Description

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

Code

Partition Equal Subset Sum 一樣是 knapsack 類型的題目。不過這題是一個 3D 的 DP。

Time Complexity: , Space Complexity:

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        int k = strs.size();
        int dp[k + 1][m + 1][n + 1];
        memset(dp, 0, sizeof(dp));
 
        for(int i = 1; i < k + 1; i++) {
            auto count = convert(strs[i - 1]);
            auto zeros = count[0];
            auto ones = count[1];
            for(int j = 0; j < m + 1; j++) {
                for(int l = 0; l < n + 1; l++) {
                    int res = dp[i - 1][j][l];
                    if((j >= zeros) && (l >= ones)) {
                        res = max(res, dp[i - 1][j - zeros][l - ones] + 1);
                    }
                    dp[i][j][l] = res;
                }
            }
        }
 
        return dp[k][m][n];
 
 
    }
 
    vector<int> convert(string s) {
        int zero = 0, one = 0;
        for(auto c: s) {
            if(c == '0') {
                zero++;
            } else if (c == '1') {
                one++;
            }
        }
 
        return {zero, one};
    }
};

Source