Formatted question description: https://leetcode.ca/all/1016.html

1016. Binary String With Substrings Representing 1 To N (Medium)

Given a binary string S (a string consisting only of '0' and '1's) and a positive integer N, return true if and only if for every integer X from 1 to N, the binary representation of X is a substring of S.

 

Example 1:

Input: S = "0110", N = 3
Output: true

Example 2:

Input: S = "0110", N = 4
Output: false

 

Note:

1. 1 <= S.length <= 1000
2. 1 <= N <= 10^9

Solution 1.

// OJ: https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/
// Time: O(N(S+logN))
// Space: O(logN)
class Solution {
private:
    string toBinary(int N) {
        string ans;
        while (N) {
            ans += '0' + (N % 2);
            N /= 2;
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
public:
    bool queryString(string S, int N) {
        for (int i = N; i >= 1 && i > N / 2; --i) {
            if (S.find(toBinary(i)) == string::npos) return false;
        }
        return true;
    }
};

Java

• Java
• C++
• Python

• class Solution {
      public boolean queryString(String S, int N) {
          Set<Integer> set = new HashSet<Integer>();
          int length = S.length();
          for (int i = 0; i < length; i++) {
              int curNum = 0;
              int end = Math.min(i + 31, length);
              for (int j = i; j < end; j++) {
                  curNum = (curNum << 1) + (S.charAt(j) - '0');
                  set.add(curNum);
              }
          }
          for (int i = 1; i <= N; i++) {
              if (!set.contains(i))
                  return false;
          }
          return true;
      }
  }
  

• // OJ: https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/
  // Time: O(N(S+logN))
  // Space: O(logN)
  class Solution {
  private:
      string toBinary(int N) {
          string ans;
          while (N) {
              ans += '0' + (N % 2);
              N /= 2;
          }
          reverse(ans.begin(), ans.end());
          return ans;
      }
  public:
      bool queryString(string S, int N) {
          for (int i = N; i >= 1 && i > N / 2; --i) {
              if (S.find(toBinary(i)) == string::npos) return false;
          }
          return true;
      }
  };
  

• # 1016. Binary String With Substrings Representing 1 To N
  # https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/
  
  class Solution:
      def queryString(self, s: str, n: int) -> bool:
          
          for i in range(n // 2 + 1, n + 1):
              b = bin(i)[2:]
              if b not in s: return False
          
          return True
  
  

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