Given a binary string s and an integer k.
Return True if every binary code of length k is a substring of s. Otherwise, return False.
Example 1:
1
2
3
| Input: s = "00110110", k = 2
Output: true
Explanation: The binary codes of length 2 are "00", "01", "10" and "11". They can be all found as substrings at indicies 0, 1, 3 and 2 respectively.
|
Example 2:
1
2
| Input: s = "00110", k = 2
Output: true
|
Example 3:
1
2
3
| Input: s = "0110", k = 1
Output: true
Explanation: The binary codes of length 1 are "0" and "1", it is clear that both exist as a substring.
|
Example 4:
1
2
3
| Input: s = "0110", k = 2
Output: false
Explanation: The binary code "00" is of length 2 and doesn't exist in the array.
|
Example 5:
1
2
| Input: s = "0000000001011100", k = 4
Output: false
|
Constraints:
- 1 <= s.length <= 5 * 10^5
- s consists of 0’s and 1’s only.
- 1 <= k <= 20
Solution:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
| class Solution {
public boolean hasAllCodes(String s, int k) {
int len = (int) Math.pow(2, k);
boolean[] bitSet = new boolean[len];
if (s.length() <= k) {
return false;
}
int counter = 0;
for (int i = 0; i < s.length(); i++) {
if (i + k <= s.length()) {
String val = s.substring(i, i + k);
int bit = Integer.valueOf(val, 2);
if (bitSet[bit] == false) {
bitSet[bit] = true;
counter++;
if (len == counter) {
return true;
}
}
}
}
return false;
}
public boolean hasAllCodes2(String s, int k) {
Set<Integer> set = new HashSet<>();
if (s.length() <= k) {
return false;
}
for (int i = 0; i < s.length(); i++) {
if (i + k <= s.length()) {
String val = s.substring(i, i + k);
Integer b = Integer.valueOf(val, 2);
set.add(b);
}
}
return set.size() == (int) Math.pow(2, k);
}
}
|