474. Ones and Zeroes

问题：

给定一组只包含'0' '1' 的字符串，给定两个正整数m和n，问，要使用m个'0' n个'1'，能最多构成多少个给出字符串组中的字符串。

Example 1:
Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3
Output: 4
Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are "10","0001","1","0".

Example 2:
Input: strs = ["10","0","1"], m = 1, n = 1
Output: 2
Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".
 
Constraints:
1 <= strs.length <= 600
1 <= strs[i].length <= 100
strs[i] consists only of digits '0' and '1'.
1 <= m, n <= 100

　　

解法：DP(动态规划)  0-1 knapsack problem（0-1背包问题）

1.确定【状态】：

• 可选择的元素：前 i 个字符串
• 和：
  • m：m个 ' 0 '
  • n：n个 ' 1 '

2.确定【选择】：

• 选择当前的字符串strs[i]
• 不选择当前的字符串strs[i]

3. dp[i][m][n]的含义：

前 i 个字符串中，用m个'0' n个'1' 最多能组成的字符串个数。

4. 状态转移：

dp[i][m][n]= OR {

• 选择 strs[i]：dp[i-1][m-strs[i].count('0')][n-strs[i].count('1')] + 1：=前 i-1 个元素中，用M个'0'N个'1'最多可组成的字符串个数 + strs[i]自己一个。（M= m-strs[i].count('0');  N= n-strs[i].count('1')）
• 不选择 strs[i]：dp[i-1][m][n]：=前 i-1 个元素中，用m个'0'n个'1'最多可组成的字符串个数。

}

5. base case：

• dp[0][m][n]=0
• dp[i][0][0]=0

代码参考：

class Solution {
public:
    //dp[i][m][n]:in first i strs, the max number of strs can be formed by m 0s and n 1s.
    //case_1.select strs[i]    : = dp[i-1][m-strs[i].count('0')][n-strs[i].count('1')]+1
    //case_2.not select strs[i]: = dp[i-1][m][n]
    //dp[i][m][n] = max(case_1, case_2)
    //base: dp[0][m][n]=0
    //dp[i][0][0]=0
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<vector<int>>> dp(strs.size()+1, vector<vector<int>>(m+1, vector<int>(n+1, 0)));
        for(int i=1; i<=strs.size(); i++) {
            int cout_1 = count(strs[i-1].begin(), strs[i-1].end(), '1');
            int cout_0 = count(strs[i-1].begin(), strs[i-1].end(), '0');
            for(int M=0; M<=m; M++) {
                for(int N=0; N<=n; N++) {
                    if(M-cout_0>=0 && N-cout_1>=0) {
                        dp[i][M][N] = max(dp[i-1][M-cout_0][N-cout_1]+1, dp[i-1][M][N]);
                    } else {
                        dp[i][M][N] = dp[i-1][M][N];
                    }
                }
            }
        }
        return dp[strs.size()][m][n];
    }
};

♻️ 优化：空间复杂度：3维->2维

去掉 i 

将 m和n 倒序遍历。

理由参考：https://www.cnblogs.com/habibah-chang/p/13581649.html

代码参考：

class Solution {
public:
    //dp[i][m][n]:in first i strs, the max number of strs can be formed by m 0s and n 1s.
    //case_1.select strs[i]    : = dp[i-1][m-strs[i].count('0')][n-strs[i].count('1')]+1
    //case_2.not select strs[i]: = dp[i-1][m][n]
    //dp[i][m][n] = max(case_1, case_2)
    //base: dp[0][m][n]=0
    //dp[i][0][0]=0
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
        for(int i=1; i<=strs.size(); i++) {
            int cout_1 = count(strs[i-1].begin(), strs[i-1].end(), '1');
            int cout_0 = count(strs[i-1].begin(), strs[i-1].end(), '0');
            for(int M=m; M>=0; M--) {
                for(int N=n; N>=0; N--) {
                    if(M-cout_0>=0 && N-cout_1>=0) {
                        dp[M][N] = max(dp[M-cout_0][N-cout_1]+1, dp[M][N]);
                    }
                }
            }
        }
        return dp[m][n];
    }
};