474. Ones and Zeroes

Given an array, strs, with strings consisting of only 0s and 1s. Also two integers m and n.

Now your task is to find the maximum number of strings that you can form with given m 0s and n 1s. Each 0 and 1 can be used at most once.

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

 class Solution {
    public static int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m + 1][n + 1];
        for (String str : strs) {
            int[] count = count(str);
            for (int i = m; i >= count[0]; i--) {
                for (int j = n; j >= count[1]; j--) {
                    dp[i][j] = Math.max(dp[i][j], dp[i - count[0]][j - count[1]] + 1);
                }
            }
        }
        return dp[m][n];
    }
    
    private static int[] count(String s) {
    int[] result = new int[2];
    char[] array = s.toCharArray();
    for (int i : array) {
        result[i - '0']++;
    }
    return result;
    }
}

0/1背包问题。。。每个string的0和1记录下来，每个string可以选或者不选，背包容量是m个0和n个1，求这个背包最多能放多少个string

dp[ i ][ j ]意思是给了i个0，j个1，最多能放多少个string，答案就应该是dp[ m ][ n ]

下面的转移方程里面左边是不算这个string，右边是算，算前先把背包腾出位置（就是string减掉需要的0和1），还是不太清楚为什么从m和n往后递减算，但感觉就应该这样做。。

dp[i - count[0]][j - count[1]],剪掉后的dp是不含当前string最多能存多少string的数量，就可能包括了前面的string

dp[i][j] = Math.max(dp[i][j], dp[i - count[0]][j - count[1]] + 1);
https://leetcode.com/problems/ones-and-zeroes/discuss/95811/Java-Iterative-DP-Solution-O(mn)-Space
https://github.com/tianyicui/pack/blob/master/V2.pdf

总结：
This is a 0/1 knapsack question, we regard the number of 0/1 as volume, and we calculate 0/1's of current string.
dp[i][j] means with i 0's and j 1's the most string we can generate.