    dp[i][j] = Math.max(dp[i][j], dp[i - zeroCount][j - oneCount] + 1);
                }
            }
        }

        return dp[maxZeroCount][maxOneCount];
    }
}

    C# # 

    public class Solution
{
    public int FindMaxForm(string[] strs, int maxZeroCount, int maxOneCount)
    {
        var dp = new int[maxZeroCount + 1, maxOneCount + 1];

        foreach (var str in strs)
        {
            var (zeroCount, oneCount) = CountZeroOne(str);

            for (var i = maxZeroCount; i >= zeroCount; i--)
            {
                for (var j = maxOneCount; j >= oneCount; j--)
                {
                    dp[i, j] = Math.Max(dp[i, j], dp[i - zeroCount, j - oneCount] + 1);
                }
            }
        }

        return dp[maxZeroCount, maxOneCount];
    }

    (int, int) CountZeroOne(string str)
    {
        var zeroCount = 0;
        var oneCount = 0;

        foreach (var bit in str)
        {
            if (bit == '0')
            {
                zeroCount++;
            }
            else
            {
                oneCount++;
            }
        }

        return (zeroCount, oneCount);
    }
}

    JavaScript # 

    var findMaxForm = function (strs, maxZeroCount, maxOneCount) {
  const dp = Array(maxZeroCount + 1).fill().map(
    () => Array(maxOneCount + 1).fill(0)
  )

  for (const str of strs) {
    const [zeroCount, oneCount] = countZeroOne(str)

    for (let i = dp.length - 1; i >= zeroCount; i--) {
      for (let j = dp[0].length - 1; j >= oneCount; j--) {
        dp[i][j] = Math.max(dp[i][j], dp[i - zeroCount][j - oneCount] + 1)
      }
    }
  }

  return dp.at(-1).at(-1)
};

function countZeroOne(str) {
  let zeroCount = 0
  let oneCount = 0

  for (const bit of str) {
    if (bit === '0') {
      zeroCount++
    } else {
      oneCount++
    }
  }

  return [zeroCount, oneCount]
}

    Go # 

    func findMaxForm(strs []string, maxZeroCount int, maxOneCount int) int {
    dp := make([][]int, maxZeroCount + 1)
    for i := range dp {
        dp[i] = make([]int, maxOneCount + 1)
    }

    for _, str := range strs {
        zeroCount, oneCount := countZeroOne(str)

        for i := len(dp) - 1; i >= zeroCount; i-- {
            for j := len(dp[0]) - 1; j >= oneCount; j-- {
                dp[i][j] = max(dp[i][j], dp[i - zeroCount][j - oneCount] + 1)
            }
        }
    }

    return dp[maxZeroCount][maxOneCount]
}

func countZeroOne(str string) (int, int) {
    zeroCount := 0
    oneCount := 0

    for _, bit := range str {
        if bit == '0' {
            zeroCount++
        } else {
            oneCount++
        }
    }

    return zeroCount, oneCount
}

    Ruby # 

    def find_max_form(strs, max_zero_count, max_one_count)
  dp = Array.new(max_zero_count + 1) do
    Array.new(max_one_count + 1, 0)
  end

  strs.each do |string|
    zero_count, one_count = count_zero_one(string)

    (zero_count...dp.size).reverse_each do |i|
      (one_count...dp[0].size).reverse_each do |j|
        dp[i][j] = [ dp[i][j], dp[i - zero_count][j - one_count] + 1 ].max
      end
    end
  end

  dp[-1][-1]
end

def count_zero_one(string)
  zero_count = 0
  one_count = 0

  string.each_char do |bit|
    if bit == '0'
      zero_count += 1
    else
      one_count += 1
    end
  end

  [ zero_count, one_count ]
end
    Solution in languages: Python C++ Java C# JavaScript Go Ruby

    Other languages

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