LeetCode 1016. Binary String With Substrings Representing 1 To N Solution in Java, C++, Python & More | Explanation + Code

Description

Given a binary string s and a positive integer n, return true if the binary representation of all the integers in the range [1, n] are substrings of s, or false otherwise.

A substring is a contiguous sequence of characters within a string.

 

Example 1:

Input: s = "0110", n = 3
Output: true

Example 2:

Input: s = "0110", n = 4
Output: false

 

Constraints:

• 1 <= s.length <= 1000
• s[i] is either '0' or '1'.
• 1 <= n <= 109

Solutions

Solution 1: Brain Teaser

We observe that the length of string s does not exceed 1000, so string s can represent at most 1000 binary integers. Therefore, if n \gt 1000, then s definitely cannot represent the binary representation of all integers in the range [1,.. n].

Additionally, for an integer x, if the binary representation of x is a substring of s, then the binary representation of \lfloor x / 2 \rfloor is also a substring of s. Therefore, we only need to check whether the binary representations of integers in the range [\lfloor n / 2 \rfloor + 1,.. n] are substrings of s.

The time complexity is O(m² × log m) and the space complexity is O(log n), where m is the length of string s and n is the positive integer given in the problem.

PythonJavaC++GoTypeScriptRust

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))(code-box)

class Solution { public boolean queryString(String s, int n) { if (n > 1000) { return false; } for (int i = n; i > n / 2; i--) { if (!s.contains(Integer.toBinaryString(i))) { return false; } } return true; } }(code-box)

class Solution { public: bool queryString(string s, int n) { if (n > 1000) { return false; } for (int i = n; i > n / 2; --i) { string b = bitset<32>(i).to_string(); b = b.substr(b.find_first_not_of('0')); if (s.find(b) == string::npos) { return false; } } return true; } };(code-box)

func queryString(s string, n int) bool { if n > 1000 { return false } for i := n; i > n/2; i-- { if !strings.Contains(s, strconv.FormatInt(int64(i), 2)) { return false } } return true }(code-box)

function queryString(s: string, n: number): boolean { if (n > 1000) { return false; } for (let i = n; i > n / 2; --i) { if (s.indexOf(i.toString(2)) === -1) { return false; } } return true; }(code-box)

func queryString(s string, n int) bool { if n > 1000 { return false } for i := n; i > n/2; i-- { if !strings.Contains(s, strconv.FormatInt(int64(i), 2)) { return false } } return true }(code-box)