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This approach uses a set to store all unique substrings of length k encountered in s. By sliding a window of size k across the string, we add each substring to the set. If the set size equals 2^k, return true because all possible binary strings of length k are present. Otherwise, return false.
Time Complexity: O(n * k) where n is the length of the string,
Space Complexity: O(min(2^k, n)), primarily due to the set storing at most 2^k elements.
1
2 def hasAllCodes(s: str, k: int) -> bool:
3 seen = set()
4 for i in range(len(s) - k + 1):
5 substring = s[i:i+k]
6 seen.add(substring)
7 if len(seen) == 2**k:
8 return True
9 return False
10 The function utilizes a set to keep track of unique substrings of length k. Iterating over the string from index 0 to len(s)-k, each k-length substring is added to the set. If the set's size becomes 2^k, we know all binary codes of length k are present in s. The operation seen.add(substring) ensures unique tracking, while len(seen) == 2**k quickly checks the condition to return true.
This approach improves efficiency by using bit manipulation instead of storing full substrings. Treat the k-length binary strings as numbers and update a rolling number using bitmasking while sliding over string s. This can detect all variations without explicitly storing them.
Time Complexity: O(n),
Space Complexity: O(2^k), primarily determined by the boolean tracking array.
1
2 using System;
3
4 public class Solution {
public bool HasAllCodes(string s, int k) {
int need = 1 << k;
bool[] seen = new bool[need];
int hash = 0;
int allOnes = need - 1;
for (int i = 0; i < s.Length; i++) {
hash = ((hash << 1) & allOnes) | (s[i] - '0');
if (i >= k - 1 && !seen[hash]) {
seen[hash] = true;
if (--need == 0) return true;
}
}
return false;
}
}
The C# solution employs a fixed-size boolean array indexed by integers representing binary substrings. Hash updates occur via bit manipulation—specifically, shifting left and combining—thus allowing new sequence detection and marking without storing full substrings.