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This problem can be approached using a dynamic programming strategy similar to the knapsack problem. For each binary string, calculate the number of 0's and 1's it contains. Then, use a DP table to keep track of the maximum subset size you can achieve with a given count of 0's and 1's. We'll fill this table by iterating over each string and attempting to include it in our subset if the current capacity allows. Update the DP table in a reverse manner to avoid overwriting results prematurely.
Time Complexity: O(strsSize * m * n) where strsSize is the number of binary strings.
Space Complexity: O(m * n) for storing the DP table.
1#include <vector>
2#include <string>
3
4using namespace std;
5
6class Solution {
7public:
8 int findMaxForm(vector<string>& strs, int m, int n) {
9 vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
10 for (auto& s: strs) {
11 int zeros = count(s.begin(), s.end(), '0');
12 int ones = s.size() - zeros;
13 for (int i = m; i >= zeros; --i) {
14 for (int j = n; j >= ones; --j) {
15 dp[i][j] = max(dp[i][j], dp[i-zeros][j-ones] + 1);
16 }
17 }
18 }
19 return dp[m][n];
20 }
21};In the C++ solution, we use a vector of vectors to implement the DP table. The logic remains similar to the C solution, where we iterate over each string, count zeros and ones, and update the DP table backward to ensure prior states are not mistakenly overwritten in one iteration. The function uses count and size functions to efficiently determine the number of zeros and ones.
The problem can also be tackled using a recursive function with memoization to store already computed results and avoid redundant calculations. This approach uses recursion to consider two choices for each string - either include it in the subset or not, based on the available capacity for zeros and ones. By storing intermediate results, we can significantly reduce the number of recursive calls needed, thus optimizing the process.
Time Complexity: O(strsSize * m * n) due to memoization.
Space Complexity: O(strsSize * m * n) for the memoization table.
1
This Python solution implements memoization using a dictionary to manage already computed results, thus eliminating redundant recursive calls. The function uses recursion to explore the potential inclusion/exclusion of each string based on the remaining capacities of zeros and ones, storing the results of each state.