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 <= S.length <= 1000
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
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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; } }
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// 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; } };
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# 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