How can dynamic programming help solve the Ones and Zeroes problem?

Dynamic programming helps by breaking the problem down into smaller subproblems, where the solution to each subproblem depends on the previous state. This is useful in optimizing the subset selection process while keeping track of the constraints on 0's and 1's.

What is the space optimization strategy for Ones and Zeroes?

Space optimization can be achieved by using a 1D DP array instead of a 2D table. This allows you to reduce the space complexity from O(m * n) to O(n) by updating the array in reverse order.

Can this problem be solved without dynamic programming?

While a brute force solution is possible, it is inefficient due to its high time complexity. Dynamic programming offers a much more efficient approach for solving the problem within the given constraints.

What is the time complexity of the Ones and Zeroes problem?

The time complexity of the solution is O(m * n * k), where `m` and `n` are the constraints on 0's and 1's, and `k` is the number of binary strings. This complexity comes from the need to iterate through each string and update the DP table.

How do I handle the edge cases for this problem?

Edge cases involve situations like empty strings, strings with only 0's or only 1's, or constraints where `m` or `n` is small. Make sure to properly handle such cases by checking the current DP state before updating it.

#474

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

一和零

给你一个二进制字符串数组 strs 和两个整数 m 和 n 。请你找出并返回 strs 的最大子集的长度，该子集中最多有 m 个 0 和 n 个 1 。如果 x 的所有元素也是 y 的元素，集合 x 是集合 y 的子集。示例 1：输入： strs = ["10", "0001", "…

数组字符串动态规划

题目描述

给你一个二进制字符串数组 strs 和两个整数 m 和 n 。

请你找出并返回 strs 的最大子集的长度，该子集中最多有 m 个 0 和 n 个 1 。

如果 x 的所有元素也是 y 的元素，集合 x 是集合 y 的子集。

示例 1：

输入：strs = ["10", "0001", "111001", "1", "0"], m = 5, n = 3
输出：4
解释：最多有 5 个 0 和 3 个 1 的最大子集是 {"10","0001","1","0"} ，因此答案是 4 。
其他满足题意但较小的子集包括 {"0001","1"} 和 {"10","1","0"} 。{"111001"} 不满足题意，因为它含 4 个 1 ，大于 n 的值 3 。

示例 2：

输入：strs = ["10", "0", "1"], m = 1, n = 1
输出：2
解释：最大的子集是 {"0", "1"} ，所以答案是 2 。

提示：

1 <= strs.length <= 600
1 <= strs[i].length <= 100
strs[i] 仅由 '0' 和 '1' 组成
1 <= m, n <= 100

lightbulb

解题思路

方法一：动态规划

我们定义 $f[i][j][k]$ 表示在前 $i$ 个字符串中，使用 $j$ 个 0 和 $k$ 个 1 的情况下最多可以得到的字符串数量。初始时 $f[i][j][k]=0$ ，答案为 $f[sz][m][n]$ ，其中 $sz$ 是数组 $strs$ 的长度。

对于 $f[i][j][k]$ ，我们有两种决策：

不选第 $i$ 个字符串，此时 $f[i][j][k]=f[i-1][j][k]$ ；
选第 $i$ 个字符串，此时 $f[i][j][k]=f[i-1][j-a][k-b]+1$ ，其中 $a$ 和 $b$ 分别是第 $i$ 个字符串中 $0$ 和 $1$ 的数量。

我们取两种决策中的最大值，即可得到 $f[i][j][k]$ 的值。

最终的答案即为 $f[sz][m][n]$ 。

时间复杂度 $O(sz \times m \times n)$ ，空间复杂度 $O(sz \times m \times n)$ 。其中 $sz$ 是数组 $strs$ 的长度；而 $m$ 和 $n$ 分别是字符串中 $0$ 和 $1$ 的数量上限。

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class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        sz = len(strs)
        f = [[[0] * (n + 1) for _ in range(m + 1)] for _ in range(sz + 1)]
        for i, s in enumerate(strs, 1):
            a, b = s.count("0"), s.count("1")
            for j in range(m + 1):
                for k in range(n + 1):
                    f[i][j][k] = f[i - 1][j][k]
                    if j >= a and k >= b:
                        f[i][j][k] = max(f[i][j][k], f[i - 1][j - a][k - b] + 1)
        return f[sz][m][n]

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for clear understanding of dynamic programming concepts.
question_mark
Check for efficient space optimization techniques in DP.
question_mark
Evaluate ability to break down a problem into manageable subproblems.

warning

常见陷阱

外企场景

error
Overcomplicating the DP state transitions or failing to properly update the table.
error
Ignoring the space complexity aspect and using an overly large DP table.
error
Misunderstanding the problem constraints, especially in counting 0's and 1's.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider variations with additional constraints, like limiting the number of binary strings that can be selected.
arrow_right_alt
Modify the problem by changing the allowed counts of 0's and 1's or adding more characters to the binary strings.
arrow_right_alt
Test how the solution handles larger inputs and evaluate performance scalability.

help

常见问题

外企场景

继续练习

#472 连接词 #691 贴纸拼词 #792 匹配子序列的单词数