Ones and Zeroes

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.

Example

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”

Solution

@DFS:超时，认识到必须在调用自身函数时改变变量才能回溯
class Solution:
    def findMaxForm(self, strs, m, n):
        """
        :type strs: List[str]
        :type m: int
        :type n: int
        :rtype: int
        """
        self.count = 0
        self.generate(strs, 0, m, n, 0, [])
        return self.count
    
    def generate(self, strs, start, m, n, num, num_list):
        if m<0 or n<0:
            num-= 1
            return
        if num>self.count:
            self.count = num
        if m==0 and n==0:
            return
        for i in range(start, len(strs)):
            num1 = strs[i].count('1')
            num2 = strs[i].count('0')
            self.generate(strs, i+1, m-num2, n-num1, num+1, num_list+[strs[i]])

@DP
class Solution:
    def findMaxForm(self, strs, m, n):
        """
        :type strs: List[str]
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [[0]*(n+1) for x in range(m+1)]
        for str in strs:
            zeros = str.count('0')
            ones = str.count('1')
            for i in range(m, zeros-1, -1):
                for j in range(n, ones-1, -1):
                    dp[i][j] = max(dp[i-zeros][j-ones]+1, dp[i][j])
        return dp[m][n]