Description

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:

  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".

Solutions

1. 3D dp-01 knapsack

　　dp[i][j][k] 表示用 j 个 1，k 个 0 能够在前 i 个子串中最多组合多少个子串。

# Time: O(lmn)
# Space: O(lmn)
class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        l = len(strs)
        # dp[l+1][m+1][n+1]
        dp = [[[0 for _ in range(n + 1)] for _ in range(m + 1)] for _ in range(l + 1)]
        for i in range(1, l + 1):
            s = strs[i - 1]
            size = len(s)
            zeros = s.count('0')
            ones = size - zeros
            for j in range(0, m + 1):
                for k in range(0, n + 1):
                    if zeros <= j and ones <= k:
                        dp[i][j][k] = max(dp[i - 1][j][k], 1 + dp[i - 1][j - zeros][k - ones])
                    else:
                        dp[i][j][k] = dp[i - 1][j][k]
        return dp[l][m][n]
# Runtime: 5136 ms, faster than 13.59%
# Memory Usage: 61.1 MB, less than 100.00%

　　优化：

# Time: O(lmn)
# Space: O(mn)
class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        # dp[m+1][n+1]
        dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
        for s in strs:
            size = len(s)
            zeros = s.count('0')
            ones = size - zeros
            for j in range(m, -1, -1):
                for k in range(n, -1, -1):
                    if zeros <= j and ones <= k:
                        dp[j][k] = max(dp[j][k], 1 + dp[j - zeros][k - ones])
        return dp[m][n]
# Runtime: 3908 ms, faster than 26.48%
# Memory Usage: 12.9 MB, less than 100.00%

References

  1. 474. Ones and Zeroes