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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.

• 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');
}
}
for (int i = 1; i <= N; i++) {
if (!set.contains(i))
return false;
}
return true;
}
}

############

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

• class Solution:
def queryString(self, s: str, n: int) -> bool:
for i in range(n, n // 2, -1):
if bin(i)[2:] not in s:
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


• func queryString(s string, n int) bool {
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 {
for (let i = n; i > n / 2; --i) {
if (s.indexOf(i.toString(2)) === -1) {
return false;
}
}
return true;
}