由字符串数组中的 A 0 和 B 1 组成的最长子集的长度 | 2套

给定一个由N 个二进制字符串和两个整数A和B组成的数组arr[] ，任务是 找到最多由A 0 s 和B 1 s 组成的最长子集的长度。

例子：

Input: arr[] = {“1”, “0”, “0001”, “10”, “111001”}, A = 5, B = 3
Output: 4
Explanation: 
One possible way is to select the subset {arr[0], arr[1], arr[2], arr[3]}.
Total number of 0s and 1s in all these strings are 5 and 3 respectively.
Also, 4 is the length of the longest subset possible.

Input: arr[] = {“0”, “1”, “10”}, A = 1, B = 1
Output: 2
Explanation: 
One possible way is to select the subset {arr[0], arr[1]}.
Total number of 0s and 1s in all these strings is 1 and 1 respectively.
Also, 2 is the length of the longest subset possible.

朴素的方法：请参阅本文的前一篇文章，了解解决问题的最简单方法。
时间复杂度： O(2 ^N )
辅助空间： O(1)

动态编程方法：有关动态编程方法，请参阅本文的前一篇文章。
时间复杂度： O(N*A*B)
辅助空间： O(N * A * B)

空间优化动态规划方法：上述方法中的空间复杂度可以基于以下观察进行优化：

• 初始化一个二维数组dp[A][B] ，其中dp[i][j]表示最长子集的长度，最多包含i个0 s 和j个1 s。
• 优化从3D表到2D表的辅助空间，思路是逆序遍历内部循环。
• 这确保无论何时dp[][] 中的值发生更改，在当前迭代中都不再需要它。
• 因此，递推关系如下所示：

dp[i][j] = max (dp[i][j], dp[i – zeros][j – ones] + 1)
where zeros is the count of 0s and ones is the count of 1s in the current iteration.

请按照以下步骤解决问题：

• 初始化一个二维数组，比如dp[A][B]并将其所有条目初始化为0 。
• 遍历给定的数组arr[]并对每个二进制字符串执行以下步骤：
  • 将当前字符串中 0 和 1 的计数分别存储在变量0和1 中。
  • 迭代在范围[A，零]使用同时迭代在范围[B，那些]使用变量j和更新到最大DP的DP [i] [j]的值的变量i和[I] [j]的和(dp[i – zeros][j –ones] + 1) 。
• 完成以上步骤后，打印dp[A][B]的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the length of the
// longest subset of an array of strings
// with at most A 0s and B 1s
int MaxSubsetlength(vector arr,
                    int A, int B)
{
    // Initialize a 2D array with its
    // entries as 0
    int dp[A + 1][B + 1];
    memset(dp, 0, sizeof(dp));
 
    // Traverse the given array
    for (auto& str : arr) {
 
        // Store the count of 0s and 1s
        // in the current string
        int zeros = count(str.begin(),
                          str.end(), '0');
        int ones = count(str.begin(),
                         str.end(), '1');
 
        // Iterate in the range [A, zeros]
        for (int i = A; i >= zeros; i--)
 
            // Iterate in the range [B, ones]
            for (int j = B; j >= ones; j--)
 
                // Update the value of dp[i][j]
                dp[i][j] = max(
                    dp[i][j],
                    dp[i - zeros][j - ones] + 1);
    }
 
    // Print the result
    return dp[A][B];
}
 
// Driver Code
int main()
{
    vector arr
        = { "1", "0", "0001",
            "10", "111001" };
    int A = 5, B = 3;
    cout << MaxSubsetlength(arr, A, B);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find the length of the
// longest subset of an array of strings
// with at most A 0s and B 1s
static int MaxSubsetlength(String arr[],
                           int A, int B)
{
     
    // Initialize a 2D array with its
    // entries as 0
    int dp[][] = new int[A + 1][B + 1];
 
    // Traverse the given array
    for(String str : arr)
    {
         
        // Store the count of 0s and 1s
        // in the current string
        int zeros = 0, ones = 0;
        for(char ch : str.toCharArray())
        {
            if (ch == '0')
                zeros++;
            else
                ones++;
        }
 
        // Iterate in the range [A, zeros]
        for(int i = A; i >= zeros; i--)
         
            // Iterate in the range [B, ones]
            for(int j = B; j >= ones; j--)
 
                // Update the value of dp[i][j]
                dp[i][j] = Math.max(
                    dp[i][j],
                    dp[i - zeros][j - ones] + 1);
    }
 
    // Print the result
    return dp[A][B];
}
 
// Driver Code
public static void main(String[] args)
{
    String arr[] = { "1", "0", "0001",
                     "10", "111001" };
    int A = 5, B = 3;
     
    System.out.println(MaxSubsetlength(arr, A, B));
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
 
# Function to find the length of the
# longest subset of an array of strings
# with at most A 0s and B 1s
def MaxSubsetlength(arr, A, B):
     
    # Initialize a 2D array with its
    # entries as 0
    dp = [[0 for i in range(B + 1)]
             for i in range(A + 1)]
 
    # Traverse the given array
    for str in arr:
         
        # Store the count of 0s and 1s
        # in the current string
        zeros = str.count('0')
        ones = str.count('1')
 
        # Iterate in the range [A, zeros]
        for i in range(A, zeros - 1, -1):
 
            # Iterate in the range [B, ones]
            for j in range(B, ones - 1, -1):
 
                # Update the value of dp[i][j]
                dp[i][j] = max(dp[i][j],
                               dp[i - zeros][j - ones] + 1)
 
    # Print the result
    return dp[A][B]
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ "1", "0", "0001", "10", "111001" ]
    A, B = 5, 3
     
    print (MaxSubsetlength(arr, A, B))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG {
 
  // Function to find the length of the
  // longest subset of an array of strings
  // with at most A 0s and B 1s
  static int MaxSubsetlength(string[] arr, int A, int B)
  {
 
    // Initialize a 2D array with its
    // entries as 0
    int[, ] dp = new int[A + 1, B + 1];
 
    // Traverse the given array
    foreach(string str in arr)
    {
 
      // Store the count of 0s and 1s
      // in the current string
      int zeros = 0, ones = 0;
      foreach(char ch in str.ToCharArray())
      {
        if (ch == '0')
          zeros++;
        else
          ones++;
      }
 
      // Iterate in the range [A, zeros]
      for (int i = A; i >= zeros; i--)
 
        // Iterate in the range [B, ones]
        for (int j = B; j >= ones; j--)
 
          // Update the value of dp[i][j]
          dp[i, j] = Math.Max(
          dp[i, j],
          dp[i - zeros, j - ones] + 1);
    }
 
    // Print the result
    return dp[A, B];
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    string[] arr = { "1", "0", "0001", "10", "111001" };
    int A = 5, B = 3;
 
    Console.WriteLine(MaxSubsetlength(arr, A, B));
  }
}
 
// This code is contributed by ukasp.

Javascript

 
// Javascript program for the above approach
 
// Function to find the length of the
// longest subset of an array of strings
// with at most A 0s and B 1s
function MaxSubsetlength(arr, A, B)
{
     
    // Initialize a 2D array with its
    // entries as 0
    var dp = Array.from(Array(A + 1),
         ()=>Array(B + 1).fill(0));
 
    // Traverse the given array
    arr.forEach(str => {
         
        // Store the count of 0s and 1s
        // in the current string
        var zeros = [...str].filter(x => x == '0').length;
        var ones = [...str].filter(x => x == '1').length;
 
        // Iterate in the range [A, zeros]
        for(var i = A; i >= zeros; i--)
 
            // Iterate in the range [B, ones]
            for(var j = B; j >= ones; j--)
 
                // Update the value of dp[i][j]
                dp[i][j] = Math.max(dp[i][j],
                    dp[i - zeros][j - ones] + 1);
    });
 
    // Print the result
    return dp[A][B];
}
 
// Driver Code
var arr = [ "1", "0", "0001",
            "10", "111001" ];
var A = 5, B = 3;
 
document.write(MaxSubsetlength(arr, A, B));
 
// This code is contributed by noob2000
 

输出：

4

时间复杂度： O(N * A * B)
辅助空间： O(A * B)

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