Question

1016. Binary String With Substrings Representing 1 To N

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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

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))

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;
  }
}

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;
  }
};

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
}

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;
}