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 ofm0sandn1srespectively. On the other hand, there is an array with strings consisting of only0sand1s.

Now your task is to find the maximum number of strings that you can form with givenm0sandn1s. Each0and1can be used at mostonce.

Note:

1. The given numbers of

   0s

   and

   1s

   will both not exceed

   100

2. 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".

分析

01背包，把value拆成0,1，变成2个循环而已。记得倒序

这里value范围必须是cnt0->m cnt1>n？？？

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        if not m or not n:
            return m|n
        ll = len(strs)
        dp = [[0]*(m+1) for _ in range(n+1)]
        def count(s):
            return sum(1 for c in s if c == '0'), sum(1 for c in s if c == '1')

        for k in range(ll):
            cnt0,cnt1 = count(strs[k])
            for x in range(n,cnt1-1,-1):#1
                for y in range(m,cnt0-1,-1):#0                    
                    #if x>=cnt1 and y>=cnt0:
                    dp[x][y] = max(dp[x][y],dp[x-cnt1][y-cnt0]+1)
        return dp[n][m]