What is the main technique used in solving the Binary String With Substrings Representing 1 To N problem?

The main technique involves using a sliding window combined with running state updates and bit manipulation for efficient substring checks.

How do we optimize the substring checking in this problem?

By limiting the length of the substrings to at most 30 bits, which corresponds to the binary representations of numbers up to 2^30, and using a hash table for efficient lookup.

Can we use any other data structures to solve this problem?

Yes, a hash set or a hash map can be used to store the binary substrings and perform fast lookups while iterating through the string.

Why is the sliding window approach particularly useful for this problem?

The sliding window allows us to efficiently check for substrings without re-checking the entire string multiple times, optimizing the solution.

What is the time complexity of this solution?

The time complexity is linear with respect to the length of the binary string, depending on the approach chosen for sliding window and hash table lookups.

#1016

Medium

auto_awesome滑动窗口（状态滚动更新）

LeetCode 题解工作台

子串能表示从 1 到 N 数字的二进制串

给定一个二进制字符串 s 和一个正整数 n ，如果对于 [1, n] 范围内的每个整数，其二进制表示都是 s 的子字符串，就返回 true ，否则返回 false 。子字符串是字符串中连续的字符序列。示例 1：输入： s = "0110", n = 3 输出： true 示例 2：输…

哈希表字符串位运算滑动窗口

题目描述

给定一个二进制字符串 s 和一个正整数 n，如果对于 [1, n] 范围内的每个整数，其二进制表示都是 s 的 子字符串 ，就返回 true，否则返回 false 。

子字符串 是字符串中连续的字符序列。

示例 1：

输入：s = "0110", n = 3
输出：true

示例 2：

输入：s = "0110", n = 4
输出：false

提示：

1 <= s.length <= 1000
s[i] 不是 '0' 就是 '1'
1 <= n <= 10⁹

lightbulb

解题思路

方法一：脑筋急转弯

我们注意到，字符串 $s$ 的长度不超过 $1000$ ，所以字符串 $s$ 能表示不超过 $1000$ 个二进制整数，因此，如果 $n \gt 1000$ ，那么 $s$ 肯定不能表示 $[1,.. n]$ 范围内的所有整数的二进制表示。

另外，对于一个整数 $x$ ，如果 $x$ 的二进制表示是 $s$ 的子串，那么 $\lfloor x / 2 \rfloor$ 的二进制表示也是 $s$ 的子串。因此，我们只需要判断 $[\lfloor n / 2 \rfloor + 1,.. n]$ 范围内的整数的二进制表示是否是 $s$ 的子串即可。

时间复杂度 $O(m^2 \times \log m)$ ，空间复杂度 $O(\log n)$ ，其中 $m$ 是字符串 $s$ 的长度，而 $n$ 为题目给定的正整数。

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class Solution:
    def queryString(self, s: str, n: int) -> bool:
        if n > 1000:
            return False
        return all(bin(i)[2:] in s for i in range(n, n // 2, -1))

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
Candidate demonstrates an understanding of sliding window techniques.
question_mark
Candidate leverages bit manipulation for optimizing binary substring checks.
question_mark
Candidate uses a hash table effectively for substring tracking and lookups.

warning

常见陷阱

外企场景

error
Failing to recognize that only substrings of length at most 30 are needed due to binary number size constraints.
error
Not efficiently checking for the presence of all required substrings by brute force.
error
Overcomplicating the problem by trying to check all possible substrings without using efficient data structures like hash tables.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the range of numbers is extremely large? Can the approach be adapted for larger ranges efficiently?
arrow_right_alt
Can this approach work if the binary string is not guaranteed to be contiguous or has interruptions?
arrow_right_alt
What happens if we need to handle non-binary strings or strings with different character sets?

help

常见问题

外企场景

继续练习

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