474. Ones and Zeroes

In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.

For now, suppose you are a dominator of m 0s and n 1s respectively. On the other hand, there is an array with strings consisting of only 0s and 1s.

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.

Note:

The given numbers of 0s and 1s will both not exceed 100
The size of given string array won't exceed 600.

Example 1:

Input: Array = {"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: Array = {“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”.
“
From:
LeetCode-Algorithm-Dynamic Programming

Hint:
这个问题算是一个0-1背包问题，对每个字符串，有两种可能：在最优解内，和不在最优解内（即选与不选）。
假设
F(i,j,k) = 在有i个0，j个1的情况下，考虑前j个字符串可以获得的最优解。
F(i,j,k) = max{F(i,j,k-1),F(i-k.zeros,j-k.ones,k-1)+1}
优化可使得空间复杂度下降到O(mn)

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        int F[101][101];
        for (int i = 0; i < 101; i++) {
            for (int j = 0; j < 101; j++) {
                F[i][j] = 0;
            }
        }
        for (int k = 0; k < strs.size(); k++) {
            int *need = getNeeds(strs[k]);
            for (int i = m; i >= 0; i--) {
                if (i < need[0]) break;
                for (int j = n; j >= 0; j--) {
                    if (j < need[1]) break;
                    F[i][j] = max(F[i-need[0]][j-need[1]]+1,F[i][j]);
                }
            }
        }
        return F[m][n];
    }

    int* getNeeds(string& s) {
        int *need = new int[2];
        need[0] = need[1] = 0;
        for (int i = 0; i < s.size(); i++) {
            if (s[i] == '0') {
                need[0]++;
            } else {
                need[1]++;
            }
        }
        return need;
    }
};