给定一个数组arr [] ，该数组由N个二进制字符串以及两个整数A和B组成，任务是 求出最多由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.
天真的方法：有关解决问题的最简单方法，请参阅本文的上一篇文章。
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，请参阅本文的上一篇文章。
时间复杂度： O(N * A * B)
辅助空间： O(N * A * B)
空间优化的动态规划方法：可基于以下观察来优化上述方法中的空间复杂度：
- 初始化一个2D数组dp [A] [B] ，其中dp [i] [j]表示最长i个子集的长度，该子集最多包含0个i数和1个j数。
- 为了优化从3D表到2D表的辅助空间，其想法是以相反的顺序遍历内部循环。
- 这样可以确保每当dp [] []中的值更改时，在当前迭代中就不再需要该值。
- 因此，递归关系如下所示：
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2D数组，例如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：
- 将当前字符串中的0和1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 –零] [j –个] + 1) 。
- 完成上述步骤后，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 Kingash4时间复杂度： O(N * A * B)
辅助空间： O(A * B)