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

  • 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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